Here is a text:
<a class="mkapp-btn mab-download" href="javascript:void(0);" onclick="zhytools.downloadApp('C100306099', 'appdetail_dl', '24', 'http://appdlc.hicloud.com/dl/appdl/application/apk/f4/f44d320c2c1b466389e6f6b3d3f5cff4/com.uniquestudio.android.iemoji.1806141014.apk?sign=portal#portal1531621480529&source=portalsite' , 'v1.1.4');">
I want to extract
http://appdlc.hicloud.com/dl/appdl/application/apk/f4/f44d320c2c1b466389e6f6b3d3f5cff4/com.uniquestudio.android.iemoji.1806141014.apk?sign=portal#portal1531621480529&source=portalsite
I use below code to extract it.
m = re.search("mkapp-btn mab-download.*'http://[^']'", apk_page)
In my opinion, I can use .* to match the string between mkapp-btn mab-download and http. However I failed.
EDIT
I also tried.
m = re.search("(?<=mkapp-btn mab-download.*)http://[^']'", apk_page)
You need to add + after exclusion ([^']) because is more than one character. Also, you need to group using parenthesis to extract only the part you want.
m = re.search("mkapp-btn mab-download.*'(http[^']+)'", apk_page)
m.groups()
And the output will be
('http://appdlc.hicloud.com/dl/appdl/application/apk/f4/f44d320c2c1b466389e6f6b3d3f5cff4/com.uniquestudio.android.iemoji.1806141014.apk?sign=portal#portal1531621480529&source=portalsite',)
Related
Consider the text structure
(Title)[#1Title-link]
(Chapter1)[#Chapter1-link]
(Chapter2)[#Chapter2-link]
(Chapter3)[#Chapter3-link]
How can i backrefence to [#Title-link] without matching it on find result. Im trying to change
(Chapter1)[#Chapter1-link] => (Chapter1)[#1Title-link-Chapter1-link]
(Chapter2)[#Chapter2-link] => (Chapter2)[#1Title-link-Chapter2-link]
(Chapter3)[#Chapter3-link] => (Chapter3)[#1Title-link-Chapter3-link]
I tried to use and find
(\(Title\)\[(.*?)])([\s\S]*?\[)#(\D.*?\])
then replace it with
$1$3$2-$4
but the problem in here it only highlight once per find and i got lots of chapter its too inefficient to replace it one by one.
Making a constant title is no good too because i have multiple files with that same structure.
Is this possible in regex? any solution or alternative is welcome.
You can first do a search to get the correct substitution string and then do a subsequent replace operation with that substitution string. You did not specify what language you were using, so here is the code in Python (where that back reference to group 1 is \1 rather than the more usual $1):
import re
text = """(Title)[#1Title-link]
(Chapter1)[#Chapter1-link]
(Chapter2)[#Chapter2-link]
(Chapter3)[#Chapter3-link]"""
m = re.search(r'(?:\(Title\)\[#([^\]]*)\])', text)
assert(m) # that we have a match
substitution = m.group(1)
text = re.sub(r'\[#Chapter([^\]]*)\]', r'[#' + substitution + r'-Chapter\1' + ']', text)
print(text)
Prints:
(Title)[#1Title-link]
(Chapter1)[#1Title-link-Chapter1-link]
(Chapter2)[#1Title-link-Chapter2-link]
(Chapter3)[#1Title-link-Chapter3-link]
See Regex Demo 1 for getting the substitution string
See Regex Demo 2 for making the subsitutions
How i can read 4th Value(inside "" i.e "vV0...." using Regex in below condition ?
I am updating a bit this part - Is it possible to first find Word "LaunchFileUploader" and then select the 4th Value, if there are multiple instance of LaunchFileUploader in the file just select 4th Value of first word found ? Attaching screenshot of file where this needs to be searched (In the file word is "LaunchFileUploader")
I tried this but it gives as - I need 4th value (Group 1 is giving me third value)
\bLaunchFileUploader\b(\:?.*?,){3}.*?\)
Match 1
Full match 11030-11428 LaunchFileUploader("ERM-1BLX3D04R10-0001", 1662, "2ecbb644-34fa-4919-9809-a5ff47594c2d", "8dZOPyHKBK...
Group 1. n/a "2ecbb644-34fa-4919-9809-a5ff47594c2d",
I am still looking for solution for this. Any help is aprreciated.
Depending on what's available to you to use, there's a couple of ways to do it.
Either way, this would work better if there were no new lines in the string, just plain ("value1","value2","value3","value4") etc. It'll still work, but you may need to clean up some new lines from the resulting string.
The easy way - use code for the hard part. Grab the inner string with:
(?<=\().*?(?=\))
This will get everything that's between the 2 parentheses (using positive lookarounds). In code, you could then split/explode this string on , and take the 4th item.
If you want to do it all in regex, you could use something along the lines of:
(?<=\()(?:.*?,){3}(.*?)(?=\))
This would a) match the entire contents of the parentheses and b) capture the 4th option in a capture group. To go even deeper:
(?<=\()(?:.*?,){3}\"(.*?)\"(?=\))
would capture the contents of the "" quotation marks only.
Some tools don't allow you to use lookarounds, if this is the case let me know and I'll see what other ways there are around it.
EDIT Ran this in JS console on browser. This absolutely does work.
EDIT 2 I see you've updated your question with the text you're actually searching in. This pattern will include the space and the new line character as per the copy/paste of the above text.
(?<=\(\")(?:.*?,\s?\n?){3}\"(.*?)\"(?=\))
See my second image for the test in console
This works for python and PHP:
(?<=\")(.*)(?:\"\);)\Z
Demo for Python and PHP
For Java, replace \Z with $ as follows:
(?:")(.*)(?:\"\);)$
Demo for JavaScript
NOTE: Be sure to look the captured group and not the matched group.
UPDATE:
Try this for your updated request:
"(.*)"(?:[\\);\] \/>}]*)$
Demo for updated input string
all the above regex patterns assume there is a line break after each comma
Auto-generated Java Code:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\"(.*)\"(?:[\\\\);\\] \\/>\\}]*)$";
final String string = "\n"
+ "}$(document).ready( function(){ PathUploader\n"
+ " (\"ERM-1BLX3D04R10-0001\", \n"
+ " 1662, \n"
+ " \"1bff5c85-7a52-4cc5-86ef-a4ccbf14c5d5\", \n"
+ "\"vV0mX3VadCSPnN8FsAO7%2fysNbP5b3SnaWWHQETFy7ORSoz9QUQUwK7jqvCEr%2f8UnHkNNVLkJedu5l%2bA%2bne%2fD%2b2F5EWVlGox95BYDhl6EEkVAVFmMlRThh1sPzPU5LLylSsR9T7TAODjtaJ2wslruS5nW1A7%2fnLB%2bljZaQhaT9vZLcFkDqLjouf9vu08K9Gmiu6neRVSaISP3cEVAmSz5kxxhV2oiEF9Y0i6Y5%2f5ASaRiW21w3054SmRF0rq3IwZzBvLx0%2fAk1m6B0gs3841b%2fw%3d%3d\"); } );//]]>";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
I have to extract some substrings, this is like an XML markup in a plain text doc, like
lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf
Can i extract this pattern in a single command?
In a case like this, I tried to use a matcher, the group command to extract this single match.
I don't want to do something like
String pattern = /<AAA>(.*)<\/AAA>/;
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher("lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf");
if (m.find( )) {
System.out.println("Found value: " + m.group(0) );
}
There must be a more elegant way.
Edit :
Thank you time_yates, i was looking for something like that.
Could you explain a little why you use [0][1] on the result of
def extract = (input =~ '<AAA>(.+?)</AAA>')[0][1]
Answer by tim_yates :
=~ returns a Matcher, and so [0] gets the first match, which is 2 groups, the first is the String that had the match in it (your whole string) the second [1] is the group you defined in your expression
Thank you so much for your help, and thanks to all the readers.
Power of a community !!!
Can't you just do:
def input = 'lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf'
def extract = (input =~ '<AAA>(.+?)</AAA>')[0][1]
assert extract == 'myString'
This is the shortest (not the best) way I can think of without external libs:
String str = "lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf";
System.out.println(str.substring(str.indexOf(">") + 1, str.lastIndexOf("<")));
Or using StringUtils (which is million times better than my previous sugestion with substring):
StringUtils.substringBetween(str, "<AAA>", "</AAA>");
Still I'd go with matcher() like you proposed among all these.
I'm trying to retrieve a filename without the extension in ColdFusion. I am using the following function:
REMatchNoCase( "(.+?)(\.[^.]*$|$)" , "Doe, John 8.15.2012.docx" );
I would like this to return an array like: ["Doe, John 8.15.2012","docx"]
but instead I always get an array with one element - the entire filename:["Doe, John 8.15.2012.docx"]
I tried the regex string above on rexv.org and it works as expected, but not on ColdFusion. I got the string from this SO question: Regex: Get Filename Without Extension in One Shot?
Does ColdFusion use a different syntax? Or am I doing something wrong?
Thanks.
Why you're not getting expected results...
The reason you are getting a one-item array with the whole filename is because your pattern matches the entire filename, and matches once.
It is capturing the two groups, but rematch returns arrays of matches, not arrays of the captured groups, so you don't see those groups.
How to solve the problem...
If you are dealing with simple files (i.e. no .htaccess or similar), then the simplest solution is to just use...
ListLast( filename , '.' )
....to get only the file extension and to get the name without extension you can do...
rematch( '.+(?=\.[^.]+$)' , filename )
This uses a lookahead to ensure there is a . followed by at least one non-. at the end of the string, but (since it's a lookahead) it is excluded from the match (so you only get the pre-extension part in your match).
To deal with non-extensioned files (e.g. .htaccess or README) you can modify the above regex to .+(?=(?:\.[^.]+)?$) which basically does the same thing except making the extension optional. However, there isn't a trivial way to get update the ListLast method for these (guess you'd need to check len(extension) LT len(filename)-1 or similar).
(optional) Accessing captured groups...
If you want to get at the actual captured groups, the closest native way to do this in CF is using the refind function, with the fourth argument set to true - however, this only gives you positions and lengths - requiring that you use mid to extract them yourself.
For this reason (amongst many others), I've created an improved regex implementation for CF, called cfRegex, which lets you return the group text directly (i.e. no messing around with mid).
If you wanted to use cfRegex, you can do so with your original pattern like so:
RegexMatch( '(.+?)(\.[^.]*$|$)' , filename , 1 , 0 , 'groups' )
Or with named arguments:
RegexMatch( pattern='(.+?)(\.[^.]*$|$)' , text=filename , returntype='groups' )
And you get returned an array of matches, within each element being an array of the captured groups for that match.
If you're doing lots of regex work dealing with captured groups, cfRegex is definitely better than doing it with CF's re methods.
If all you care about is getting the extension and/or the filename with extension excluded then the previous examples above are sufficient.
#Peter's response is great, however the approach is perhaps a bit longer-winded than necessary. One can do this with reMatch() with a slight tweak to the regex.
<cfscript>
param name="URL.filename";
sRegex = "^.+?(?=(?:\.[^.]+?)?$)";
aMatch = reMatch(sRegex, URL.filename);
writeDump(aMatch);
</cfscript>
This works on the following filename patterns:
foo.bar
foo
.htaccess
John 8.15.2012.docx
Explanation of the regex:
^ From the beginning of the string
.+? One or more (+) characters (.), but the fewest (?) that will work with the rest of the regex. This is the file name.
(?=) Look ahead. Make sure the stuff in here appears in the string, but don't actually match it. This is the key bit to NOT return any file extension that might be present.
(?: Group this stuff together, but don't remember it for a back reference.
. A dot. This is the separator between file name and file extension.
[^.]+? One or more (+) single ([]) non-dot characters (^.), again matching the fewest possible (?) that will allow the regex as a whole to work.
? (This is the one after the (?:) group). Zero or one of those groups: ie: zero or one file extensions.
$ To the end of the string
I've only tested with those four file name patterns, but it seems to work OK. Other people might be able to finetune it.
A few more ways of achieving the same result. They all execute in roughly the same amount of time.
<cfscript>
str = 'Doe, John 8.15.2012.docx';
// sans regex
arr1 = [
reverse( listRest( reverse( str ), '.' ) ),
listLast( str, '.' )
];
// using Java String lastIndexOf()
arr2 = [
str.substring( 0, str.lastIndexOf( '.' ) ),
str.substring( str.lastIndexOf( '.' ) + 1 )
];
// using listToArray with non-filename safe character replace
arr3 = listToArray( str.replaceAll( '\.([^\.]+)$', '|$1' ), '|' );
</cfscript>
I have a string
IsNull(VSK1_DVal.RuntimeSUM,0),
I need to remove IsNull part, so the result would be
VSK1_DVal.RuntimeSUM,
I'm absolute new to RegEx, but it wouldn't be a problem, if not one thing :
VSK1 is dynamic part, can be any combination of A-Z,0-9 and any length. How to replace strings with RegEx? I use MSSQL 2k5, i think it uses general set of RegEx rules.
EDIT : I forgot to say, that I'm doing replacement in SSMS Query window's Replace Box (^H) - not building RegEx query
br
marius
here's a regex that should work:
[^(]+\(([^,]+),[^)]\)
Then use $1 capture group to extract the part that you need.
I did a sanity check in ruby:
orig = "IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /[^(]*\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => VSK1_DVal.RuntimeSUM,
It gets trickier if you have a prefix that you want to retain. Like if you have this:
"somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
In this case, you need someway to identify the start of the pattern. Maybe you can use '=' to identify the start of the pattern? If so, this should work:
orig = "somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /=\s*\w+\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
But then the case where you don't have an equals sign will fail. Maybe you can use 'IsNull' to identify the start of the pattern? If so, try this (note the '/i' representing case insensitive matching):
orig = "somestuff = isnull(VSK1_DVal.RuntimeSUM,0),"
regex = /IsNull\(([^,]+),[^)]\)/i
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
/IsNULL\((A-Z0-9+),0\)/
Then pick group match number 1.
Here's a very useful site: http://www.regexlib.com/RETester.aspx
They have a tester and a cheatsheet that are very useful for quick testing of this sort.
I tested the solution by Dave and it works fine except it also removes the trailing comma you wanted retained. Minor thing to fix.
Try this:
IsNULL\((.*,)0\)
You say in your question
I use MSSQL 2k5, i think it uses
general set of RegEx rules.
This is not true unless you enable CLR and compile and install an assembly. You can use its native pattern matching syntax and LIKE for this as below.
WITH T(C) AS
(
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,0),' UNION ALL
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,123465),' UNION ALL
SELECT 'No Match'
)
SELECT SUBSTRING(C,8,1+LEN(C)-8-CHARINDEX(',',REVERSE(C),2))
FROM T
WHERE C LIKE 'IsNull(%,_%),'