I have CSV file with breadcrumb (prestastop products)
I want to delete the content after last separator (product name), my structure is:
col1|col2|col3|product
I can delete with simple regex, problem is that number of separators is not always the same so for example ([^|]+/[^|]+/[^|]+/[^|]+|).* wont work.
Is there any way to do it with one regex?
I want:
col1|col2|col3|product
col1|col2|col3|col4|product
col1|col2|col3|col5|product
to become
col1|col2|col3
col1|col2|col3|col4
col1|col2|col3|col5
I think the simple way would be to read from right to left and not left to right...
Use the following regex to match the last part including the | symbol. Just replacing the matched characters with an empty string will give you the desired output.
Regex:
\|[^|]*$
REplacement string:
Empty string
DEMO
^(.*)\|.*
Try this.Replace by
$1
See demo.
http://regex101.com/r/jT3pG3/12
Related
I have a problem that I really hope that somebody could help me. So, I want to delete some parts of text from a notepad++ document using Regex. If there's another software that I can use to delete this part of text, let me know please, I am really really noob with regex
So, my document its like this:
1
00:00:00,859 --> 00:00:03,070
text over here
2
00:00:03,070 --> 00:00:09,589
text over here
3
00:00:09,589 --> 00:00:10,589
some numbers here
4
00:00:10,589 --> 00:00:12,709
Text over here
5
00:00:12,709 --> 00:00:18,610
More text with numbers here
What I want to learn is how can I delete the first 2 lines of numbers in all the document? So I could get only the text parts (the "text over here" parts)
I would really appreciate any kind of help!
My solution:
^[\s\S]{1,5}\d{1,3}:\d{1,3}:\d{1,3},\d{1,5}\s-->\s*?\d{1,3}:\d{1,3}:\d{1,3},\d{1,5}\s
This solution match both types: either all data in one line, or numbers in one line and data in the second.
Demo: https://regex101.com/r/nKD0DQ/1/
Simplest solution;
\d+(\r\n|\r|\n)\d{2}:\d{2}.*(\r\n|\r|\n)
Get line with some number \d+ with its line break (\r\n|\r|\n)
Also the next line that starts with two 2-digit numbers and a colon \d{2}:\d{2} with the rest .* and its line break. No need to match all since we already are in the correct line, since subtitle file is defined well with its predictable structure.
Put this as Find what: value in Search -> Replace.. in Notepad++, with Seach Mode: Regular Expression and with replace value (Replace with:) of empty space. Will get you the correct result, lines of expected text with empty line in between each.
to see it on action on regex101
Subtitles, for accuracy you can use this:
\d+(\r\n|\n|\r)(\d\d:){2}\d\d,\d{3}\s*-->\s*(\d\d:){2}\d\d,\d{3}(\r\n|\n|\r)
Check Regular Expression, Find what with this and Replace with empty would do.
Regxe Demo
srt subtitles are basically ordered. And it's better accurate than lose texts.
\d : a single digit.
+ : one or more of occurances of the afore character or group.
\r\n: carriage and return. (newline)
* : zero or more of occurances of the afore character or group.
| : Or, match either one.
{3}: Match afore character or group three times.
I'm going for a less specific regex:
^[0-9]*\n[0-9:,]*\s-->\s[0-9:,]*
Demo # regex101
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]+)
for example I have txt with content
qqqqaa
qqss
ss00
I want to replace only one q at the beginning of line, that is to get
qqqaa
qss
ss00
I tried replace ^q in notepad++. But After I click replaceAll, I got
aa
ss
ss00
What is wrong? Is my regex wrong? What is the correct form?
The issue is that Notepad++ Replace All functionality replaces in a loop using the modified document.
The solution is to actually consume what we need to replace and keep within one regex expression like
^q(q*)
and replace with $1.
The pattern will find a q at the beginning of the line and then will capture into Group 1 zero or more occurrences of q after the first q, and in the replacement part the $1 will insert these qs inside Group 1 back into the string.
You can use ^q(.+) and replace with $1 if you also want to replace single q's.
I'm trying to select text between ></ . Example below I want "text"
>text</
but I'm unable to do so.
tried the following but it doesn't like the slash at the end of the regex
\>(.*?)\<\
I'm trying to do this in TextPad. How is this supposed to be done?
I'm ultimately wanting to delete all text between these two characters so all I'm left with is something like: <element></element>
RegEx wise, you can use 3 groupings and for the replace only use the first and 3rd group: \1\3.
Find: (>)(.*)(</)
Replace: \1\3
Try doing:
\>(.*?)\<\/
The regex that you were trying would actually have given error because you had a \ and nothing after that.
You are close.. use the following:
(>).*?(<\/)
And replace with \1\2
See DEMO
OR
You can use lookbehind and lookaheads:
(?<=>)(.*?)(?=<\/)
And replace with '' (empty string)
See DEMO
Before somebody points me to that question, I know that one can't parse html with regex :) And this is not what I am trying to do.
What I need is:
Input: a string containing html.
Output: replace all opening tags
***<tag>
So if I get
<a><b><c></a></b></c>, I want
***<a>***<b>***<c></a></b></c>
as output.
I've tried something like:
(<[~/].+>)
and replace it with
***$1
But doesn't really seem to work the way I want it to. Any pointers?
Clarification: it's guaranteed that there are no self closing tags nor comments in the input.
You just have two problems: ^ is the character to exclude items from a character class, not ~; and the .+ is greedy, so will match as many characters as possible before the final >. Change it to:
(<[^/].+?>)
You can also probably drop the parentheses and replace with $0 or $&, depending on the language.
Try using: (<[^/].*?>) and replace it with ***$1