I want to introduce a backspace character at the beginning of the line where a particular pattern is not found. Please advise.
Thanks,
Sagar
If you mean that you want to "remove the first character" then you can do this:
1) Write your regex pattern of what you want to find. For example, if you want to match Remove me at the start of the line, use:
^R\(emove me\)
Here we use ^ to assert the position to the start of the string. We also capture everything apart from the string we wish to keep in a backreference so it can be used later.
2) Replace the matches we find with whatever we grabbed in our backreference, in this case emove me, in effect backspacing the first character.
3) Make sure regular expression is checked and the cursor is at the start of the file, and hit Replace All.
Before
After:
Related
I have a main string as below
"/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
From the main string i need to extract a substring starting from the uuid part
"/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
I tried
string.match("/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/", "/[a-fA-F0-9]{8}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{12}/(.)/(.)/$"
But noluck.
if you want to obtain
"/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
from
"/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
or let's say 7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0, output and 9999.317528060546245771146821638997525068657 as this is what your pattern attempt suggests. Otherwise leave out the parenthesis in the following solution.
You can use a pattern like this:
local text = "/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
print(text:match("/([%x%-]+)/([^/]+)/([^/]+)"))
"/([^/]+)/" captures at least one non-slash-character between two slashs.
On your attempt:
You cannot give counts like {4} in a string pattern.
You have to escape - with % as it is a magic character.
(.) would only capture a single character.
Please read the Lua manual to find out what you did wrong and how to use string patterns properly.
Try also the code
s="/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
print(s:match("/.-/.-(/.+)$"))
It skips the first two "fields" by using a non-greedy match.
My current regex is only picking up part of my string. It creates a match as soon as one if found, even though I need the longer version of that match to hit. For example, I am creating matches for both:
SSS111
and
SSS111-L
The first SSS111 matches fine with my current regex, but the SSS111-L is only getting matched to the SSS111, leaving the -L out.
How can I create a greedy regex to read the whole line before matching? I am currently using
[-A-Z0-9]{3,12}
to capture the numbers and letters, but have not had any luck outside of this.
Regex are allways greedy. This ist mostly the Problem.
Here i think you have only to escape the '-'
#"[-A-Z]{3-12}"
I'm trying to create a RegEx Validator that checks the file extension in the FileUpload input against a list of allowed extensions (which are user specified). The following is as far as I have got, but I'm struggling with the syntax of the backward slash (\) that appears in the file path. Obviously the below is incorrect because it just escapes the (]) which causes an error. I would be really grateful for any help here. There seems to be a lot of examples out there, but none seem to work when I try them.
[a-zA-Z_-s0-9:\]+(.pdf|.PDF)$
To include a backslash in a character class, you need to use a specific escape sequence (\b):
[a-zA-Z_\s0-9:\b]+(\.pdf|\.PDF)$
Note that this might be a bit confusing, because outside of character classes, \b represents a word boundary. I also assumed, that -s was a typo and should have represented a white space. (otherwise it shouldn't compile, I think)
EDIT: You also need to escape the dots. Otherwise they will be meta character for any character but line breaks.
another EDIT: If you actually DO want to allow hyphens in filenames, you need to put the hyphen at the end of the character class. Like this:
[a-zA-Z_\s0-9:\b-]+(\.pdf|\.PDF)$
You probably want to use something like
[a-zA-Z_0-9\s:\\-]+\.[pP][dD][fF]$
which is same as
[\w\s:\\-]+\.[pP][dD][fF]$
because \w = [a-zA-Z0-9_]
Be sure character - to put as very first or very last item in the [...] list, otherwise it has special meaning for range or characters, such as a-z.
Also \ character has to be escaped by another slash, even inside of [...].
I'm trying to complete a regular expression that will pull out matches based on their opening and closing characters, the closest I've gotten is
^(\[\[)[a-zA-Z.-_]+(\]\])
Which will match a string such as "[[word1]]" and bring me back all the matches if there is more than one, The problem is I want it to pick up matchs where there may be a space in so for example "[[word1 word2]]", now this will work if I add a space into my pattern above however this pops up a problem that it will only get one match for my entire string so for example if I have a string
"Hi [[Title]] [[Name]] [[surname]], How are you"
then the match will be [[Title]] [[Name]] [[surname]] rather than 3 matches [[Title]], [[Name]], [[surname]]. I'm sure I'm just a char or two away in the Regex but I'm stuck, How can I make it return the 3 matches.
Thanks
You just need to make you regex non-greedy by using a ? like:
^(\[\[)[a-zA-Z.-_ ]+?(\]\])
Also there is a bug in your regex. You've included - in the char class thinking of it as a literal hyphen. But - in a char class is a meta char. So it effectively will match all char between . (period) and _ (underscore). So you need to escape it as:
^(\[\[)[a-zA-Z.\-_ ]+?(\]\])
or you can put is in some other place in the regex so that it will not have things on both sides of it as:
^(\[\[)[a-zA-Z._ -]+?(\]\])
or
^(\[\[)[-a-zA-Z._ ]+?(\]\])
You need to turn off greedy matching. See these examples for different languages:
asp.net
java
javascript
You should use +? instead of +.
The one without the question mark will try to match as much as possible, while the one with the question mark as little as possible.
Another approach would be to use [^\]] as your characters instead of [a-zA-Z.-_]. That way, a match will never extend over your closing brackets.
I am using the following regex
/[a-zA-Z0-9]+/i.test(value)
If I enter a space in the word, it passes.
I don't see where spaces are aloud in the regex, why is it passing?
You need to set the beginning and end bounderies so that the entire string must match the regular expression, otherwise it'll look for any match (which in this case is one or more of the characters specified).
Try this:
/^[a-zA-Z0-9]+$/i.test(value)
Because you haven't anchored it.
For these sorts of tests, it's typically safer to make sure you don't have the negated character class:
/[^a-zA-Z0-9]/