Display the name change for the story?
Hi all
I was wondering, if I change my display name (not username, of course), which would change my nickname for ALL past and present message and call history/records (not only recent but all THE records taken)?
A quick response would be greatly appreciated.
Thank you.
Hi, Narpoo,
Yes, it is a change of 'global '.
See you soon!
Elaine
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Need to find the version number referring to the last name change for each emplid.
emplID Name age Version --- ----- --- -------- 1 ABC 25 2 1 ABC 26 5 1 def 27 9 1 def 28 15 1 def 29 18 2 JKL 15 5 2 MNO 16 8 2 MNO 17 11 2 PQR 18 20 Need to find the version number referring to the last name change for each emplid.
As indicated by the version number change when there is change in age as well, but I need a query that gives the version number referring to the last name change for each emplid.
with
DATA_TABLE (EmplId, Name, Age, version) as
(select 1, 'abc', 25, 2 double Union all
Select 1, 'abc', 26, 5 Union double all the
Select 1, 'def', 27, 9 double Union all
Select 1, 'def', 28, 15 double Union all
Select 1, 'def', 29, 18 double Union all
Select 2, 'jkl', 15 5 Union double all the
Select 2, 'mno', 16, 8 double Union all
Select 2, 'mno', 17, 11 double Union all
Select 2, 'pqr', 18, 20 double
)
Select emplid,
Max (Name) name of Dungeon (dense_rank last order by version).
Max (Age) age of Dungeon (dense_rank last order by version).
version Max (version)
from (select emplid, name, version,)
-case where name! = lag(name,1,name) on (emplid version order partition)
then "renamed".
end change
of data_table
)
where the change is not null
Emplid group
EMPLID NAME AGE VERSION 1 def 27 9 2 PQR 18 20 Concerning
Etbin
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The worksheet name change for the interactive user role
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I have a question about DRM security for users with access add to the sheet and only read access to the members. The requirement is that the user should be able to add a sheet and change all the properties associated with the leaves but cannot add or change the Member or any property associated with a branch. To do this, I created a group of node (NAG1) and assigned categories of goods (PC1) associated with the hierarchy of the NAG1 with editing access to PC1. The NAG1 for journal access ADD and NAG1 for branch had read access. The user has only one role which is the interactive user. This way, the user cannot add spreadsheets, edit the properties associated with the leaves, but don't can't add limb or change all the properties associated with limb, HOWEVER, the user is not able to change the name of an existing journal. If I give the user role 'Director of Application', while they are able to change the name of the system, but then they see the section Administration on the left and everything related which we want give...
Is it possible to give the user the ability to change the name of the worksheet without giving ""Application Administrator ' role? "
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Have you checked RenameLeaf and RenameLimb system preferences? I think that by default, only the administrator can change the name of the node, but you can grant this possibility of additional roles.
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Find the layer with the name "Name1" a name change for 'name2' and then save PDF
Hi there together, we have the following problem:
We need a Javascript script to rename all the layers named 'name1' to 'name2' and then save the PDF file.
Any help would be greatly appreciated.
Thank you very much
Patrick
This code will rename the layer:
var ocgArray = this.getOCGs(); for (var i=0; i
Save the file afterwards is more complicated. You can use the Save command sub in a context of trust to do it silently, or you can call the dialog box save as, and let the user manually, by using the command app.execMenuItem ("SaveAs").
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Here's how to uninstall the Google toolbar: http://www.google.com/support/toolbar/bin/answer.py?hl=en&answer=9231
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How can I active names change for images exchanged in duplicate symbols?
I have what seems to be a bug in cc2014 driven on-Board under the following circumstances:
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You can assign IDS and classes of symbols with attr ('id', 'newIDName') and addClass().
So let's say I have a symbol that contains other symbols - I had to first give an id to the outermost symbol, then drill down to other internal symbols. So since I can remedy directly, then I can do what I want with them.
When I want to create several symbols, then I run a loop so that they have the same name with an index. Here, I'm not add an id to the internals, but you could use the same principle. You can also assign names of classes.
example:
create buttons in a div. navigation bar
for (m = 0; m<>
var buttons = sym.createChildSymbol ("button", "navBar");
Buttons.Element.attr ("ID", "Button" + m);
SYM. $("#button" + m) .css ({'position': 'absolute', 'top': 5, 'left': (distance += 50)}); Make sure you have the # in front of the name of the new
sym.getSymbol("#button"+m).$('bg').css ({'background-color':'white'});})
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How can I change this request, so I can display the name and partitions in a r
How can I change this request, so I can add the ID of the table SPRIDEN
from now on gives me what I want:
in a single line, but I would like to add the id and the name that is stored in the SPRIDEN table1,543 A05 24 A01 24 BAC 24 BAE 24 A02 20 BAM 20
SELECT sortest_pidm, max(decode(rn,1,sortest_tesc_code)) tesc_code1, max(decode(rn,1,score)) score1, max(decode(rn,2,sortest_tesc_code)) tesc_code2, max(decode(rn,2,score)) score2, max(decode(rn,3,sortest_tesc_code)) tesc_code3, max(decode(rn,3,score)) score3, max(decode(rn,4,sortest_tesc_code)) tesc_code4, max(decode(rn,4,score)) score4, max(decode(rn,5,sortest_tesc_code)) tesc_code5, max(decode(rn,5,score)) score5, max(decode(rn,6,sortest_tesc_code)) tesc_code6, max(decode(rn,6,score)) score6 FROM (select sortest_pidm, sortest_tesc_code, score, row_number() over (partition by sortest_pidm order by score desc) rn FROM (select sortest_pidm, sortest_tesc_code, max(sortest_test_score) score from sortest,SPRIDEN where SPRIDEN_pidm =SORTEST_PIDM AND sortest_tesc_code in ('A01','BAE','A02','BAM','A05','BAC') and sortest_pidm is not null GROUP BY sortest_pidm, sortest_tesc_code)) GROUP BY sortest_pidm;
Hello
That depends on whether spriden_pidm is unique, and you want to get the results.
Whenever you have a problem, post a small example of data (CREATE TABLE and INSERT, relevamnt columns only instructions) for all the tables and the results desired from these data.
If you can illustrate your problem using tables commonly available (such as in the diagrams of scott or HR) so you need not display the sample data; right after the results you want.
Whatever it is, explain how you get these results from these data.
Always tell what version of Oracle you are using.Looks like you are doing something similar to the following.
Using the tables emp and dept of the scott schema, producing a line of production by Department showing the highest salary for each job, for a set given jobs:DEPTNO DNAME LOC JOB_1 SAL_1 JOB_2 SAL_2 JOB_3 SAL_3 ------ -------------- ------------- ------- ----- ------- ----- ------- ----- 20 RESEARCH DALLAS ANALYST 3000 MANAGER 2975 CLERK 1100 10 ACCOUNTING NEW YORK MANAGER 2450 CLERK 1300 30 SALES CHICAGO MANAGER 2850 CLERK 950
On each line, jobs are listed in order by the highest salary.
This seems to be similar to what you are doing. The roles played by the sortest_pidm, sortest_tesc_code and sortest_test_score in your table sortest are played by deptno, job and sal in the emp table. The roles played by the spriden_pidm, id and the name of your table spriden are played by deptno, dname and loc in the dept table.Looks like you already have something like the query below, which produces a correct output, except that it does not include the dname and loc of the dept table columns.
SELECT deptno , MAX (DECODE (rn, 1, job)) AS job_1 , MAX (DECODE (rn, 1, max_sal)) AS sal_1 , MAX (DECODE (rn, 2, job)) AS job_2 , MAX (DECODE (rn, 2, max_sal)) AS sal_2 , MAX (DECODE (rn, 3, job)) AS job_3 , MAX (DECODE (rn, 3, max_sal)) AS sal_3 FROM ( SELECT deptno , job , max_sal , ROW_NUMBER () OVER ( PARTITION BY deptno ORDER BY max_sal DESC ) AS rn FROM ( SELECT e.deptno , e.job , MAX (e.sal) AS max_sal FROM scott.emp e , scott.dept d WHERE e.deptno = d.deptno AND e.job IN ('ANALYST', 'CLERK', 'MANAGER') GROUP BY e.deptno , e.job ) ) GROUP BY deptno ;
Dept.DeptNo is unique, it won't be a dname and a loc for each deptno, so we can modify the query by replacing "deptno" with "deptno, dname, loc" throughout the query (except in the join condition, of course):
SELECT deptno, dname, loc -- Changed , MAX (DECODE (rn, 1, job)) AS job_1 , MAX (DECODE (rn, 1, max_sal)) AS sal_1 , MAX (DECODE (rn, 2, job)) AS job_2 , MAX (DECODE (rn, 2, max_sal)) AS sal_2 , MAX (DECODE (rn, 3, job)) AS job_3 , MAX (DECODE (rn, 3, max_sal)) AS sal_3 FROM ( SELECT deptno, dname, loc -- Changed , job , max_sal , ROW_NUMBER () OVER ( PARTITION BY deptno -- , dname, loc -- Changed ORDER BY max_sal DESC ) AS rn FROM ( SELECT e.deptno, d.dname, d.loc -- Changed , e.job , MAX (e.sal) AS max_sal FROM scott.emp e , scott.dept d WHERE e.deptno = d.deptno AND e.job IN ('ANALYST', 'CLERK', 'MANAGER') GROUP BY e.deptno, d.dname, d.loc -- Changed , e.job ) ) GROUP BY deptno, dname, loc -- Changed ;
In fact, you can continue to use just deptno in the analytical PARTITION BY clause. It may be slightly more efficient to just use deptno, as I did above, but it won't change the results if you use all 3, if there is only 1 danme and 1 loc by deptno.
Moreover, you don't need so many subqueries. You use the internal subquery to calculate the MAX and the outer subquery to calculate rn. Analytical functions are calculated after global fucntions so you can do both in the same auxiliary request like this:
SELECT deptno, dname, loc , MAX (DECODE (rn, 1, job)) AS job_1 , MAX (DECODE (rn, 1, max_sal)) AS sal_1 , MAX (DECODE (rn, 2, job)) AS job_2 , MAX (DECODE (rn, 2, max_sal)) AS sal_2 , MAX (DECODE (rn, 3, job)) AS job_3 , MAX (DECODE (rn, 3, max_sal)) AS sal_3 FROM ( SELECT e.deptno, d.dname, d.loc , e.job , MAX (e.sal) AS max_sal , ROW_NUMBER () OVER ( PARTITION BY e.deptno ORDER BY MAX (sal) DESC ) AS rn FROM scott.emp e , scott.dept d WHERE e.deptno = d.deptno AND e.job IN ('ANALYST', 'CLERK', 'MANAGER') GROUP BY e.deptno, d.dname, d.loc , e.job ) GROUP BY deptno, dname, loc ;
It will work in Oracle 8.1 or more. In Oracle 11, however, it is better to use the SELECT... Function PIVOT.
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Hi, back back can check files you opened with "recent files"? With my current setup, I can only see the last 10 files. Can I change the setting to display the last 20 files for example? Thank you
File management preferences is where you can change the number. I don't know the maximum.
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connect display mode database using dreamweaver user name
LionelLionel wrote:
OK, sorry 4 that; I use dreamweavercs6 and the script server is php.i as accurate must I set up a server (wampserver) local on my laptop for this level of competence of project.my with scripts php is very limited. and now, after you have created the database (in phpmyadmin), I want to display the user name of the user after login... I need help please.thx
You want to display it where? On the login page of "success"?
Add the below to each top of the page:
Then add the following in the code where you want the ' user name to appear:
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Hello
I have an application that is in talks with the VC to get the events from the virtual computer.
I have a virtual machine on which I install VMWare tools to get the host name of the virtual machine. From now, the first time that the virtual machine gets up, my request to get the host name for this virtual machine. Now, if the host name of this virtual machine changes, is there any event that will be generated by VMWare or VC tools that my application can subscribe to in order to choose the host name change?
Thank you
There is no integrated case that will capture this. The virtual machine has the dns name property that can be extracted from the db, or using one of the vi toolkits. You can run gets against these data and validate it against the previous value, but no direct case is only triggered when the host inside the virtual machine name is changed.
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VMware vExpert
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Description not changed for KEY, even if the name changes applied patch
EBS 12.1.3 installed, and I try to install Russian localization following < R12 EMEA locations - CIS [ID 472418.1] module > document.
So, first of all, I installed 8640014 (KEY: Adsplice for R12.1 (All RUPs of R12.1)), as is the condition sine qua non of 9350941.
Then I installed 10164748 (change of name of Consulting EMEA locations' 'EMEA locations (9034879 replacement) module), as this is a postrequired of 8640014 patch.
Both went ok without any error.
However, when I continue with the installation of 9350941, GEN (R12.1 - KEY: Company Info package), he always complains about mismatching of the description below.
- - - - -
...
Reading patch file of the driver.
Determine the version target...
The current version of target is 12.1.3
Caution:
The description of the entity 'key' in the driver patch
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The current description is: "EMEA Consulting locations.
The description in the patch is: 'EMEA locations module.
The current description will not be changed.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
This basic patch contains files that may require translation
According to the language currently installed.
Oracle Corporation recommends that provide you translated versions of this
patch for each of your languages other than the United States before applying this basic patch.
The translated version of the patch should be applied immediately AFTER
apply this hotfix to base.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
You do not want to apply this fix now [no]?
You must check the file
/D01/Oracle/SID/apps/apps_st/appl/Admin/SID/log/adpatch_9350941_GEN.log
to find errors.
- - - - -
Then, I stopped by answering NO to the question.
How does this happen? Shouldn't the name changed after the application of the patch 10164748?Hello
This warning can be ignored.
Please check
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If I want to use Thunderbird for all of my emails in one place, I'm tired of constantly switch tabs in google to see all the... But google rocks because of how it displays everything is so beautiful.
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Mike.
No, which is combining them into one Inbox, which I don't want to do. Thanks anyway. But it seems that this program does not have what I want what it more so I'll find something else.
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When I visit a Web site that requires SSL I displays the message "this connection is untrusted". Any Web site that I visit, it's always exactly the same message and the same SSL certificate that she is no longer valid for www.thawte.com
support.Mozilla.org uses an invalid security certificate.
The certificate is not approved, because no sender string has been provided.
The certificate is valid for www.thawte.com
The certificate expired on 11/11/2011 23:59. The time now is 11:46 28/01/2012.When I click "Add the Exception" on a Web site and view the certificate, it is exactly the same certificate with the exact same serial number.
I had a similar problem with Internet Explorer showing a 404 error when I visited SSL protected pages but to do a restore of the system a month ago to correct this. All other bowsers are / were very good.
I installed Firefox 3.x month last to test something that is when the problem started. I have since uninstalled Firefox 3.x and reinstalled the latest version. I deleted all the preferences/settings, disabled modules and reinstalled many times. I did a Windows system restore to before that the problem started with no luck.
The time / Date on my computer are correct. I have no firewall other than the windows one. I had no antivirus (netbook) until I installed a (Avast) yesterday to see if a virus was causing issues (found nothing). This problem arises on any internet connection (tested to work and home).
or try to use the module Skip Cert error (to jump to the SSL/TLS certificate error page)
Thank you
Please check 'Resolved' the answer really solve the problem, to help others with a similar problem.
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Change the path where Windows search displays directory name
Dear Sir
I'm regularly using Windows search and I am facing some problems with the way the folder names appear.
Imagine the following situation: I have three files, called text.txt, located in the following directories:
D:\Folder_A\SubFolder_1
D:\Folder_A\SubFolder_2
D:\Folder_B\SubFolder_1
When I launch a Windows search to find these files, I will get answers in the column 'file ':
SubFolder_1 (D:\Folder_A)
SubFolder_2 (D:\Folder_A)
SubFolder_1 (D:\Folder_B)
When I then click on the "Folder" column header to sort, I will get following result:
SubFolder_1 (D:\Folder_A)
SubFolder_1 (D:\Folder_B)
SubFolder_2 (D:\Folder_A)
This way of sorting mix main directories "D:\Folder_A" and "D:\Folder_B", which is false.
This problem is due to the fact that the names of directories are not displayed as '
' but as ' ( )", which is not what I want. Is it possible (as a registry key) where I can configure my Windows 7 to display the names of directory as '
' (as in previous versions of Windows)? Thank you
Right-click a column header and choose "More"... "since the menu drop down.
In the choice, find the path and turn it on. When you click OK, the new column 'path' will be the one you like (showing the full path). Example:
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Firefox for Android only displays the desktop version of gmail rather than mobile.
Android only question.
I prefer to access gmail via a browser rather than the gmail app. Tests of other browsers on Android (Chrome and Dolphin) I see that they display a mobile-specific version of the gmail Web site.
Firefox for Android only displays the version of office on my Nexus 7 (5.1 runs Android 2012) - and therefore more difficult to read and use. Any help or ideas would be greatly appreciated.
Thank you
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Well, Yes, it's that Word annoying Tablet.
What I'm going to describe is difficult to type, so I hope that you can open this page on your tablet.
Select (0) and copy the following text on your Tablet if possible to avoid the excesses of type:
general.useragent.override.mail.google.com Android; Tablet;#Android; Mobile;
The forum linkifies this first part, but it is not supposed to be a link.
(1) open a new tab to Subject: config (type about: config the Awesomebar and hit the arrow go or tap Enter).
You can bookmark it for easy future access.
(2) in the search box, type agen and make a pause while Firefox filter list
(3) scroll down to the bottom and look for a preference named general.useragent.site_specific_overrides and confirm that it has its default value true. If it is set to false, press the reset button for it back to true.
(4) scroll back up and press the big button "+" to the left of the search box to add a new parameter
(A) the name of the preference must be general.useragent.override.google. com (for all Google sites) general.useragent.override.mail.google. com (for Gmail only)
If you type this, beware of the correction automatic insertion of spaces and be sure to remove those.
If paste, remove the part after that.
(B) on the right side, type Boolean and in the popup, change it to a String
(C) the value of the preference should be Android; Tablet; #Android; Mobile; -in other words, substituting Mobile for Tablet.
If you type, pay attention to the excess spaces.
If gluing, remove the preference name and extra space before this part.
(D) then click on create
If you open https://mail.google.com/ must now use the mobile layout.
Success?
Edit: To clarify the name of preference options.
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