I try take of two word in this string, the first one and the the end:
radius-bottom-left-bgc
radius-bottom-left-l
radius-bottom-left-bgc
radius-bottom-left-l
etc...
I need keep just the middle ?
top-left
$display_prop= preg_replace('/^(\w{6})\-(\w{3,6})\-(\w{4,5})\-(\w{6})$/','$2-$3', $display_prop);
echo $display_prop.'<br>';
my echo display
radius-top-left-bgc
radius-top-right-bgc
radius-bottom-right-bgc
radius-bottom-left-bgc
radius-top-left-l
radius-bottom-right-l
What' wrong with my regex ?
Thanks
Your last capturing group doesn't match your example..
Your last capturing group is (\w{6}), but for your example data it should be (\w{1,3}).
Fix that and it should work. :)
Related
I'm new at regular expressions and wonder how to phrase one that collects everything after the last /.
I'm extracting an ID used by Google's GData.
my example string is
http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123
Where the ID is: p1f3JYcCu_cb0i0JYuCu123
Oh and I'm using PHP.
This matches at least one of (anything not a slash) followed by end of the string:
[^/]+$
Notes:
No parens because it doesn't need any groups - result goes into group 0 (the match itself).
Uses + (instead of *) so that if the last character is a slash it fails to match (rather than matching empty string).
But, most likely a faster and simpler solution is to use your language's built-in string list processing functionality - i.e. ListLast( Text , '/' ) or equivalent function.
For PHP, the closest function is strrchr which works like this:
strrchr( Text , '/' )
This includes the slash in the results - as per Teddy's comment below, you can remove the slash with substr:
substr( strrchr( Text, '/' ), 1 );
Generally:
/([^/]*)$
The data you want would then be the match of the first group.
Edit Since you’re using PHP, you could also use strrchr that’s returning everything from the last occurence of a character in a string up to the end. Or you could use a combination of strrpos and substr, first find the position of the last occurence and then get the substring from that position up to the end. Or explode and array_pop, split the string at the / and get just the last part.
You can also get the "filename", or the last part, with the basename function.
<?php
$url = 'http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123';
echo basename($url); // "p1f3JYcCu_cb0i0JYuCu123"
On my box I could just pass the full URL. It's possible you might need to strip off http:/ from the front.
Basename and dirname are great for moving through anything that looks like a unix filepath.
/^.*\/(.*)$/
^ = start of the row
.*\/ = greedy match to last occurance to / from start of the row
(.*) = group of everything that comes after the last occurance of /
you can also normal string split
$str = "http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123";
$s = explode("/",$str);
print end($s);
This pattern will not capture the last slash in $0, and it won't match anything if there's no characters after the last slash.
/(?<=\/)([^\/]+)$/
Edit: but it requires lookbehind, not supported by ECMAScript (Javascript, Actionscript), Ruby or a few other flavors. If you are using one of those flavors, you can use:
/\/([^\/]+)$/
But it will capture the last slash in $0.
Not a PHP programmer, but strrpos seems a more promising place to start. Find the rightmost '/', and everything past that is what you are looking for. No regex used.
Find position of last occurrence of a char in a string
based on #Mark Rushakoff's answer the best solution for different cases:
<?php
$path = "http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123?var1&var2#hash";
$vars =strrchr($path, "?"); // ?asd=qwe&stuff#hash
var_dump(preg_replace('/'. preg_quote($vars, '/') . '$/', '', basename($path))); // test.png
?>
Regular Expression to collect everything after the last /
How to get file name from full path with PHP?
I have a regex but it's not quite working the way i want
page[0-9]*
/pages/search.aspx?pageno=3&pg=232323&hdhdhd/page73733/xyz
In the above example, the only thing I want to match is page73733. But my regex matches the page in /pages and it matches page in pageno=3
i also tried page[0-9].*, then it matches page73733 but it also matches everything that comes after it so that it actually matches page73733/xyz
page[0-9].*[^a-zA-Z&?/=]
That seems to do what i want, but that also seems like a ugly way to do it. Plus if i had something like /page123/xyz/page456 it'll match that whole string.
So is there a better way to do this? I want to match ONLY the string page when it is followed by any number of digits, and if anything comes after the digits it should stop.
* means 0 or more occurrences. + means 1 or more occurrences.
page[0-9]+ should work.
page[0-9]*
Will match page followed by zero or more numbers. What you want is:
page[0-9]+
Which will match page followed by one or more numbers.
You almost got it. Just use + instead of * as that will force a match that has numbers after it.
Another way to type that expression would be
/page[0-9]+
note the / , this would be helpful because without it you might get a match with something like "notApage123"
The regex page[0-9]* will match [0-9] 0 or more times. + would match it 1 or more times, and ? would match it 0 or 1 times. An equivalent method to ?+* is as follows:
?={0,1}
*={0,}
+={1,}
This may be helpful for if you wanted to match a date:\\d{4}(-\\d{1,2}){2} which would match 2013-5-31
-
That said, the resulting Regex for your particular problem would be:
page\\d+
page\\d{1,}
page[0-9]+
or page[0-9]{1,}
In your example "/page123/xyz/page456" you may want to match all occurrences, so don't forget the g or global modifier.
If I understand your problem correctly, you only need to add $ to your original regex to specify that after page you want the string to end. So the regex would be
page[0-9]*$
Also, this will match strings that end in page too, if you want only strings that end in page followed by any number, use this regex
page[0-9]+$
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:
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.
some text I want to capture. <tag> junk I don't care about</tag> more stuff I want.
Is there a easy way to write a regex that captures the first and third sentences in one capture?
You could also consider stripping out the unwanted data and then capturing.
data = "some text to capture. <tag>junk</tag> other stuff to capture".
data = re.replace('<tag>[^<]*</tag>', data, "")
data_match = re.match('[\w\. ]+', data)
Not to my knowledge. Usually that's why regex search-and-replace functions allow you to refer to multiple capturing groups in the first place.
Unfortunately No, its not possible. The solution is to capture into two seperate captures and then contactenate after the fact.
According to this older thread on this site:
Regular expression to skip character in capture group
A group capture is consecutive so you cant. You can do it in one parse with regex like below and join the line in code
^(?<line1>.*?)(?:\<\w*\>.*?\</\w*\>)(?<line3>.*?)$
here's a non regex way, split on </tag>, go through the array items, find <tag>, then split on <tag> and get first element. eg
>>> s="some text I want to capture. <tag> junk I don't care about</tag> more stuff I want. <tag> don't care </tag> i care"
>>> for item in s.split("</tag>"):
... if "<tag>" in item:
... print item.split("<tag>")[0]
... else:
... print item
...
some text I want to capture.
more stuff I want.
i care
Use the split() function of asp.net to do the same.