on security, the withdrawal of restrictedsites all entries at once?
on security, the withdrawal of restrictedsites all entries at once?
Hi parakkatmenon,
What version of IE are you using?
It is not possible to delete all entries at once. You can remove sites one by one in the list of sensitive sites.
For more information, refer to this link: security zones: adding or removing websites
Hope the helps of information.
Please post back and we do know.
Tags: Windows
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Josh,
Welcome to the forum.
The best way to handle things is to choose a pre-defined sequence that matches your footage source of the layer at 100%.
In light of the where you are now, you can go to your first Clip and adjust the fixed effect > Motion > adapted for her greatness. Then, Rt-click on it, choosing the copy. Now select all the other Clips (Lasso with the slider will work) and then Rt click, choose the attributes of the dough.
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adding all entries from one collection to another?
Hello
my database is a 11g
I have a function (in a box) than returnes a collection.
I want to call the function multiple times and the caller must combine all entries selected in a single collection (new?):
create or replace package my_test is procedure public_proc; end my_test; create or replace package body my_test is cursor test_data(selected_type varchar2) is select (1+ABS(MOD(dbms_random.random,100000))) as id, selected_type||level as my_value from dual connect by level<5; type my_type is table of test_data%row_type; function private_function(selected_type varchar2) return my_type is v_my_type my_type := my_type(); begin open test_data(selected_type=> selected_type); fetch test_data bulk collect into v_my_type; close test_data; return v_my_type; end private_function; procedure public_proc is v_my_type_a my_type := my_type(); v_my_type_b my_type := my_type(); v_my_type_ab my_type := my_type(); begin v_my_type_a := private_function(selected_type=>'A'); v_my_type_b := private_function(selected_type=>'B'); -- simplest solution to combine v_my_type_a and v_my_type_b to v_my_type_ab, possible without explicit loop? for i in v_my_type_ab.first..v_my_type_ab.last loop dbms_output.put_line('id: '||v_my_type_ab(i).id||', value:'||v_my_type_ab(i).my_value); end loop; end public_proc; end my_test;
desired output (x instead of random number):
id:x, value:A1 id:x, value:A2 id:x, value:A3 id:x, value:A4 id:x, value:B1 id:x, value:B2 id:x, value:B3 id:x, value:B4
Good bye
DPT
I can also do a big collect in against a collection of nested type?
OK – what a trick question?
You ask if you can do what the code you first published already does?
Hmmm - well, I'll say YES - since your code is already doing it.
What type of collection, exactly, do you think that you have agreed with this line?
type my_type is % row_type test_data table;
Of course, 'row_type' is not valid; It is % ROWTYPE.
If you are unsure what type of collection which is then I suggest that you only consider "with the help of the Collections of PL/SQL and Records" in the language PL/SQL doc
http://docs.Oracle.com/CD/B28359_01/AppDev.111/b28370/Collections.htm
He explained each of the collection types, explain how to choose one that suits and examples of their use.
DECLARE
cursor emp_data (my_job emp.job%type) is select *.
from emp where job = my_job;
type my_emp_data is table of the emp_data % rowtype;
my_emp_data v1.
my_emp_data v2.
my_emp_data v3;
BEGIN
Open emp_data ("REGISTRAR");
extraction emp_data bulk collect in v1;
close emp_data;
Open emp_data ("SELLER");
extraction emp_data bulk collect in v2;
close emp_data;
v3: = v1 v2 of MULTISET UNION;
for me in v3.first... loop v3. Last
dbms_output.put_line (v3 (i) .ename |) ' - ' || V3 (i) .job);
end loop;
END;
/
SMITH - CLERK
ADAMS - CLERK
JAMES - CLERK
MILLER - CLERK
ALLEN - SELLER
WARD - SELLER
MARTIN - SELLER
TURNER - SELLER
Also note that I do NOT use a query to combine collections:
v3: = v1 v2 of MULTISET UNION;
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Hi carmol,.
Your question of Windows is more complex than what is generally answered in the Microsoft Answers forums. It is better suited to the audience on TechNet forum.
Please post your question in the Sub forum. Link: http://social.technet.microsoft.com/Forums/en-us/winserverManagement/threads
With regard to:
Samhrutha G S - Microsoft technical support.
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By default, event viewer produced a considerable number of newspapers.
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Hi tgblueblade,
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2 are. what device or software you using Sync?
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http://Windows.Microsoft.com/en-us/Windows-Vista/managing-your-contacts
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Hi all
I have Apple devices for a few years now, but I still find it difficult to understand the system. I want to put an end to that, so I finally registered.
I try to look at the tutorials and read the forums, but I can't find concrete answers to my situation.
First of all, in my application of Photos of my MacBook, I remove hundreds of photos and videos, then I go to recently deleted screen and I delete them all permanently.
However my disk space is the same before under 'Storage' summary when I check "to about this Mac". I restarted the Mac and waited for days, the space does not release upward.
Then I read in a forum and found the following answer:
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If you can live without ever using iPhoto/Aperture once again, go to your
Pictures
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andAperture Library
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(1) after reading this, I decided to look it up in my pictures folder in the Finder to locate the iPhoto library, however it is not there on the left side, after reading the forums, I still cannot locate the iPhoto library anywhere.
(2) in case I finally find my iPhoto library and delete it, are the images that I see now in the deleted Application Photos?
(3) another curiosity, when I try to open IPHOTO, the following message appears:
And then after trying the 'search App Store':
I did not need to actually use my iPhoto or App Photos in particular, I do not yet fully understand the difference. I want to just my photos to be stored in ONE place and when I take hours to remove them permanently, get my rear space and do not learn, they still exist elsewhere (iPhoto library).
Also please help me understand what am I supposed to do with the iPhoto application now that gives me these error messages.
Thank you very much!
(1) after reading this, I decided to look it up in my pictures folder in the Finder to locate the iPhoto library, however it is not there on the left side, after reading the forums, I still cannot locate the iPhoto library anywhere.
The images folder must be manually added to the sidebar.
Open the Finder preferences:
- Click anywhere on the empty office to bring the Finder to the front, then go to the Finder menu in the main menu bar, open "Preferences".
Click the tab of the sidebar and add all the items to the sidebar, you want to see.
Now, open the images in the sidebar folder in the Finder. Is your photo library in iPhoto there?
Then drag it to an external drive to make a backup copy, then delete it and empty the trash. Accidentally delete the photo library.
(2) in case I finally find my iPhoto library and delete it, are the images that I see now in the deleted Application Photos?
No, the photos in your library of Photos and the iPhoto library are separate files. It was created by hard links.
Hard links are new files, which use the same entry in the table. They store the contents in blocks of same on the disks, so the entries are not duplicated, but if you remove one of the linked files, the others will remain. The storage will be released, when the last linked file has been deleted.
See this link: Six colors: the (hard) link between iPhoto and photo
- Click anywhere on the empty office to bring the Finder to the front, then go to the Finder menu in the main menu bar, open "Preferences".
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Hello!
After the end of my second day of test error, I put this question on the table:
I use ArraytoChannels function to store ADO recordsets as strings. What is strange, is that for the first Recordset, it works; but for the next time through the loop, it always fails with the error message 'cannot be added because the channels of the target are not all the same length.
I confirmed that:
the RowData sizes and the ChannelNames are equal,
both spend the isarray = true test,
I change the order of the ChannelNames,.
I have reconnected/disconnected from the oConnexion every time, nothing has changed.
Apparently I'm missing something - but crazy to know what! -If anyone can share his opinion I'll so much appriciate. Here is my code:
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Call OpenSQLConnection
Set oRecordset = CreateObject ("ADODB. Recordset')
Call SelectWell
Call GetWellStateIDsData.Root.Clear
for j = 0 to ubound(oTables,1)sSQLSting = "select * []" & oTables (j) & "] where [WellStateID] between" & WellStateIDFirst & "and" & WellStateIDLast ".
oRecordset.Open sSQLSting, oConnexion
Protected oFieldNames: table: ReDim oFieldNames (orecordset. Fields.Count - 1).
for i = 0 to orecordset. Fields.Count - 1
oFieldNames (i) = orecordset. Fields.Item (i) .name
next
oArray = oRecordset.GetRows (-1, 0, oFieldNames)
Set oGroup = Data .root .ChannelGroups .Add (oTables (j))oArray, oFieldNames arraytochannels
oRecordset.close
oConnection.Close
nextSub GetWellStateIDs
sSQLSting = "select * from [WellStates] where [wellid] =" & WellID
oRecordset.Open sSQLSting, oConnexion
oArray = oRecordset.GetRows)
WellStateIDFirst = oArray (0,0)
WellStateIDLast = oArray (0, ubound(oArray,2))
oRecordset.close
EndSubSub OpenSQLConnection
Set WshNetwork = CreateObject
oComputerName = WshNetwork.ComputerName
oDB = "MX2. Player.DB ".
Set oConnexion = CreateObject ("ADODB. Connection")
oProvider = "Provider = SQLOLEDB.1; Integrated Security = SSPI; PeoExecuteist Security Info = True; Data Source ='
oProvider = oProvider & oComputerName & "\MX; Use procedure for prepare = 1; Machine translation = True; The packet size = 4096; Workstation ID ="
oProvider = oProvider & oComputerName & " Use encryption for data = False; Tag with column collation when possible = False; Initial Catalog ='
oProvider = oProvider & oDB
oConnection.ConnectionString = oProvider
oConnection.Open
EndSubAnother clue. If you check using DIAdem
Microsoft Windows Script Debugger
you are able to install the debugger in DIAdem.
It would potentially have shown that the command does not work as expected.
Sorry for the inconveniance
Andreas
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