Error-4702, Regular Expressions

Similar Questions

• CS6 - error in search: Syntaxerror in regular expression: invalid quantifier ++

  Hello

  I get an error when you search for a string containing ++ in Dreamweaver on Mac OS X 10.9.1 CS6.

  The error message reads:

  SyntaxError in regular expression:

  invalid quantifier ++

  Same error when I search a +.

  No search no results when I search open + without, but there are two in the source code...

  What's not here? What can I do? Thank you.

  Michael

  OK, it is resolved: there was an option selected by default in the search box: use regular expressions. Unchecked this and it works...

• Complex regular expressions without multiple passes

  Does anyone know of a tool that can handle more complex regular expressions without chaining of multiple copies of the VI regular expressions?

  For example, if I have a XML string as

  Power supply error has occurred.

  Sorensen SGA166/188

  and I am interested in the tag method to retry only, I could write a regular expression something like

  .*.*

  to parse the string inside the tag.

  kc64 wrote:

  For example, if I have a string like

  My email address is [email protected].  Please no spamming not me.

  and I am interested in the domain name of the email only address, I could write a regular expression something like

  @(\w)*\. (com: net | org)

  to parse the string 'gmail '.

  Forgive me if I am away from base, but I'm flying blind at the moment (not LV to test what I say).  You can add to the power of a regular expression using submatches or capture groups.  The regular expression you wrote will grab (I think) @gmail.com for the entire game.  Let's say you want to get 'gmail' without a second function call.  You can do the first group of a little dishonest selection by moving the * inside the parentheses.  Then, on the BD pull down on the bottom of the function of regular expression matching to expose a variable number of submatches (both should be in this case).  The first should be 'gmail '.   The second one should be "com."

  In summary, @(\w*)\.) (com: net | org) should give you gmail in the first submatch.  Of course, my Perl is a little rusty and LV cannot apply in the same way.

• Is the function of the regular Expression

  Hi guys, using this model, I got this error:

  -4600 error has occurred during the regular expression match.

  I have attached the VI.

  can you help me? Thank you

  inuyasha84 wrote:

  Hi well I want to save (create) a file and do a check to see if the new file that I want to create already exists or not. If the idea was to see if the path of the new file is equal to the old path

  Why not just use 'check if the file or folder exists? (E/s files-> Adv file functions)

  Cameron

• regular expression to remove the zeros on the right

  I need a regular expression to remove the zeros after the decimal point. I tried (?.)<=\.\d+?)0+(?=\D|$) but="" i="" get="" a="" error="" about="" look="" behind="" not="" a="" fixed="" length="" or="" something="" like="" that.="" i="" am="" not="" a="" regex="" expert="" and="" i="" was="" wondering="" just="" how="" to="" do="" this="" with="" regular="" expression="" or="" some="" other="">

  Z.K. wrote:

  [...] or some other way.

  

  I tried and I tried but I couldn't crack with a regular expression, so I took the easy way.  The first match found pattern the comma and the other removes the zeros to the right of the rest.  It is not discriminate between numbers and all the rest, though.

• Regular expression help please. (extraction of a subset of the string between two markers)

  I haven't used regular expressions before, and I can't find a regular expression to extract a subset of the string between two markers.

  The chain;

  Stuff of header I want
  Stuff of header I want
  Stuff of header I want

  Stuff of header I want
  Stuff of header I want
  Stuff of header I want

  6 ERRORS
                       
  Info I want to line 1
  Info I want line 2
  Info I want line 3
  Info I want to line 4
  Info I want to line 5
  Info I want line 6
  END_ERRORS

  From the string above (it is read from a text file), I try to extract the subset of string between ERRORS 6 and END_ERRORS. The number of errors (6 in this case) can be any number from 1 to 32, and the number of lines I want to extract will correspond with this number. I can provide this number of a caller VI if necessary.

  My current solution, which works, but is not very elegant;

  (1) using Match Regular Expression for the return of the string after you have synchronized the 6 ERRORS

  (2) uses the Regular Expression matches to return all characters before game END_ERRORS of the string returned by (1)

  Is there a way this can be accomplished using 1 Regular Expression Match? If so someone could suggest how, as well as an explanation of the work of the given regular expression.

  Thank you very much

  Alan
  

  

  I used a character class to catch any word or whitespace characters.  This put inside parentheses a substring matching the criteria that you can get by developing the node for regular expression matching.  The \d matches the numbers and the two * s repetition of the previous term.  So, \d* will find the '6', as well as "123456".

• Oracle regular expressions - splits the string into words for

  Hello

  Nice day!

  My requirement is to split the string into words.

  So I need to identify the new line character and the semicolon (;), comma and space like terminator for string entry.

  Please note that I am currently embedded blank and the comma as separator, as shown below.

  Select regexp_substr('test)
  TO
  string in words, "([^, [: blanc:]] +) (', 1, 1) double;"

  How to integrate the semicolons and line break characters in regular expression Oracle?

  Please notify.

  Thanks and greetings

  Sree

  This has nothing to do with REGEXP. Is SQL * more parser does not not a semicolon at the end of the line:

  SQL > select ' testto, mm\;
  ERROR:
  ORA-01756: city not properly finished chain

  SQL >

  Just break the chain:

  SQL > select regexp_substr ('testto, mm\;' |) '
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<= regexp_count('testto,mm\;'="" ||="">
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;' |') STRINGIN
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  Or modify SQL * more the character of endpoints:

  SQL > set sqlterm.
  SQL > select regexp_substr ('testto, mm\;)
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<=>
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;) STRINGINTOWO
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  SY.

• Regular expression breaks script

  I have the following statement of regular expressions in my JSX file:

  var regex = /[^/]*(?=\.[^.]+($|\?))/; // Doesn't work while this does: var regex = /[\u0400-\u04FF]+/;
  

  
  .. .and it breaks execution of my script for unknown reasons. What's not here? I thought the syntax / theRegEx.
  (comment to the statement of regex solves the problem and the script runs normally)

  I tested the regex here:
  RegExr: Learn, build, & Test of RegEx

  .. .on this string: "c:/Users/martin.dahlin/Desktop/Test/lolFile.lol".
  gives me "lolFile" without the quotes.

  xbytor is right, there is really no "regex-standard/compliance. Basic things work quite universally, but there are the specifics of the platform and conventions and you just test & work around them.

  That said, get into your regular expression to Tester online regex and a debugger: JavaScript, Python, PHP, and PCRE gives and error "/ separator unescaped ', i.e. must be ' / [^\/] * (?)» = \. [ ^.] +($|\?)) /". Also the use of the explicit constructor is a tad safer, typos lead more reasonable errors.

• Validation of regular expression delimiter alphanumeric and hose

  I have a header in the file as below

  EMP_ID | EMP_NAME | DEPARTMENT | SALARY | ACTIVE1

  past to a string

  String test = ' EMP_ID | EMP_NAME | DEPARTMENT | SALARY | ACTIVE1;

  I need to check if the header is only of alphanumeric characters and pipedelimiter is allowed.

  other than these, I need to trigger an error.

  Any suggestions.

  had the regular expression pattern

  It's the string pattern = "^ ([a-zA-Z0-9 |]) '. +)$";

• pattern by using regular expressions match

  I'm playing (and wrong) with regular expressions

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + > ') twice;

  give < PSN > - I understand that

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.2 ') twice;

  give < 231 > - I understand that

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.3 ') twice;

  give < ABc > which confused me until I realized that < 3 25 > has not been matched, because it has a space inside

  so I changed it to

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ', 1.3) twice;

  who gave a syntax error, so I changed it again to

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ') twice;

  that works, but gives < PSN > / # < 231 > # < 3 25 3 > / < ABc >

  I guess because the. * corresponds to anyting that all but the last closure >

  The question I have is how can I retrieve the text of mounting bracket included (even if it has multiple spaces)?

  Hello

  9c5dfde3-EAAE-45A7-80a1-bba8a71c826c wrote:

  Thanks for that - is there a way to return all 4 surrounded by extracts of <> without resorting to PL/SQL?

  Of course;

  REGEXP_REPLACE (str

  , '(^|>)[^<>

  , '\1'

  )

  Returns a copy of str with all outside rafters removed, for example

  <231><3 25="">

  .  It doesn't matter how many pairs of sharp hooks - there is.

• Help in regular Expression for the beaches of limitation

  Hi, I'm working on the provision of a text field is limited to dates, it's just a part of the code. I already have the validation of the dates, but I am now limiting what the user enters using a regular expression. This code works a little however, it does not limit me for example I can enter more than 2 digits, but then he limits based on the total amount allowable so for example 8 digits are allowed if I just type. I need to stop after 2 digits then have a - then 2 other numbers then one - and then followed by 4 digits. I tried to limit each section and grouping as well. Any help would be greatly appreciated. Thank you.

  It is in the format code and I am the appellant in the key sequence.

  function DateKS () {}

  var value = AFMergeChange (event);

  If (! event.willCommit) {}

  Allow only characters that match the regular expression

  Event.RC = /^([0]{0,1}[1-9]{0,1}|[_1]{0,1}[012]{0,1}) ([-] {0,1}) ([0] {0,1} [1-9] {0,1} |) [12] {0,1} [0-9] {0,1} | ([3] {0,1} [01] {0,1}) ([-] {0,1}) ([0-9] {0,4}) $/ .test (value);

  }

  }

  I decided that control for 100 and 400 was not necessary because this event does occur that all 400 years. But I'm working on it further and changed even more. Here is my code to work.

  function isLeapYear (year) {}
  year return % 4 = 0;
  }

  function checkDaysInMonth (day, month, year) {}
  daysInMonth var = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];

  If (month = 2) {}
  If (isLeapYear (year)) {}
  daysInMonth [1] += 1;
  }
  }

  return daysInMonth [month - 1] > = day;
  }

  function checkDateFormat (dateStr) {}
  var errorMsg = ",
  maxYear = (new Date()) .getFullYear (),
  minYear = maxYear - 1,.
  match = dateStr.match(/^(\d{2})-(\d{2})-(\d{4})$/),
  months,
  day,
  year;

  If {(matches)
  month = parseInt (matches [1], 10);
  day = parseInt (matches [2], 10);
  year = parseInt (matches [3], 10);

  If (month < 1="" ||="" month=""> 12) {}
  errorMsg = "invalid value for the month: ' + matches [1];"
  } ElseIf (day = 0) {}
  errorMsg = "invalid value for the day:" + match [2];
  } else if (! checkDaysInMonth (day, month, year)) {}
  errorMsg = "number of days for invalid month: ' + match [2];"
  } ElseIf (year < minyear="" ||="" year=""> maxYear) {}
  errorMsg = "invalid value for the year:" + match [3] + "-must be between" + minYear + "and" + maxYear;
  }
  } else {}
  errorMsg = "invalid date format: ' + dateStr + ' \r\nPlease use format: dd-mm-yyyy ';"
  }

  return errorMsg;
  }

  function checkReceivedDate() {}
  var value = AFMergeChange (event),
  errorMsg = ";
     
  ignore control if the value is blank, because this field is not mandatory
  If (! value) {}
  return;
  }

  If {(event.willCommit)
  errorMsg = checkDateFormat (value);

  If (errorMsg) {}
  App.Alert (errorMsg, 0, 0, "error");
  Event.value = ";

  Returns false;
  }
  } else {}
  Allow only characters that match the regular expression
  Event.RC = /^(?:0) [1-9]? 1 [012]?) ? -? ( ? : 0 [1-9] ? | [12] [0-9]? 3 [01]?) ? - ? 2? 0? [0-9] {0,2} $/ .test (value);
  Event.RC = / ^ \d{0,2}-?\d{0,2}-?\d{0,4}$/.test(value);
  }

  Returns true;
  }

• Regular expression for invalid number

  Hello world

  I use version oracle as follows:

  SQL > select * from v version $;

  BANNER
  ----------------------------------------------------------------
  Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - production
  PL/SQL Release 10.2.0.4.0 - Production
  CORE 10.2.0.4.0 Production
  AMT for 32-bit Windows: release 10.2.0.4.0 - Production
  NLSRTL Version 10.2.0.4.0 - Production

  ------------------------------------------------------------------------

  I use a regular expression to replace invalid values in a table.

  I got oracle error 'ORA-01722 invalid number '.

  My query looks like this:

  SELECT DISTINCT

  MRC_KEY,

  PURPOSE_CD,

  RESIDENCE_DESC,

  TO_NUMBER (regexp_replace (ICAP_GEN_MADAPTIVE, "..?")) 0? 0? (\d+) [-.] ((?', '\1')) as ICAP_GEN_MADAPTIVE,

  Of

  MRRC_INT

  I don't know what are invalid values in the table, so I can write regexp accordingly.

  Any guidance is appreciated!

  Thanks in advance

  J

  Hello

  Whenever you have a problem, please post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and also publish outcomes from these data.
  Explain, using specific examples, how you get these results from these data.  View of the code which is not what you want can be useful, but there niot enough by itself.
  See the FAQ forum: https://forums.oracle.com/message/9362002

  Let's look at what you do:

  REGEXP_REPLACE (icap_gen_madaptive

  , '[+. ]' || -exactly 1 of these characters

  '?0?0?' || -up to 2 ' 0 '

  '(\d+)' || -some numbers (it comes to \1)

  '[-.]?'    -0 or 1 of these characters

  , '\1'

  )

  Is it really what you want?  A group of figures don't match, as the patten must start with a sign more, a point or a space.  You want to allow a decimal point at the beginning and at the end, as in "." 9. '?  If there are numbers before and after the decimal point, as in "1.2", you want only the digints before the comma?  You never said what you want, but I'm guessing that these are things that you don't want.

  If the pattern is found, this changes only the schema, so a string like "FU + 123 BAR" more than the model + 9 "" changed to "9" and the resulting string. "" "" FU9BAR"would be passed TO_NUMBER.

  If the pattern is not found, as in "FOO", nothing is changed, and the entire string is passed to TO_NUMBER.

  I suspect you want to use instead of REGEXP_REPLACE REGEXP_SUBSTR, or, if there is stuff to ignore before and/or after the number you want to include in the model, so that REGEXP_REPLACE will be replaced by nothing.

  This looks like a good argument to use the NUMBER of columns to store numbers.

• Chain problem: not being able to escape a bracket for a possible regular expression

  So, I have barely any idea of how to use regular expressions except a little knowledge of it in PHP.
  
  What I'm trying to do is to separate the first part of a string of qualifier, so I need a regular expression since ideally I wish I could. The problem I have is that my attempt of separation does not work. I tried to do what I think the model or class Matcher, I tried to use String.split, and I tried using the regular String.replace method. What I'm trying to do is illustrated (I hope) in the following NBS:
  

  public class SSCCE
  {
      public static void main(final String args[])
      {
          final String testString = "I want to keep this sentence. (I wish to "+
              "discard this sentence and the space in between the sentences.)";
          final String testExtract = testString.replace(" \(.+\)","");
          System.out.println(testExtract);
      }
  }

  From my understanding, the parentheses are supposed to be a wild card of sorts, so they would need to be escaped in order to be taken as a literal, which means that line 7 is supposed to correspond to any of the characters inside the brackets where the opening is preceded by a space and then delete them. The problem I have is that escaping brackets generates an error and does not escape their product the original string. So I don't know how to tell Java I use actually literal parentheses.
  
  I tried to configure my preferences of error to ignore everything, but it does not work.
  
  I don't know if it's important, but I use Java 1.6 and Eclipse for PHP 1.3.2.20110218 - 0812.

  rp0428 wrote:
  Shouldn't be using replaceAll if you use a regular expression?

  http://docs.Oracle.com/javase/6/docs/API/Java/lang/string.html

  and, on the assumption that you actually use a replaceAll(), you need to escape to the '-' characters as they are themselves i.e. Java escape characters

  " \\(.+\\)"
  

  In addition, the use of ".". +"can lead lead to important feedback. In this simple example you probably won't notice it, but in general, you'd better avoid the return back by using the reluctant modifier or by replacing the '.' by a character set that does not include the character terminal.

• Help... Regular expression for characters [special] /: ^ &amp; * () @# $

  Regular expression to parse the string in square brackets:
  I am trying to parse a string in square brackets, but as [] are special characters used in regular expressions to start a character class, I want to remove its special meaning. This regular expression **_user=[a-zA-Z]*@[a-zA-Z]{2}+.abc.com_** works for the analysis of [email protected]* but I want to analyze _user = [[email protected]] _ *, I use the term regular *user=\[[a-zA-Z]*@[a-zA-Z]{2}+.abc.com\]*_* but is in error because it makes special sense of character class. I tried to use \ backslash before *------[]] * but it not working giving error * "INVALID ESCAPE SEQUENCE."
  
  [JAVA CODE: user = "user=\[[a-zA-Z]*@[a-zA-Z]{2}+.abc.com\"];
  Pat pattern is Pattern.compile (user);.
  Carpet to match = pat.matcher ("2011-03-11 02:08:44, 653: User = [[email protected]], Doing report: account.server.regprocmemunix.daily");
  If (mat.find ()) {}
  User1 = mat.group ();
  }
  This room code gives error _ _ "SEQUENCE EXHAUST" INVALID and throw a PatternSyntaxException*.
  
  Please help me parse the string within large brackets [].

  You realize that, although ' [' is a special character in a regular expression, ' \' is a special character in Java, right?] You also need to escape to the ' \'.

  So, if you want to use ' [' in your regular expression, you must use something in the sense of]

  \\[
  

  Published by: almightywiz on April 25, 2011 14:05

  DOH... too slow...

• AFExactMatch and regular Expressions

  Hello

  I'm working on something similar to this http://forums.adobe.com/message/4215772#4215772.

  My first shot at it.  Found it useful messages of try67 and George.  I'm almost there and now, I'm stumped.

  My Custom Format Script is the following:

  Custom Format script for the text field

  If (event.value) event.value += '% ';

  My typing a custom Script is the following:

  Custom keystroke script

  DigOnlyKS();

  My custom calculation script is the following:

  If (event.value == "") event.value = 0;

  My Javascript Document is:

  function DigOnlyKS() {}

  Get everything that is currently in the field

  var val = AFMergeChange (event);

  Refuse entry if other thing that the figures

  Event.RC = AFExactMatch (/ \d* /, val);

  Event.RC = AFExactMatch (/^100\.0| [1-9] {1,2}\.\d{1}|0\.\d{1}$/, val);

  }

  Valid entries must be: 100.0, 1.0 to 99.9, and 0.1 to 0.9.

  The regular Expression seems to work on the Regex testers online.

  However, I can't enter anything in the field with the foregoing.

  If replace my Regex with the link above - ^ \d+\. ? \d{0,2}$ - I can enter data in the field.

  However, if I remove the? the - ^\d+\.\d{0,2}$ - regex to require (I think) the decimal places, I can't enter anything in the field.

  Clearly, Miss something.

  Thank you.

  Regular expression is not appropriate for a typo script until the value has been committed, and even in this case, it may be too restrictive (for example, does not allow for ".") 9 »).

  The way I would be inclined to implement would be to put the strike and Validate Format (not calculate) scripts to call the following functions, respectively.

  // Custom Keystroke script calls this
  function percent_ks() {
  
      AFNumber_Keystroke(1, 1, 0, 0, "", false);
  
  }
  
  // Custom Format script call this
  function percent_fmt() {
  
      AFNumber_Format(1, 1, 0, 0, "", false);
      if (event.value) event.value += "%";
  
  }
  
  // Custom Validate script calls this
  function percent_val() {
  
      if (event.value === "") event.value = 0;
  
      // Only allow numbers from 0 to 100
      if (+event.value >= 0 && +event.value <= 100) {
          // round to nearest tenth
          event.value = util.printf("%.1f", event.value);
      } else {
          event.rc = false;
          app.alert("Your error message goes here.", 3);
      }
  
  }
  

  It lets you enter values outside the valid range, but the validation script takes care of it. It also allows more than one digit to the right of the decimal, but the validation script it rounds up to the tenth closest. Make sense?

  Edit: correction of fault