I have a big sql file with thousand user something like this:
('someone1#mydomain.com','{SSHA512}JWHCqHzazH2vGneLPfhMKkoAamzvxdNCWYOlhZ+uDx36jHdoMXwQmbEemvUMn7ZG6c9+22noXjjb2hAb99/5A/slscDJPKav','','en_US','maildir','Maildir','/home/vmail','vmail1','mydomain.com/someone1/',0,'mydomain.com','','','normal','',0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,NULL,'1970-01-01 01:01:01',0,'',NULL,NULL,'2020-03-19 13:15:58','2015-08-03 06:11:53','2020-03-19 13:15:58','9999-12-31 00:00:00',1'someone1'),
('someone2#mydomain.com','{SSHA512}UoMeyocmdC2DxM0S7B4WFdjnCNuvkngzzLus33h9nugKVlvdhlcboKmMDDuAkCHEyLBUgf8DicKWFPJVS7EOF/ytv27MQ3Ch','','en_US','maildir','Maildir','/home/vmail','vmail1','mydomain.com/someone2/',0,'mydomain.com','','','normal','',0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,NULL,'1970-01-01 01:01:01',0,'',NULL,NULL,'2015-12-17 12:27:35','2015-08-03 06:44:10','2021-06-08 06:55:33','9999-12-31 00:00:00',1'someone2'),
('someone3#mydomain.com','{SSHA512}A6ToCf4OfP3XNEU9ngEmGN/LDquH9+s9Qxme3SoJaDyVvxiWpnwwTiAALSdnmhIxDB2VQK0zhdF+jP8ARvh0N3IDL0Xv/KmL','','en_US','maildir','Maildir','/home/vmail','vmail1','mydomain.com/someone3/',0,'mydomain.com','','','normal','',0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,NULL,'1970-01-01 01:01:01',0,'',NULL,NULL,'2018-04-03 12:31:09','2015-08-03 06:50:01','2018-04-03 12:31:18','9999-12-31 00:00:00',1'someone3'),
('someone4#mydomain.com','{SSHA512}t7/JbUPQ+rtKeRTgWRH6KlETr2JsqYORBOZouzOzs4Wo6YfHYLoy0m+U4kZXk+AeNgMep2hGZSodPZdK2l2bn9MhOKHOuF/L','','en_US','maildir','Maildir','/home/vmail','vmail1','mydomain.com/someone4/',0,'mydomain.com','','','normal',''0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,NULL,'1970-01-01 01:01:01',0,'',NULL,NULL,'2020-03-18 07:48:26','2016-11-14 06:59:04','2021-06-08 05:54:28',9999-12-31 00:00:00',1'someone4')
And now I need to delete the last word ('someone1' , 'someone2' , 'someone3' , 'someone4') for every user which adjoining to 1. It will be looks like
....9999-12-31 00:00:00',1)
not like in original
....9999-12-31 00:00:00',1'someone1')
....9999-12-31 00:00:00',1'someone2')
etc
But don't forget they are not in different lines. All this is in one big line and this makes me to ask you help. Thanks a lot.
It seems that (from your examples) the rows do not contain any parentheses except their start and end characters. So you can search for one quotation mark ', and a number of letters and/or digits, and one quotation mark ', and than ).
To do this;
Open Replace window in Notepad++ by using ctrl+h shortcut
From Search Mode section select Reqular expression
Write '[a-zA-Z0-9]*?[-,_,.]*?[a-zA-Z0-9]*?[-,_,.]*?[a-zA-Z0-9]*?[-,_,.]*?[a-zA-Z0-9]*?'\) to Find what box
Write '\) to Replace with box
Click Replace All button.
This works if user names consist of letters or digits and _, -, . at most 3 times.
Be Sure that you have a copy of original file as a backup. And also be aware of that the regular expression that we use may find unrelated parts if any row contains closing parentheses except end of it.
All.
I am used to programming VBA in Excel, but am new to the structures in Word.
I am working through a library of text files to update them. Many of them are either OCR documents, or were manually entered.
Each has a recurring pattern, the most common of which is unnecessary carriage returns.
For example, I am looking at several text files where there is a double return after each line. A search and replace of all double carriage returns removes all paragraph distinctions.
However, each line is approximately 30 characters long, and if I manually perform the following logic, it gives me a functional document.
If there is a double carriage return after 30+ characters, I replace them with a space.
If there were less than 30 characters prior to the double return, I replace them with a single return.
Can anyone help me with some rudimentary code that would help me get started on that? I could then modify it for each "pattern" of text documents I have.
e.g.
In this case, there are more than
thirty characters per line. And I
will keep going to illustrate this
example.
This would be a new paragraph, and
would be separated by another of
the single returns.
I want code that would return:
In this case, there are more than thirty character returns. And I will keep going to illustrate this example.
This would be a new paragraph, and would be separated by another of the single returns.
Let me know if anyone can throw something out that I can play with!
You can do this without code (which RegEx requires), simply using Word's own wildcard Find/Replace tools, where:
Find = ([!^13]{30,})[^13]{1,}
Replace = \1^32
and, to clean up the residual multi-paragraph breaks:
Find = [^13]{2,}
Replace = ^p
You could, of course, record the above as a macro...
Here is a RegEx that might work for you:
(\n\n)(?<!\.(\n\n))
The substitution is just a plain space, you can try it out (and modify / tweak it) here: https://regex101.com/r/zG9GPw/4
This 'pattern' tells the RegEx engine to look for the newline character \n which occurs x2 like this \n\n (worth noting this is from your question and might be different in your files, e.g. could be \r\n) and it assumes that a valid line break will be proceeded by a full stop: \..
In RegEx the full stop symbol is a single character wild card so it needs to be escaped with the '\' (n and r are normal characters, escaping them tells the RegEx engine they represent newline and return characters).
So... the expression is looking for a group of x2 newline characters but then uses a negative look-behind to exclude any matches where the previous character was a full stop.
Anyway, it's all explained on the site:
Here is how you could do a RegEx find and replace using NotePad++ (I'm not sure if it comes with RegEx or if a plugin is needed, either way it is easy). But you can set a location, filters (to target specific file types), and other options (such as search in sub-directories).
Other than that, as #MacroPod pointed out you could also do this with MS Word, document by document, not using any code :)
So I have a field that's basically storing an entire XML file per row, complete with line breaks, and I need to remove some text from close to three hundred rows. The replace() function doesn't find the offending text no matter what I do, and all I can find by searching is a bunchy of people trying to remove the line breaks themselves. I don't see any reason that replace() just wouldn't work, so I must just be formatting it wrong somehow. Help?
Edit: Here's an example of what I mean in broad terms:
<script>...</script><dependencies>...</dependencies><bunch of other stuff></bunch of other stuff><labels><label description="Field2" languagecode="1033" /></labels><events><event name="onchange" application="false" active="true"><script><![field2.DataValue = (some equation);
</script><dependencies /></event></events><a bunch more stuff></a bunch more stuff>
I need to just remove everything between the events tags. So my sql code is this:
replace(fieldname, '<events><event name="onchange" application="false" active="true"><script><![field2.DataValue = (some equation);
</script><dependencies /></event></events>', '')
I've tried it like that, and I've tried it all on one line, and I've tried using char(10) where the line breaks are supposed to be, and nothing.
Nathan's answer was close. Since this question is the first thing that came up from a search I wanted to add a solution for my problem.
select replace(field,CHAR(13)+CHAR(10),' ')
I replaced the line break with a space incase there was no break. It may be that you want to always replace it with nothing in which case '' should be used instead of ' '.
Hope this helps someone else and they don't have to click the second link in the results from the search engine.
Worked for me on SQL2012-
UPDATE YourTable
SET YourCol = REPLACE(YourCol, CHAR(13) + CHAR(10), '')
If your column is an xml typed column, you can use the delete method on the column to remove the events nodes. See http://msdn.microsoft.com/en-us/library/ms190254(v=SQL.90).aspx for more info.
try two simple tests.
try the replace on an xml string that has no double quotes (or single quotes) but does have CRLFs. Does it work? If yes, you need to escape the quote marks.
try the replace on an xml string that has no CRLFs. Does it work? Great. If yes use two nested replace() one for the CRLFs only, then a second outter replace for the string in question.
A lot of people do not remember that line breaks are two characters
(Char 10 \n, and Char 13 \r)
replace both, and you should be good.
SELECT
REPLACE(field , CHR(10)+CHR(13), '' )
FROM Blah..