I am trying to parse out some data using notepad++ macro. Here is the example of the data I have
<abcdefghkdadajsdkdjg><hhDate>2019-12-31 <dklajdlajdkjasd>
I want hhDate 2019-12-31 from the above data. I am very new to RegEx so I didn't try anything but I used notepad++ techniques to select and delete the unnecessary text but didn't work out.
Any help is appreciated.
Thanks
Assuming each of the strings are on a new line because you have to capture the whole line to remove the 'junk' and leave the good stuff, find the start of the line (^), then find first bit you want to capture and wrap it in () then find the second bit and wrap it in (), then proceed on to the end of the line ($).
So in Notepad++ work to get all the strings on separate lines first if they are not already. Then find/replace with 'regex' mode selected:
Find:
^.*?<.*?<(hhDate)>(\d+-\d+-\d+).*$
Replace:
$1 $2
https://regex101.com/r/BKha4m/1
If you don't want < to be removed before hh ? Then try this short code.
Find what: \s<.*?>
Replace with: nothing
Otherwise use this \s<.*?><|<.*>
Uncheck match-case
Related
I would like to append _OLD to the end of each strings that starts with SR_ but before the symbol ' or without it
For example my text is the following:
SR_Apple
When the 'SR_APPLE' rotten, we must discard it.
I would like the find and replace to do:
SR_Apple_OLD
When the 'SR_APPLE_OLD' rotten, we must discard it.
I have tried (SR_*)+$.*(?='\s) based on what i Learned but no luck so far. Please help. Thx in Adv
For simple cases you should be able to use
Find: (\bSR_[\w]+)
Replace: $1_OLD
(\bSR_.+?)('|$) and $1_OLD$2 could also work if the text after SR_ is more complex
The lookbehind you're using is only matching the string if it ends with a ' so it won't find the text not in quotes.
regex101 is a useful tool for debugging expressions
I don't know anything about Notepad++ Regex.
This is the data I have in my CSV:
6454345|User1-2ds3|62562012032|324|148|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|0|0|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|1534|51564|411b0fdf54fe29745897288c6ad699f7be30f389
How can I use a Regex to remove the 5th and 6th column? The numbers in the 5th and 6th column are variable in length.
Another problem is the User row can also contain a |, to make it even worse.
I can use a macro to fix this, but the file is a few millions lines long.
This is the final result I want to achieve:
6454345|User1-2ds3|62562012032|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|411b0fdf54fe29745897288c6ad699f7be30f389
I am open for suggestions on how to do this with another program, command line utility, either Linux or Windows.
Match \|[^|]+\|[^|]+(\|[^|]+$)
Repalce $1
Basically, Anchor to the end of the line, and remove columns [-1] and [-2] (I assume columns can't be empty. Replace + with * if they can)
If you need finer detail then that, I'd recommend writing a Java or Python script to manual parse and rewrite the file for you.
I've captured three groups and given them names. If you use a replace utility like sed or vimregex, you can replace remove with nothing. Or you can use a programming language to concatenate keep_before and keep_after for the desired result.
^(?<keep_before>(?:[^|]+\|){3})(?<remove>(?:[^|]+\|){2})(?<keep_after>.*)$
You may have to remove the group namings and use \1 etc. instead, depending on what environment you use.
Demo
From Notepad++ hit ctrl + h then enter the following in the dialog:
Find what: \|\d+\|\d+(\|[0-9a-z]+)$
Replace with: $1
Search mode: Regular Expression
Click replace and done.
Regex Explain:
\|\d+ : match 1st string that starts with | followed by number
\|\d+ : match 2nd string that starts with | followed by number
(\|[0-9a-z]+): match and capture the string after the 2nd number.
$ : This is will force regex search to match the end of the string.
Replacement:
$1 : replace the found string with whatever we have between the captured group which is whatever we have between the parentheses (\|[0-9a-z]+)
In the example below, is there any way to place a string like ("1one1") before {",} at the end of all lines which contain {ī}?
īn:"ZZin",
ín:"FFin",
ǐn:"QQin",
ìn:"TTin",
ie:"XXie",
iē:"TTie",
ié:"GGie",
Thanks
Using Notepad++ regex search for ^(.*ī.*)(",)$ and replace with \11one1\2.
You will need to use regex regex for notepad++.
so, mark "Regular Expression" in the final of Replace box.
in your fields to search:
find what: ī.[^"]"([A-Za-z0-9]*)
replace with: īn:"\11one1
i think it will do what you want. Let me know if it doesn't to edit the regex.
I have 58K files where I need to find this expression
()">A Random sentence.</A></P>
and i need to replace A Random Sentence by nothing.
I was trying on Notepad++ something like
Find What: ()">[[:alnum:][:punct:][:space:]]</A></P>
Replace: <empty>
Not even gettng results from the search...
Waiting for some feedback.
Try to find
(\(\)">).*(<\/A><\/P>)
and replace it with
$1\<empty\>$2
The idea is to save left part and right part, placing essential parts in brackets ().
The ".*" means every character in between.
In replace statement we call $1 and $2 to access saved parts.
You also can try :
(?<=\(\)">)[a-z \.-]+(?=</A></P>)
here [a-z \.-] you put everything what you want to search
Also parenthesis in Notepad++ should be mark with \
This should work for you:
Find: (?<=\(\)">)A Random sentence.(?=<\/A><\/P>)
Replace: <empty>
If A Random sentence. is not the actual sentence you can replace the find with:
(?<=\(\)">).*?(?=<\/A><\/P>)
I have a large logfile (+100 000 lines) in XML like so:
<container>
<request:getApples xml="...">
...
</request:getApples>
<request:getOranges xml="...">
...
</request:getOranges>
</container>
...
I want to extract the :getXXXX part to
getApples
getOranges
by doing a regex find & replace in Sublime Text 2.
Something like
Find: [^(request:)]*(.*) xml
Replace: $1\n
Any regex masters that can assist?
Correcting mart1n's answer and actually using ST2 and your sample input, I came up with the following:
First, CtrlA to select all. Then, CtrlH,
Search: .*?(get\w+) .*
Replace: $1
Replace All
Then,
Search: ^[^get].*$
Replace: nothing
Replace All
Finally,
Search: ^\n
Replace: nothing
Replace All
And you're left with:
getApples
getOranges
Not familiar with Sublime Text but you can do in two parts:
Find .*?\(get\w+\) .* and replace with \1. Now those get* strings are on separate lines with nothing else. All that remains is to remove the cruft.
So, many ways to do this. Easy one: find ^[^g][^e][^t].*$ and replace with nothing (an empty string).
Now you have your file that contains just the string you want and some empty lines, which (I hope) Sublime can get rid of with some delete-empty-lines function.
You can quickly throw all of the above in a macro and execute at will for any input following the same format ;-)
If you're willing to take the problem out of sublime text, you can use the dotall flag along with lazy matching to extract only the getXXX parts.
Replacing
.*?(get\w*) .*?
with
$1\n
should get you most of the way, only leaving a bit of easily removeable closing tags at the end of the file that I can't figure out at present.
You can check this solution here.
Maybe someone could take this and figure out a way to remove the extra closing tags.
Try this
Find what: :(\w+)>|.\s?
Replace with: $1
And if didn't work as intended, then let me know?