I am trying to run some regular expressions(grep) on a text file of about 4K lines. The main portion that I need replaced looks like this:
1,"An Internet-Ready Resume",1,2,"","
And I need it to look like this:
<item>
<title>An Internet-Ready Resume</title>
<category>1</category>
<author>2</author>
<content>
So far, this is what I was trying to no avail:
[0-9]{1}\,\"*\"\,[0-9]\,[0-9]\,\"\"\,\"
You should start with doing a little reading on regular expressions. There are tons of useful resources online. Then you would see that:
you needn't escape everything (such as commas or quotes)
the asterisk * doesn't mean anything, but zero or more times
the any character is the . character. .* means any character any number of times (or anything)
if you need to make substitutions where you need atoms of what you're searching, you have to set those atoms by using (<atom content>) where <atom content> is a bit of a regexp.
A tip to start: instead of \"*\" try ".*"; Check the reference.
Also note that the part regarding the replacement will depend on the text editor/tool you're using. Usually a regexp such as (a)(b) (where a,b are regexp atoms) being replaced by x\1y\2z would produce xaybz.
The error is the \"*\" part. When you use the * operator you need to tell it what is to be repeated. As written it is going to repeat the previous quote character. Instead of that you should tell it to repeat any character (.), thus: \".*\"
A secondary comment is that you have a lot of unnecessary backslashes. In fact, none of them are necessary as far as I can tell. Without them your regex looks like:
[0-9],".*",[0-9],[0-9],"","
Related
I have a project that demands extracting data from XML files (values inside the <Number>... </Number> tag), however, in my regular expression, I haven't been able to extract lines that had multiple data separated by a newline, see the below example:
As you can see above, I couldn't replicate the multiple lines detection by my regular expression.
If you are using a script somewhere, your first plan should be to use a XML parser. Almost every language has one and it should be far more accurate compared to using regex. However, if you just want to use regex to search for strings inside npp, then you can use \s+ to capture multiple new lines:
<Number>(\d+\s)+<\/Number>
https://regex101.com/r/MwvBxz/1
I'm not sure I fully understand what you are trying to do so if this doesn't do it then let me know what you are going for.
You can use this find+replace combo to remove everything which is not a digit in between the <Number> tag:
Find:
.*?<Number>(.*?)<\/Number>.*
Replace:
$1
finally i was able to find the right regular expression, I'll leave it below if anyone needs it:
<Type>\d</Type>\n<Number>(\d+\n)+(\d+</Number>)
Explanation:
\d: Shortcut for digits, same as [1-9]
\n: Newline.
+: Find the previous element 1 to many times.
Have a good day everybody,
After giving it some more thought I decided to write a second answer.
You can make use of look arounds:
(?<=<Number>)[\d\s]+(?=<\/Number>)
https://regex101.com/r/FiaTKD/1
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I have issues to perform a mass change in a huge logfile.
Except the filesize which is causing issues to Notepad++ I have a problem to use more than 10 parameters for replacement, up to 9 its working fine.
I need to change numerical values in a file where these values are located within quotation marks and with leading and ending comma: ."123,456,789,012.999",
I used this exp to find and replace the format to:
,123456789012.999, (so that there are no quotation marks and no comma within the num.value)
The exp used to find is:
([,])(["])([0-9]+)([,])([0-9]+)([,])([0-9]+)([,])([0-9]+)([\.])([0-9]+)(["])([,])
and the exp to replace is:
\1\3\5\7\9\10\11\13
The problem is parameters \11 \13 are not working (the chars eg .999 as in the example will not appear in the changed values).
So now the question is - is there any limit for parameters?
It seems for me as its not working above 10. For shorter num.values where I need to use only up to 9 parameters the string for serach and replacement works fine, for the example above the search works but not the replacement, the end of the changed value gets corrupted.
Also, it came to my mind that instead of using Notepad++ I could maybe change the logfile on the unix server directly, howerver I had issues to build the correct perl syntax. Anyone who could help with that maybe?
After having a little play myself, it looks like back-references \11-\99 are invalid in notepad++ (which is not that surprising, since this is commonly omitted from regex languages.) However, there are several things you can do to improve that regular expression, in order to make this work.
Firstly, you should consider using less groups, or alternatively non-capture groups. Did you really need to store 13 variables in that regex, in order to do the replacement? Clearly not, since you're not even using half of them!
To put it simply, you could just remove some brackets from the regex:
[,]["]([0-9]+)[,]([0-9]+)[,]([0-9]+)[,]([0-9]+)[.]([0-9]+)["][,]
And replace with:
,\1\2\3\4.\5,
...But that's not all! Why are you using square brackets to say "match anything inside", if there's only one thing inside?? We can get rid of these, too:
,"([0-9]+),([0-9]+),([0-9]+),([0-9]+)\.([0-9]+)",
(Note I added a "\" before the ".", so that it matches a literal "." rather than "anything".)
Also, although this isn't a big deal, you can use "\d" instead of "[0-9]".
This makes your final, optimised regex:
,"(\d+),(\d+),(\d+),(\d+)\.(\d+)",
And replace with:
,\1\2\3\4.\5,
Not sure if the regex groups has limitations, but you could use lookarounds to save 2 groups, you could also merge some groups in your example. But first, let's get ride of some useless character classes
(\.)(")([0-9]+)(,)([0-9]+)(,)([0-9]+)(,)([0-9]+)(\.)([0-9]+)(")(,)
We could merge those groups:
(\.)(")([0-9]+)(,)([0-9]+)(,)([0-9]+)(,)([0-9]+)(\.)([0-9]+)(")(,)
^^^^^^^^^^^^^^^^^^^^
We get:
(\.)(")([0-9]+)(,)([0-9]+)(,)([0-9]+)(,)([0-9]+\.[0-9]+)(")(,)
Let's add lookarounds:
(?<=\.)(")([0-9]+)(,)([0-9]+)(,)([0-9]+)(,)([0-9]+\.[0-9]+)(")(?=,)
The replacement would be \2\4\6\8.
If you have a fixed length of digits at all times, its fairly simple to do what you have done. Even though your expression is poorly written, it does the job. If this is the case, look at Tom Lords answer.
I played around with it a little bit myself, and I would probably use two expressions - makes it much easier. If you have to do it in one, this would work, but be pretty unsafe:
(?:"|(\d+),)|(\.\d+)"(?=,) replace by \1\2
Live demo: http://regex101.com/r/zL3fY5
Apologies in advance for the confusing title. My issue is as follows, I have the following text in about 600 files:
$_REQUEST['FOO']
I would like to replace it with the following:
$this->input->post('FOO')
To clarify, I am matching against the following:
$_REQUEST any number of A-Za-z\d followed by a ]
and replacing it with:
$this->input->post( the alphanumeric word from above followed by a )
Or in general:
Anchor token TEXT TO KEEP end anchor token
This differs from standard find/replace as I want to retain text inside of two word boundaries.
Is this functionality present in any text editors (Eclipse,np++,etc). Or am I going to need to write some type of program to parse these 600 files to make the replacement?
s/\$__REQUEST\[(.*?)]/$this->input->post(\1)/
The .*? will match everything from [ to the first ] rather than the last although it's unlikely that it will matter in this case.
By the way the PHP superglobal is $_REQUEST rather than $__REQUEST
You can do this in Notepad++ using regular expressions. Replace
\$_REQUEST\['([^']*)'\]
with
$this->input->post('$1')
If you ever have double-quotes too, you can do use a more complex expression to handle both cases, though I'm not sure Notepad++ supports backreferences; replace
\$_REQUEST\[(['"])(.*?)\1\]
with
$this->input->post($1$2$1)
Note that I've reverted to using #ExplosionPills' suggested (.*?) here—it may be better, actually.
Let me preface this by saying I'm a complete amateur when it comes to RegEx and only started a few days ago. I'm trying to solve a problem formatting a file and have hit a hitch with a particular type of data. The input file is structured like this:
Two words,Word,Word,Word,"Number, number"
What I need to do is format it like this...
"Two words","Word",Word","Word","Number, number"
I have had a RegEx pattern of
s/,/","/g
working, except it also replaces the comma in the already quoted Number, number section, which causes the field to separate and breaks the file. Essentially, I need to modify my pattern to replace a comma with "," [quote comma quote], but only when that comma isn't followed by a space. Note that the other fields will never have a space following the comma, only the delimited number list.
I managed to write up
s/,[A-Za-z0-9]/","/g
which, while matching the appropriate strings, would replace the comma AND the following letter. I have heard of backreferences and think that might be what I need to use? My understanding was that
s/(,)[A-Za-z0-9]\b
should work, but it doesn't.
Anyone have an idea?
My experience has been that this is not a great use of regexes. As already said, CSV files are better handled by real CSV parsers. You didn't tag a language, so it's hard to tell, but in perl, I use Text::CSV_XS or DBD::CSV (allowing me SQL to access a CSV file as if it were a table, which, of course, uses Text::CSV_XS under the covers). Far simpler than rolling my own, and far more robust than using regexes.
s/,([^ ])/","$1/ will match a "," followed by a "not-a-space", capturing the not-a-space, then replacing the whole thing with the captured part.
Depending on which regex engine you're using, you might be writing \1 or other things instead of $1.
If you're using Perl or otherwise have access to a regex engine with negative lookahead, s/,(?! )/","/ (a "," not followed by a space) works.
Your input looks like CSV, though, and if it actually is, you'd be better off parsing it with a real CSV parser rather than with regexes. There's lot of other odd corner cases to worry about.
This question is similar to: Replace patterns that are inside delimiters using a regular expression call.
This could work:
s/"([^"]*)"|([^",]+)/"$1$2"/g
Looks like you're using Sed.
While your pattern seems to be a little inconsistent, I'm assuming you'd like every item separated by commas to have quotations around it. Otherwise, you're looking at areas of computational complexity regular expressions are not meant to handle.
Through sed, your command would be:
sed 's/[ \"]*,[ \"]*/\", \"/g'
Note that you'll still have to put doublequotes at the beginning and end of the string.