I'm new to Regular expressions and can't seem to find out how I have to solve this:
I need a regular expressions that "allows" only numbers, letters and /. I wrote this:
/[^a-zA-Z0-9/]/g
I think it's possible to strip the first / off, but don't know how.
so #/register/step1 becomes register/step1
Who knows how I could get this result?
Thanks!
You can use a non-global match, if the pattern is contiguous in the string:
var rx=/(([a-zA-Z0-9]+\/*)+)/;
var s='#/register/step1';
var s1=(s.match(rx) || [])[0];
alert(s1)>>> returned value: (String) "register/step1"
"/register/step1".match(/[a-zA-Z0-9][a-zA-Z0-9/]*/); // ["register/step1"]
\w is Equivalent to [^A-Za-z0-9_], so:
"/register/step1".match(/\w[\w/]*/); // ["register/step1"]
edit: Don't know why i didn't suggest this first, but if you're simply enforcing the pattern rather than replacing, you could just replace that slash (if it exists) before checking the pattern, using strpos(), substr(), or something similar. If you are using a preg_replace() already, then you should look at the examples on the function docs, they are quite relevant
Related
I have a problem with a regular expression.
I'm working with tokens and I have to parse a text like this:
Just some random text
#IT=AB|First statement# #xxxx=xxx|First statement|Second statement#
More text
I use preg_replace_callback since I have to use the first statement or the second one, depending on the first expression is true or not; it's a sort of IF...ELSE... statement.
What I expect are 2 elements like this:
#IT=AB|First statement#
#xxxx=xxx|First statement|Second statement#
So I can start manipulating them inside my callback function.
I tried with this regex /#.*#/, but i get the entire string, it's not parsed into elements.
How can I achieve that? I'm sorry but regex aren't my thing :(
The quantifier * is greedy by default. So a .* will match as much as it can and as a result it'll match a # as well. To fix this you can make the * non-greedy by adding a ? after it. Now a .*? will try to much as little as it can.
/#.*?#/
or you can look for only non # characters between two #:
/#[^#]*#/
I'm having this issue with a regular expression in PHP that I can't seem to crack. I've spent hours searching to find out how to get it to work, but nothing seems to have the desired effect.
I have a file that contains lines similar to the one below:
Total','"127','004"','"118','116"','"129','754"','"126','184"','"129','778"','"128','341"','"127','477"','0','0','0','0','0','0
These lines are inserted into INSERT queries. The problem is that values like "127','004" are actually supposed to be 127,004, or without any formatting: 127004. The latter is the actual value I need to insert into the database table, so I figured I'd use preg_replace() to detect values like "127','004" and replace them with 127004.
I played around with a Regular Expression designer and found that I could use the following to get my desired results:
Regular Expression
"(\d+)','(\d{3})"
Replace Expression
$1$2
The line on the top of this post would end up like this: (which is what I am after)
Total','127004','118116','129754','126184','129778','128341','127477','0','0','0','0','0','0
This, however, does not work in PHP. Nothing is being replaced at all.
The code I am using is:
$line = preg_replace("\"(\d+)','(\d{3})\"", '$1$2', $line);
Any help would be greatly appreciated!
There are no delimiters in your regex. Delimiters are required in order for PHP to know what is the pattern to match and what is a pattern modifier (e.g. i - case-insensitive, U - ungreedy, ...). Use a character that doesn't occur in your pattern, typically you'll see a slash '/' used.
Try this:
$line = preg_replace("/\"(\d+)','(\d{3})\"/", '$1$2', $line);
You forgot to wrap your regular expression in front-slashes. Try this instead:
"/\"(\d+)','(\d{3})\"/"
use preg_replace("#\"(\d+)','(\d+)\"#", '$1$2', $s); instead of yours
I have the following possible string:
'', or '4.', or '*.4' or '4.35'
all the above format are valid, others are all invalid.
basically, if I don't care the digit or word character, this is what I used in PHP for the validation:
else if ( !ereg('^\*|.*\..*$',$bl_objver) )
Now, I would like to add some clientside validation, so I just translate it into javascript:
var ver_reg = new RegExp("^\*|.*\..*$");
if (ver_reg.test(obj_ver) == false)
but firebug always shows some error, like: "invalid quantifier |...*$" etc..
any suggestions?
(I'm not convinced your expression is correct, but for the moment just going with what you have.)
Using the RegExp object, you need to escape the slashes:
var ver_reg = new RegExp("^\\*|.*\\..*$");
Alternatively you can use regex literal notation:
var ver_reg = /^\*|.*\..*$/;
That answers your direct question, but...
As for the expression, well, what you definitely want to correct is the start/end anchors each applying to one side of the alternation.
i.e. you're saying <this>|<that> where <this> is ^\* and <that> is .*\..*$
What you want is ^(?:<this>|<that>)$ to ensure the start/end markers are not part of the alternatives (but using ?: since we're not capturing the group).
So /^(?:\*|.*\..*)$/ using the second example above - this fix would also need applying to the PHP version (which can use the same syntax).
I'd also question your use of . instead of \w or [^.] or similar, but without knowing what you're actually doing, I can't say for sure what makes most sense.
Hope this helps! :)
Here is the subject:
http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov
What I need using regex is only the bit before the last / (including that last / too)
The 937IPiztQG string may change; it will contain a-z A-Z 0-9 - _
Here's what I tried:
$code = strstr($url, '/http:\/\/www\.mysite\.com\/files\/get\/([A-Za-z0-9]+)./');
EDIT: I need to use regex because I don't actually know the URL. I have string like this...
a song
more text
oh and here goes some more blah blah
I need it to read that string and cut off filename part of the URLs.
You really don't need a regexp here. Here is a simple solution:
echo basename(dirname('http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov'));
// echoes "937IPiztQG"
Also, I'd like to quote Jamie Zawinski:
"Some people, when confronted with a problem, think 'I know, I'll use regular expressions.' Now they have two problems."
This seems far too simple to use regex. Use something similar to strrpos to look for the last occurrence of the '/' character, and then use substr to trim the string.
/http:\/\/www.mysite.com\/files\/get\/([^/]+)\/
How about something like this? Which should capture anything that's not a /, 1 or more times before a /.
The greediness of regexp will assure this works fine ^.*/
The strstr() function does not use a regular expression for any of its arguments it's the wrong function for regex replacement.
Are you thinking of preg_replace()?
But a function like basename() would be more appropriate.
Try this
$ok=preg_match('#mysite\.com/files/get/([^/]*)#i',$url,$m);
if($ok) $code=$m[1];
Then give a good read to these pages
http://www.php.net/preg_match
preg_replace
Note
the use of "#" as a delimiter to avoid getting trapped into escaping too many "/"
the "i" flag making match insensitive
(allowing more liberal spellings of the MySite.com domain name)
the $m array of captured results
im new to regular expressions in php.
I have some data in which some of the values are stored as zero(0).What i want to do is to replace them with '-'. I dont know which value will get zero as my database table gets updated daily thats why i have to place that replace thing on all the data.
$r_val=preg_replace('/(0)/','-',$r_val);
The code im using is replacing all the zeroes that it finds for eg. it is even replacing zero from 104.67,giving the output 1-4.56 which is wrong. i want that data where value is exact zero that must be replaced by '-' not every zero that it encounter.
Can anyone please help!!
Example of the values that $r_val is having :-
10.31,
391.05,
113393,
15.31,
1000 etc.
This depends alot on how your data is formatted inside $r_val, but a good place to start would be to try:
$r_val = preg_replace('/(?<!\.)\b0\b(?!\.)/', '-', $r_val);
Where \b is a 0-length character representing the start or end of a 'word'.
Strange as it may sound, but the Perl regex documentation is actually really good for explaining the regex part of the preg_* functions, since Perl is where the functionality is actually implemented.
Again, it would be more than helpful if you could supply an example of what the $r_val string really looks like.
Note that \b matches at word boundaries, which would also turn a string like "0.75" into "-.75". Not a desirable result, I guess.
Whilst the other answer does work, it seems overly complex to me. I think you need only to use the ^ and $ chars either side of 0.
$r_val = preg_replace('/^0+$/', '-', $r_val);
^ indicates the regex should match from the beginning of the line.
$ indicates the regex should match to the end of the line.
+ means match this pattern 1 or more times
I altered the minus sign to it's html code equivalent too. Paranoid, yes, but we are dealing with numbers after all, so I though throwing a raw minus sign in there might not be the best idea.
Why not just do this?
if ( $r_val == 0 )
$r_val = '-';
You do not need to use a regex for this. In fact, I'd advise against doing so for performance reasons. The operation above is approximately 20x faster than the regex solution.
Also, the PHP manual advises against using regexes for simple replacements:
If you don't need fancy replacing rules (like regular expressions), you should always use this function instead of ereg_replace() or preg_replace().
http://us.php.net/manual/en/function.str-replace.php
Hope that helps!