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 have file with rows like this:
"B4P(6-3,5)-VH(LF)(SN)",JST,2018+,34000,SMD
893D226X0016C8W,VISHAY,2018+,"30,000",SMD
BL-BUF1V4V-AT-L,FOXLINK,2018+,1890,CONN
"TLP721F(D4-GR,M,F)",NSC,2001+,114,AUCDIP-16
How can i find all commas inside quotes? For example, i need to find this:
"B4P(6-3 >>,<< 5)-VH(LF)(SN)",JST,2018+,34000,SMD
893D226X0016C8W,VISHAY,2018+,"30 >>,<< 000",SMD
BL-BUF1V4V-AT-L,FOXLINK,2018+,1890,CONN
"TLP721F(D4-GR >>,<< M >>,<< F)",NSC,2001+,114,AUCDIP-16
Now I can only find text in quotes, tell me how to select only commas from it, using one regular expression?
("(?:\[??[^\[]*?"))
Regex101 - online regex editor and debugger
Here is a simplistic solution that works with your example:
It match only quoted strings having one or more , inside.
grep '"[^,]*,[^"]*"'
Hope it works for you.
Explanation
"[^,]* match " and following non , chars
, match the first , char
[^"]*" match following non " till find the next"
I've been trying to find a way to find a single comma between inverted commas without much luck. Example: "text , text " - how do I isolate the "," between the inverted commas line by line in a flat file?
My attempt .["].[,].["].
Thanks in advance
this regex will work
(?<=truck).*(?=car)
finds e.g. "plane" in the string
truckplanecar
so for test,test the regex would be
(?<=test).*(?=test)
PS. can you please provide an more detailed example what you would like to do
Try using 2 group at the start and end of the string, the following regex should work:
(".*),(.*")
it does match the example you've shared:
"text , text "
Furthermore, using groups, you can isolate the string before the comma and afterwards, in case you'll be needed it.
Suppose I have these strings:
'akshay ' ,
' ankur'
I want to remove trailing and leading spaces present under quotation mark like this:
'akshay',
'ankur'
How can I achieve this in Notepad++ using RegEX?
Use this to find: '(\s*)(.*?)(\s*)' and this to replace: '\2'.
Find this regexp: (^')\s+(\w+)\s*(')|(^')\s*(\w+)\s+(')
Replace with : \1\2\3\4\5\6
Like this:
All the RegExp given above solve my problem in a better way. I appreciate to all who helped me to solve this problem. Thank you everyone.
Here i am going to add a RegExp given below which can be used to count total number of spaces inside single quotation. It can be used to remove all spaces also.
Find--> \s(?!(?:[^']*'[^']*')*[^']*$) and ReplaceWith-->Nothing to do(leave blank) then click 'ReplaceAll'.
all! Have CSV files coming in with with text inside double quotes that contain one or more commas and I am wondering if there is a regex form for Notepad++ that would remove any number of commas inside a CSV file.
For example I need to go from the this:
text,text1,"interesting, text,"
To this:
text,text1,"interesting text"
There can be 1,2 or more commas inside the quotes.
Anyone know a of a way to make this happen using regex form in Notepad++?
Thanks in advance!
use this pattern
,(?!(([^"]*"){2})*[^"]*$)
and replace with nothing
it is looking for a comma , that does not see an optional even number of double quotes " to the end of the string