What expression can I use in Notepad ++ to remove all the words containing http??Not all the line but only the word!
Because I have a list of sentences and some of them contain a link that I would like to remove. Can someone help me??
thank you!!!
Search -> Replace (Ctrl + H).
In the Find what field type http.
In the Replace with field leave it blank.
Click Replace All
To replace a word containing http, then you'll need to do the same steps but in Search Mode select Regex, and in Find What change http to \bhttp\b.
Related
I am tryng to get rid of shortcodes inside a Google Sheet column. I have many items such as [spacer type="1" height="20"][spacer] or [FinalTilesGallery id="37"] I just would like to cancel them. Is there any simple way to do it?
Thanks !
For in-place replacement, the quick option would be to use the Find and Replace dialog (Ctrl + H) with Search Using Regular Expressions turned on, which is more powerful than your standard Find and Replace.
Find: \[.*?\] - Match anything within an open-bracket up to the very next close-bracket. This should work assuming you have no nested brackets, e.g. [[no][no]].
If you do have nested brackets, you'll have to change this to \[[^\[\]]*\]. And continue to Replace All until all the codes are gone.
Replace: Nothing.
Replace All. If you don't want to affect other sheets that may be in your document, make sure you select the right range to work with, too.
This just erases everything within the brackets.
If you want to erase any redundant spaces left by this, simply Find and Replace again (with Regular Expressions) on + (space and plus), which will match 1 or more spaces and replace with (single space).
E.g.:
string [] [] string2 -> string string2 after the shortcode replacement.
After replacing spaces, it will become string string2.
Let's say your original strings are in the range A2:A. Place the following into B2 of an otherwise completely empty Column B (or the second cell of any other empty column):
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(A2:A,"\[[^\[\]]+\]",""))))
I can't see your data, so I don't know what kind of information is between these shortcodes. If you find that this leaves you with concatenated pieces of data where there should be spaces between them, replace the above with this version:
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(SUBSTITUTE(SUBSTITUTE(A2:A,"["," ["),"]","] "),"\[[^\[\]]+\]",""))))
I can't teach regular expression language here. But I will note that, since square brackets have specific meaning within regex, your literal square brackets must be indicated with the escape character: the backslash.
Here is the regex expression alone:
\[[^\[\]]+\]
The opening \[ and the closing \], then, reference your actual opening and closing bracket sets. If we remove those, we have this left:
[^\[\]]+
Again, you see the escaped opening and closing square brackets, which I'll replace with the word these:
[^these]+
What remains there are opening and closing brackets with regex meaning, i.e., "anything in this group." And the circumflex symbol ^ as the first character within this set of square brackets means "anything except." The + symbol means "in any string length of one or more characters."
So that whole regex expression then reads: "A literal open square bracket, followed by one or more characters that are anything except square brackets, ending with a literal closing square bracket."
And we are REGEXREPLACE-ing any instance of that with "" (i.e., nothing).
Hi, I have this data of Bricks which are separated by Next line, I want all to be in the from of Last Line. I can't do it manually one by one because it's very huge data. Can anyone please guide me the easiest way or built in functionality of notepad or notepad++. Thank you
Click the link to see my file and data.
Regards,
Mubasher Aqeel
Use "Line Operations" to remove line breaks
You just need to remove the new lines. Using Notepad++ that is as simple as:
Ctrl + A (to select everything)
Press Ctrl + J (which is the shortcut for "Edit → Line Operations → Join Lines")
Use search/replace
Alternatively you can do it more manually using:
Ctrl + H (Replace)
Choose "extended" under "Search Mode" and replace occurrences of "\r\n" (line breaks) with a blank.
I do have one large text file with lot of the following patterns;
because of,this
or that,has
or,not
Of course I want to change the following
because of, this
or that, has
or, not
To make myself clear: i would like to insert a space after each ,
How can i do that with BBEdit Find/Replace/Grep?
Find works ok with
[\,](\w)
but i can't figure out the coresponding part for replace.
Find: (\,)(\w)
Replace: \1 \2
Note: You need to hit the spacebar between \1 and \2. Works on my computer with BBedit v13
I complicates matters.
Pattern like Letter,Letter can also be found with ,\b.
\b is at the beginning or end of a word. \b in a regular expression means "word boundary".
The replacement is then done with ,_
Nota Bene: _ is a "Space" after ,
you could just replace all commas with a comma-space and then replace all space-space with space
Here is a simple lists where I like to change the numbers: the entries are as below and it got over 300 entries like it
tom112
smith113
harry114
linda115
cindy106
samantha147
It need to be changed to
tom212
smith213
harry214
...and so on.
Please assist using notepad++ regular expression.
Thanks.
Assuming it's just a matter of replacing a name followed by a number starting with 1 with the same number but starting with 2 instead:
Ctrl + H for search & replace.
Check Regular expression under Search Mode.
Next to Find what type or copy in ([a-zA-Z]+)1([0-9]+).
Next to Replace with type or copy in \12\2.
Click Replace All and that should do it.
Add any other characters that might appear in the name before the number inside the first set of brackets with a-zA-Z.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.