PHP Notice: Undefined offset?

Similar Questions

• Notice: Undefined index:

  I have that works well on another site and have replicated just it but on this site, it does not work, anyone can see what I'm doing wrong?

  I'm having an error come on my page success.php when returning to a paypal payment...

  Notice: Undefined index: in E:\Domains\b\website.com\user\htdocs\success.php on line 248 OrderID

  Basically when the transaction is successful the return page is return.php, this page tells

  <? PHP

  session_start();

  $colname_rsReturn = "-1";

  If (isset($_SESSION['OrderID'])) {}

  $colname_rsReturn = $_SESSION ["OrderID"];

  }

  @mysql_select_db ($database_beau, $beau);

  $query_rsReturn = sprintf ("SELECT filled WITH beauSS13_orders WHERE the beauSS13_orders.") (OrderID = %s", GetSQLValueString ($colname_rsReturn,"int"));

  $rsReturn = mysql_query ($query_rsReturn, $beau) or die (mysql_error ());

  $row_rsReturn = mysql_fetch_assoc ($rsReturn);

  $totalRows_rsReturn = mysql_num_rows ($rsReturn);

  If ($row_rsReturn ['fill'] == 1) {}

  Header ("' Location: success.php '");

  } another header ("' rental: order.php failed '");

  ? >

  so the value of the order is 1 then it send to success.php to update the stock as follows

  <? PHP

  Update the stock *.

  $details_table = "beauSS13_orderdetails";

  $ID_column = "OrderID";

  $details_prodID = "ProductID";

  $details_qty = "amount";

  $XStock_TableName = "beauSS13_products";

  $XStock_FieldName = "stock";

  $XStock_unID = "ProductID";

  If (! session_id()) session_start();

  If (isset($_SESSION["OrderID"])) {}

  @mysql_select_db ($database_beau, $beau);

  $details_Source = "select * from".  $details_table. 'where '. $ID_column. " = " . $_SESSION ["OrderID"];

  $detailsRS = mysql_query ($details_Source, $beau) or die (mysql_error ());

  $row_detailsRS = mysql_fetch_assoc ($detailsRS);

  {}

  $XStock_qtySource = "select". $XStock_FieldName. "from".  $XStock_TableName. 'where '. $XStock_unID. " = " . $row_detailsRS [$details_prodID]. "";

  $XStock_rsUpd = mysql_query ($XStock_qtySource, $beau) or die (mysql_error ());

  $row_XStock_rsUpd = mysql_fetch_assoc ($XStock_rsUpd);

  If ($row_XStock_rsUpd [$XStock_FieldName] > 0) {}

  $XStock_new = $row_XStock_rsUpd [$XStock_FieldName] - $row_detailsRS [$details_qty];

  If ($XStock_new < 0) $XStock_new = 0;

  $XStock_UpdSource = "update". $XStock_TableName. 'together '. $XStock_FieldName. " = " . $XStock_new. 'where '. $XStock_unID. " = " . $row_detailsRS [$details_prodID]. "";

  $XStock_rsUpd = mysql_query ($XStock_UpdSource, $beau) or die (mysql_error ());

  }

  } While ($row_detailsRS = mysql_fetch_assoc ($detailsRS));

  $XStock_rsUpd = null;

  $detailsRS = null;

  session_unregister ("OrderID");

  }

  ? >

  but when I get to this page, I get the error

  Notice: Undefined index: in E:\Domains\b\website.com\user\htdocs\success.php on line 248 OrderID

  and the stock is not updated

  line 248 is order < p > <? PHP echo $_SESSION ["OrderID"];? > success < br / >

  then he tells me that the session is not respected on the whole?

  It is not clear from your code where line 248, but the last line of code you posted here using this:

  session_unregister ("OrderID");

  This destroys $_SESSION ["OrderID"]. Thus, if line 248 comes after this code, that would explain why you get notice of undefined index.

  Furthermore, session_unregister() is obsolete and has been removed from PHP 5.4. You must use unset() instead:

  unset($_SESSION['OrderID']);

• The PHP variable undefined [was: Dreamweaver CC Access database]

  A few error messages popping up due to a variable, that I created to establish a connection with a database. The variable was created on another page that I linked to the page that it is in use. Why these errors occur baffles me, please help. See below.

  Error messages:

  1. unset variable - Variable. »

  2 mysqli_real_escape_string() expects parameter 1 to mysqli (* insert Variable in parameter 1)

  The encoding is used:

  Variable = mysqli_connect ($hostname_localhost, $username_localhost, $password_localhost) or trigger_error (mysql_error (), E_USER_ERROR);

  (Variable transferred via 'require_once ('Link') ;)

  What's more confusing is that I used the Variable on other functions via the function "require_once" on the same page without error message.

  The reason why you get "undefined variable" will be because you are trying to access the variable before it is created. Without seeing your code, it is impossible to identify the problem.

• Error notice question

  Hello

  My website (development) php code now everything works as it should, and tests on the database are all producing correct results, however after you set error_reporting to show all errors, I have several cases of:

  Notice: undefined constant...

  Notice: undefined variable...

  Notice: undefined index...

  I am aware that I can set error_reporting to not display all records, but my concern is that my code is safe.

  How important are these opinions regarding a secure website from php? any information would be greatly appreciated, I don't want to ignore the significant safety issues before going live.

  Thanks in advance for your time.

  Those who are good to get rid of.  Show all is not normally what is shown, but those errors mean that you have not tested whether the variables are defined (isset php - http://php.net/manual/en/function.isset.php ).  It is always good when writing scripts to make sure that the variables are set before trying to use them in the code.  A simple:

  If (isset ($var))

  will do the trick.

• Send variables from flash to PHP

  Hello world
  I have a problem when sending variables from flash to PHP
  Here is the code:
  
  submit.onPress = function)
  {
  c = new LoadVars ();
  c.NME = "Myname"
  c.sendAndLoad ("getVars.php", reply, 'POST');
  }
  
  PHP code:
  
  <? PHP
  $name = $_POST ['nme'];
  Print 'your name is. " $name. «. < br / > ';
  ? >
  
  What I'm doing wrong here.
  I get an error message saying notice: Undefined index: name in \xxx\getVars.php on line 2
  
  Please help me. Thanks in advance

  2 things that I notice immediately

  1. you never initialize the variable 'Reply' to your sendAndLoad.
  2. you must pass a string encoded in URL variables which is ampersand delimited. Usually the returned string would look like echo ("" & result = 1 & msg = your % 20name % 20is %20 '. ") $name. ». 3rd % 3Cbr % &");

  Your mistake is that the information is not get passed correctly. I usually use $_REQUEST communication flash-php, although this should not make a difference. I've seen people use $_REQUEST for testing purposes and will then to $_POST when finished.

• Dreamweaver 8 with ASP, Coldfusion, &amp; amp; PHP

  I'm new to php and I'm working in Dreamweaver 8 with ASP, Coldfusion and PHP training from the Source. I have a real simple form. A user enters their name and name and then click on the submit button. The form is then treated with a script using the $_GET [name]; $_GET [name]; and it should display "Thank you, name,." using my form But the page shows the HTML with a notice: Undefined index: first_name in C:\htdocs\newland\test_form_processor.php on line 9. Line 9 is - thank you < p >, <? PHP echo $_GET ['name'];? > <? PHP echo $_GET ['name'];? >, for my application. < /p >. This was accomplished using the link Panel in DW. It is not any kind of troubleshooting in the book. Can someone tell me what this means? It worked as it is supposed to work when I had just the first name.

  Thank you, David. Yes my form looks like as you suggest. The problem is resolved. I have problems with DW not to upload files to the test server on my machine. And the problems with the browser doesn't not refreshing pages. Between the two, I was about to lose my mind.

  I have read and worked through your book "Foundation PHP for DW 8" last year and I just got your new book "PHP Solutions" today. PHP for DW 8 was really easy to follow and everything worked as it should. I hope I like "PHP Solutions" as well.

  Charles Burnell

• Problem in mysqli_real_escape_string

  Hi all

  I used mysqli_real_string in my project. But it shows the error like that.

  Kindly help me to fix it.

  Notice: Undefined variable: conn in C:\xampp\htdocs\BHCNET\admin\insert_event.php on line 224

  Warning: mysqli_real_escape_string() expects parameter 1 to mysqli, null given in C:\xampp\htdocs\BHCNET\admin\insert_event.php the line 224

  Here is my code:

  <? PHP

  If (isset($_POST['add_new']))

  {

  Event $= $_POST ['event'];

  $cat = $_POST ['category'];

  / * $department = $_POST ['category'] ;*/

  $department = $_POST ['State'];

  $date_of_event = trim($_POST['event_date']);

  $date_of_event = mysqli_real_escape_string($conn,$date_of_even);

  $date_of_event = strtotime ($date_of_event);

  $date_of_event = date('Y-m-d',$date_of_event);

  $time_of_event = trim($_POST['timepicker1']);

  $place = $_POST ['halles'];

  include("include/Connection.php");

  If ($Évévement == "| $department ==" | $date_of_event == "| $time_of_event ==" | $place == ")

  {

  echo"< script > alert ('Please enter all values!'); < /script > ';

  Exit();

  }

  on the other

  {

  $query = "insert into add_event (category, title, start, date, time, come)

  values('$cat','$event_title','$department','$date_of_event','$time_of_event','$place') «;»

  If (! mysqli_query ($conn, $Query))

  {

  echo "query Failed".mysqli_error ($conn);

  }

  on the other

  {

  echo "< script > alert (" new event has been added successfully!'");" < /script > ';

  echo "< script > window.open('events.php?view=view','_self')";

  Header("Location:events.php?) Insert = Insert.php");

  }

  }

  mysqli_close ($conn);

  }

  ? >

  Thank you in advance.

  You cannot use mysqli_real_escape_string() before the creation of the database connection. Include connection.php before you define your variables.

• Column sum based on a single value of another column

  Hi all

  I've searched and searched, but just can't find if this is even possible, let alone how to do it.  What I'm trying to do, it's the sum of a column.  However, only summarize the results of this column if a specific value of another column is present.  So, for example, the following is a sample database entry:

  Rate of pay hours

  2.3                                  $14.00

  1.2                                  $14.00

  2.4                                  $12.50

  0.75                                $12.50

  So for that, I want the sum of hours worked column but only the sum of hours worked if the wage rate is $14.00 and $ 12.50.  Because these numbers are generated in a database (with the various people who enter multiple rates of pay), I won't be able to set the where statement to have a specific amount (since I don't know the specific amount each registrant), the amount must be a variable in the where the statement.  With MySQLI, I know not how to make the sum of hours worked, but I do not know how the sum of hours worked if and only if the line corresponds to the unique value in the rate column.

  Then have it automatically loop through and total hours worked for other unique values in the column salary rate.  Finally, out all unique values in the column rate in variables so that I can do the following for each game:

  $total1 = $total_hours1 * $wage_rate1
  $total2 = $total_hours2 * $wage_rate2
  $overall_total = $total1 + $total2
  

  where $wage_rate1 = $14.00 and $total_hours1 = 2.3 + 1.2: etc.

  Is it still possible?  If so, how?  If possible, can someone please let me know a good reference so I can learn how to be the best way to implement?

  Thank you

  So with that, I get the correct $final_total printed, but I still get the error, 'notice: Undefined variable: final_total. Why is it not defined, but prints the total exact? How to remove the error?

  You get this notice is because you use the assignment combined, operator that is normally used to add a value to a variable and assign the new value to it.

  $final_total += $total;

  is a shortcut for this:

  $final_total = $final_total + $total;

  The undefined $final_total is on the right side of the assignment operator. The first time through the loop, you add $total to a non-existent value. Because of the type of PHP juggling, this becomes:

  $final_total = 0 + $total;

  The next time the loop runs, $final_total has a value, so that you get the view variable not defined only once.

  To avoid advice, initialize $final_total to zero before the start of the loop.

  $final_total = 0;

  While ($row = {$sql-> fetch_array())}

• Connection difficulties

  Hello

  I used behavior several times (DW CC) Server user authentication but today he gives me. It allows me to connect to a page that has restricted access, but it is also back this ugly error at the top of the page:

  Notice: Undefined index: MM_UserGroup in C:\wamp\www\2010 new site\search.php line 46

  
  

  Here's the code that surround the restricted access:

  
  

  $MM_restrictGoTo = "loginfail.php";

  If (!) {(((isset($_SESSION['MM_Username']))) & & (isAuthorized ("", $MM_authorizedUsers, $_SESSION ['MM_Username'], {$_SESSION ['MM_UserGroup']})))}

  $MM_qsChar = ';?

  $MM_referrer = $_SERVER ['PHP_SELF'];

  If (strpos ($MM_restrictGoTo, "?")) $MM_qsChar = "&";

  If (isset($_SERVER['QUERY_STRING']) & & strlen($_SERVER['QUERY_STRING']) > 0)

  $MM_referrer. = « ? ». $_SERVER ['QUERY_STRING'];

  $MM_restrictGoTo = $MM_restrictGoTo. $MM_qsChar. "accesscheck =". UrlEncode ($MM_referrer);

  Header ("Location:".) $MM_restrictGoTo);

  "exit";

  Can anyone place what I need to do?

  gbergan wrote:

  That is absolutely right. And my login.php page is already working with MySQLi coding. I can not get the coding directly on my search.php page, which is where I need to restrict access. Again, as I said in a previous post, login.php use MySQLi. Search.php does not because DwZone products are not yet updated. The coding is already displayed in the above login page.

  But like I said help the Web should provide in their connection solution "limit access solution to apply to other pages" in the contrary case, it makes their connection solution a waste of time? This does not have any mysqli require just a few lines of php coding.

  Can you not apply it directly to the search php page?

  The login page takes you to the search page?

  This created at the top of the extension of research Web page help or DwZone?

  <>

  If (! isset($_SESSION['Member'])) {}

  Header ("Location: login.php");

  }

  ?>

  It looks not to $_SESSION ['Member'] has been on your login page so you get there will pass if you try and go to the search page.

  As a stab in the dark because this authentication code is not familiar to me trying to change:

  $_SESSION ['username'] = $Member;

  
  

  TO

  
  

  $_SESSION ['Member'] = 'true ';

  
  

  
  

  then substituting success2.php with search.php
  

  
  

  
  

• help with forms

  I'm quite new to Dreamewaver CS4 and I'm currently working my way through the book by Jeffery Bardzell "training from the source. I got to Lesson 5 'crossing of data between pages. I created the fields first name and last name on the form, and then run the form via the server. I works very well and will display the first and last names on the form processor IE "thank you, john smith, to complete my form" the book then tells me to open the form processor and it should look to "thank you for my application, but it displays:

  Thank you, notice: Undefined index: firstName in c:\xampp\htdocs\newland\test_form_processor.php on line 9.

  lastName undefined index in c:\xampp\htdocs\newland\test_form_processor.php on line 9

  for my application.

  Could someone tell me what I did wrong. I followed the book Word for Word and I also checked the code on my page with the code on the lesson of "completed" on the CD. theyr'e the same two.

  Any help would be apreciated

  Thank you much for the help. Both methods work very well.

  Paul

• session_start();

  I want to send information from one page to another by using the session_start();

  Lets say that page1.php has a session that precedes the <! DOCTYPE HTML > doctype.

  <? PHP

  session_start();

  $_SESSION ['myaccess'] = $row_mymeeting ["congremtid"];

  ? >

  And in my body tag the following...

  < body >

  < a href = "page2.php" > there < /a >

  < / body >

  When I look at page1.php in my browser I get an error

  Notice: Undefined variable: row_mymeeting in/Applications/MAMP/htdocs/phptest/www/index.php on line 5

  However if I put $_SESSION ['myaccess'] = ' Hello, welcome to Adobe! " When I click on the link to the page page2.php it shows.

  And on page2.php I want to echo what ever the $row_mymeeting ["congremtid"] has in my database.

  A page 2 I...

  <? PHP

  session_start();

  ? >

  <? PHP

  echo $_SESSION ['mylink'];

  ? >

  But I have a white.

  How can I make my $_SESSION ['myaccess'] = $row_mymeeting ["congremtid"]; ECHO in my second page, that I am I missing?

  Thank you.

  Kind regards

  AdonaiEchad

  You have a recordset on the page that allows you to SELECT 'congremtid' from the database? AND is the line $_SESSION ['myaccess'] = $row_mymeeting ["congremtid"]; AFTER the sql query? If you do not have, or that the line is NOT AFTER the sql query, you will get an error message about undefined variables.

• Flex $_POST does not

  OK, so I'm working on an application that allows users to connect, when they connected it redirects to another page. The page that they are redirected to must however know their ID card, which I suppose can be done safely through the post OFFICE or the SESSION, I prefer the POST.  I have this code so far:

  Flex/AS3:

  <fx:Script>
            <![CDATA[
                 import flash.net.navigateToURL;
                 
                 import mx.controls.Text;
                 import mx.rpc.events.ResultEvent;
                 
                 private var loginID:String;
                 
                 private function checkLogin(event:ResultEvent):void
                 {
                      loginID = event.result.loginsuccess.toString();
                      if(event.result.loginsuccess != "no")
                      {
                           var data:Object = new Object();
                           data.id = loginID;
                           var request:URLRequest = new URLRequest("http://myurl.com/startPage.php");
                           request.contentType = "text";
                           request.data = data;
                           request.method = URLRequestMethod.POST;
                           
                           navigateToURL(request, "_self");
                      }
                      if(event.result.loginsuccess == "no")     
                           navigateToURL(new URLRequest("http://mybox.dyndns-web.com/Petty/incorrectLogin.php"), "_self");
                 }          
                 private function register(evt:ResultEvent):void
                 {
                      if(evt.result.regsuccess == "yes")
                           navigateToURL(new URLRequest("http://mybox.dyndns-web.com/Petty/registerSuccess.php"), "_self");
                      if(evt.result.regsuccess == "no")
                           navigateToURL(new URLRequest("http://mybox.dyndns-web.com/Petty/registerFail.php"), "_self");
                 }
            ]]>
       </fx:Script>
  

  PHP - I want to just display the ID at the moment:

  <?php     
  
  echo $_POST["id"];
  
  ?>
  

  But at the time the user arrives on the PHP page, they see an error:

  Notice: Undefined index: id in C:\xampp\htdocs\startPage.php on line 3

  Where I'm going wrong? Please help, I need this problem as soon as POSSIBLE!

  Try this

  var url:URLRequest = new URLRequest("http://myurl.com/startPage.php");
  var uv:URLVariables = new URLVariables();
  url.method = "POST";
  uv.id = loginID;
  url.data = uv;
  navigateToURL(url,"_self");
  

  Ref:

  http://help.Adobe.com/en_US/Flex/using/WS2db454920e96a9e51e63e3d11c0bf626ae-7FE8.html#WS2d b454920e96a9e51e63e3d11c0bf69084-7ea7

• Licence DW has does not not on remote server

  Hi all,

  
  

  A couple of things about my registration system are baffling me:

  
  

  The registration form, I have a list to select the country of. When I left the list as a manual list in the code on the remote server, everything works fine. It works very well in Dreamweaver on the local test server.

  
  

  I use a < CONSTANTS > file to establish the database connection.

  
  

  When I now convert the manual list to a dynamic list by converting it to a set of folder in Dreamweaver, it does not and shows several errors:

  
  

  ( ! ) Notice: Undefined variable: database_mysqli_connect in C:\wamp\www\website\Connections\mysqli_connect.php on line 34

  
  

  ( ! ) Notice: Undefined variable: mysqli_connect in C:\wamp\www\website\Connections\mysqli_connect.php on line 34

  
  

  ( ! ) Warning: mysql_select_db() expects the parameter 2 resource, null in C:\wamp\www\ website \Connections\mysqli_connect.php on the 34 line

  
  

  ( ! ) Notice: Undefined variable: mysqli_connect in C:\wamp\www\ \Connections\mysqli_connect.php on line 36 website

  
  

  ( ! ) Warning: mysql_query() expects the parameter 2 resource, null in C:\wamp\www\ \Connections\mysqli_connect.php on line 36 website

  
  

  If I now connect normal (NON CONSTANT) to the database, it works, but only in Dreamweaver on the local test server.

  
  

  On the remote server, it throws an error:

  
  

  Warning : require_once (.. / Connections/mysqli_connect.php) [function.require - once]: failed to open stream: no such file or directory in /xxx/xxx /domain/registrar/register.php on line 1
  
  Fatal error: require_once() [function.require]: failed to achieve "... / Connections/mysqli_connect.php ' (include_path ='.: / usr/lib/php: / usr/local/lib/php ') in /xxx//domain/registrar/register.php xxx on line 1 "

  
  

  Is a short coming in Dreamweaver or is it the man in front of the computer?

  
  

  Please throw some wisdom on the subject...

  
  

  Thank you

  
  

  zabberwan

  zabberwan wrote:

  The problem seems to be the generated code from Dreamweaver recordset and the "constant" file

  Your constants file using MySQLi (improved MySQL). Dreamweaver recordsets using the original MySQL extension. They are not compatible.

• Navigation to search pages

  Jin
  
  I made a page of research having 2 fields for entering text, the other is a menu with 2 opcions... Author = title 1 = 2
  Send message
  The idea is that people with a word search and select from the menu if you want to search by author, title of any book.
  
  the first result of end page... I add a navigation links - the problem is when I try to go to the first... .next - last, whit this link I've lost the result and show me this message =
  
  Notice: Undefined variable: temp in D:\www\editorial\respuesta0.php on
  line 53
  Notice: Undefined variable: temp in D:\www\editorial\respuesta0.php on
  line 56
  Notice: Undefined variable: query_GetRecordset_libros in \editorial\respuesta0.php on line 60 D:\www, you have an error in your SQL syntax. Consult the manual for your version of the MySQL server for the right syntax to use near 'LIMIT 10, 10' at line 1
  No idea how to solve this problem?
  Thanks for any response... Saludos
  Here is the code that I used with dreamweaver
  $colname_GetRecordset_libros = "a";
  If (isset($_GET['busqueda'])) {}
  $colname_GetRecordset_libros = $_GET ['search'];
  }
  
  If (isset($_POST['criterio'])) {}
  $temp = $_POST ['sort'];
  }
  
  If ($temp == '1') {}
  @mysql_select_db ($database_editorialconn, $editorialconn);
  $query_GetRecordset_libros = sprintf ("SELECT * FROM lb_libros WHERE the")
  lb_libros_autor AS %s ", GetSQLValueString ("%"."
  $colname_GetRecordset_libros. '%', 'text'));
  } else if ($temp == '2') {}
  @mysql_select_db ($database_editorialconn, $editorialconn);
  $query_GetRecordset_libros = sprintf ("SELECT * FROM lb_libros WHERE the")
  lb_libros_titulo AS %s ", GetSQLValueString ("%"."
  $colname_GetRecordset_libros. '%', 'text'));
  }
  
  $query_limit_GetRecordset_libros = sprintf ("%s LIMIT %d, %d",)
  $query_GetRecordset_libros, $startRow_GetRecordset_libros,.
  $maxRows_GetRecordset_libros);
  $GetRecordset_libros = mysql_query ($query_limit_GetRecordset_libros,)
  $editorialconn) or die (mysql_error ());
  $row_GetRecordset_libros = mysql_fetch_assoc ($GetRecordset_libros);
  
  
  Thank you very much!! for your help!
  
  Saludos
  
  Caroline

  

  Yes, I got the solution, jeje I send the var by get method...
  And a code that help me get the error was make a to the query I see that I have a '$temp' in the query so that makes the result no work...

  so it's the final solution, I did..., sorry for my English if any body understand me! Saludos

  If (($_GET['criterio']) == 1) {}
  $temp = "lb_libros_autor";
  } else if (($_GET['criterio']) == 2) {}
  $temp = "lb_libros_titulo";
  }

  @mysql_select_db ($database_editorialconn, $editorialconn);
  $query_GetRecordset_libros = sprintf ("SELECT * FROM lb_libros WHERE $temp LIKE %s", GetSQLValueString ("%". "") $colname_GetRecordset_libros. '%', 'text'));
  $query_limit_GetRecordset_libros = sprintf ("%s LIMIT %d, %d", $query_GetRecordset_libros, $startRow_GetRecordset_libros, $maxRows_GetRecordset_libros);
  $GetRecordset_libros = mysql_query ($query_limit_GetRecordset_libros, $editorialconn) or die (mysql_error ());
  $row_GetRecordset_libros = mysql_fetch_assoc ($GetRecordset_libros);

• Fatal error: Call to undefined function get_header() in /home/rascal98/public_html/index.php on line 1.

  I received this message: Fatal error: Call to undefined function get_header() in /home/rascal98/public_html/index.php on line 1.  This has happened after that I tried to load a model of Web site using FTP Filezilla.  I asked advice of HostGator support, but offered only man I should have a backup.  Not much help at the moment.  I may be in the last two days charged that more than just a Web site template to the same file or directory public_html, it is called. I tried to reverse all this leaving the Site Manager to erase all files.  That did not help.  I'm stuck because I can't establish a few areas I need.  The terminology "fatal error" was quite worried.

  Hi rascal98,

   

  Welcome to Microsoft Answers Forums.

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