How can I replace characters on the hexadecimal level in Notepad++?
I have a file with some characters that can not be displayed in UTF-8 (problem with ANSI to UTF-8 conversion) and would like to replace these with the correct characters.
Notepad++ replace (Ctrl + H) can not distinguish these characters.
You can use a hexadecimal code to find and replace in the Find dialog box, with Search Mode is set to Regular expression. Use the following search expressions as an example:
Find what: \x{0A} # Finds a line feed
Find what: \x{0D} # Finds a carriage return
Related
In Eclipse, if I search for charCode == 44in File Search>Containing Text, I miss out on all of the results that might have different spaces e.g. charCode== 44, charChode ==44 etc
Is there a way (I imagine with would require the Regular Expression option) to include the whitespace combinations in the search text?
You could use the regular expression
charCode\s*==\s*44
Explanation:
\s means any space character (space, tab, ...). The * means zero or more occurrences of the previous literal.
Note: Be sure to check the Regular expression checkbox to the right of the search field.
I am viewing a CSV file which has LF characters in the middle of a field and CRLF character to actually denote a new line. I am viewing the file in hexadecimal in Sublime Text 3 and I want to do a simple find and replace where I search for LF characters but NOT CRLF and replace it with a space.
I've gotten as far as to search for LF but NOT CRLF, I could use the regular expression
[^0d]0a. Problem with this is that it doesn't capture the case where you could have XX0d 0aXX and I don't know how to capture this with regular expressions. I would then want to replace this with '20' which is space in hexadecimal.
Use a negative lookbehind that matches 0d with optional whitespace.
(?<!0d\s*)0a
However, some regexp engines won't allow quantifiers in lookbehinds. So you may need to put the whitespace check after the lookbehind, and then capture it to use it in the replacement.
(?<!0d)(\s*)0a replace with ${1}20
It would probably be easier if you did this in text mode instead of hex. Replace
(?<!\r)\n
with space.
I have a problem to find a character, enlarge it by constant number of characters around and return it.
Example of text:
Contrary to popular belief, (Lorem Ipsum) is not simply random text. It (has) roots in a piece of ...
Expected result:
r belief, (Lorem Ipsu
text. It (has) roots
How it should work:
find position of "(" - 10 characters
find position of "(" + 10 characters
extract text with start position of point 1. and end position of point 2. (and store it in a new row)
Please is it possible to do this in Notepad++ or similar software with function Find and Replace?
I believe this can be done with regex, but I am not able to write it.
Thank you very much!
Do a regular expression find/replace like this:
Open Replace Dialog
Find What: (.{10}\(.{10})
Replace With: \r\n\1\r\n
check regular expression
click Replace or Replace All
Depending on your line endings, you may need to change the \r\n to \n in the replacement.
Explanation:
the regular expressin centers at a literal ( (it has to be escaped as \( due the regex rules)
it captures the 10 character before and after it with the two .{10} sections
all the 21 character are captured into \1 (by putting the whole regular expression in unescaped parenthesis)
the replacement inserts \1 surrounded by linebreaks (either \r\n or \n, adopt what you need)
I have a plain text file with content like this:
prežrať/RN
prežrieť/Z
prežrúc/zZ
prežuť/c
...
Q: How can I remove all strings after / symbol in every row in Notepad++?
Desired output:
prežrať
prežrieť
prežrúc
prežuť
...
I am doing this with Find-and-Replace for every different string after /, but there are too many combinations.
Search for: /.*, replace with nothing.
The character / matches just /. ., however, matches any character except newlines, so .* will match a sequence of characters up until the first newline. You can find a demonstration here: http://regex101.com/r/kT0uE3.
If you want to remove characters only after the last on the line /, you should use the regex /[^/]*$. You can find an explanation and demonstration here: https://regex101.com/r/sZ6kP7/74.
In regular expression mode
Find:
/.*
Replace:
(empty)
Set find and replace to regular expression mode.
Find string: /.*
Replace String: (empty string)
Notepad++ find and replace is by default line ended (it won't go over multiple lines)
Using find and replace:
Hit CTRL-H to open the Replace dialogue box
enter /.* into "Find what"
leave "Replace with" empty
Select "Regular expression" (and .matches newline if it is single line)
Click on Replace
Here we go... You are done.
How to replace all the words coming before the character '=' in every line of the code in Notepad++?
It can be done with regular expressions.
Find what: ^[\w\s]+\=(.*)$
Replace with: newword\=\1.
See the screenshot below for more information:
You should enter your desired replacement word instead of "newword" in the regular expression.