I am setting up a Zend_Route (but it is still just a regex) and I wish to match a url like
/en/experience/this-is-my-name-and-the-last-is-1-of-id-123456.html
So I want to grab the
this-is-my-name-and-the-last-is-1-of
and the
123456
I tried
\w{2}/experience/(.+)?-(\d+)\.html
but that doesn't seem to work.
It would be easy if the other way around e.g. if it was id the name
/en/experience/123456-this-is-my-name-and-the-last-is-1-of-id.html
I could use
\w{2}/experience/(\d+)-(.+)\.html
But that is a cop out - so any advice on how to match original format?
Try this one:
/\w{2}/experience/(.+?)-(\d+)\.html
try this:
/\w{2}/experience/(.+)?-(\d+)\.html
zend route internally does this:
preg_match('#^/\w{2}/experience/(.+)?-(\d+)\.html$#i', '/en/experience/this-is-my-name-and-the-last-is-1-of-id-123456.html', $matches);
so, your pattern only matches with a slash on the beginning.
Related
i am doing a script in php, and i need to use preg_replace or something similar to add some tags in front of and behind matches. For example i have this pattern (regular expression which i am parsing from a file) and text:
$pattern = aa*
$string = "Example, exaaample"
Basicly, what i need is to add some tags in front of and behind all matches, so it will look like this:
"Ex<t>a</t>mple, ex<t>aaa</t>mple
Is there any way how to make this happen? I am pretty sure it's not that complicated but I am stuck on this for quite a while. Thanks
Sure. You can do it like this:
preg_replace("/(aa*)/", "<t>$1</t>")
$1 will be replaced by the matched pattern.
I am using PHP, and I have been trying to create a regular expression pattern to capture part of URL path, but to no avail.
The possible URL path could be any of these:
"product/zzz"
"yyyyyyyy/product/zzz"
"xxxxx/yyyyyyyy/product/zzz"
"xxxxx/yyyyyyyy/.../product/zzz" (... means other possible words)
what I need to capture is the part before "product".
for the first case, the result should be an empty string.
for the rest, they are "yyyyyyyy", "xxxxx/yyyyyyyy" and "xxxxx/yyyyyyyy/..."
Can anyone here give me hint? thanks!
PS.
It looks like the part I wanted is a repetition of same pattern "xxxx/". but I am not good at using group of regex.
Update:
I probably found a solution, by capturing pattern "xxx/" with zero or more repetitions: "([^/]+/)*"
so the full regex should be "(([^/]+/)*)product/([^/]+)"
#SERPRO: it passed the test in your "Live RegExp".
Hope it is helpful.
I would use parse_url():
$path = parse_url($url, PHP_URL_PATH);
// Deal with $path to figure out what's after '/product/'
This should work for you:
#(.*?)/?product.*\b#
You can see an example of result strings here:
http://xrg.es/#5awa10
This should do it:
^(.*[^/]|)/*product/[^/]+/*$
It will also allow an arbitrary number of slashes at the end of the path.
The part inside parentheses is your result.
I'm trying to create a snippet of regex that will match a URL route.
Basically, if I have this route /users/:id I want /users/100 to match, but /users/100/edit not to match.
This is what I'm using now: users/(.*)/ but because of the greedy match it's matching regardless of what's after the user ID. I need some way of "breaking" the match if there's an /edit or something else on the end of the route.
I've looked into the Regex NOT operator but with no luck.
Any advice?
Are you just trying to collect digits?
You could use users/(\d*)/
And this one is how you would do it if you wanted to collect everything until a /, and it uses a NOT, ^/users/[^/]*$
You can use negative lookahead:
users/(.*)/(?!edit)
This will always require a trailing slash however. Maybe a better solution would be:
users/(\d+)(?!/edit)
See this post for more information.
Need to get all URLs like:
http://domain.name/novostroyki/novyy_petergof/
http://domain.name/novostroyki/novyy_petergof/?var1=value1&val2=value2=...
but not the following ones:
http://domain.name/novostroyki/novyy_petergof/flats/
http://domain.name/novostroyki/novyy_petergof/flats/?var1=value1&val2=value2=...
Tried something like that, but it doesn't work as I wish:
/novostroyki/((?!flats)[a-z_0-9A-Z\.])*/?\??(.*)/
Try this regex:
/novostroyki/((?!flats)[\w.]*/?)*(\?.*)?
Not sure if it will be fine in all cases - it certainly should be in the ones listed above.
check if this suggestion is true:
if an url matches the regex on the list, it doesn't continue to the next item on the list.
then you can use:
the regex with all novostroyki after the regex with all novostroyki/flats/
I need to replace certain user-entered URLs with embedded flash objects...and I'm having trouble with a regex that I'm using to match the url...I think mainly because the URLs are SEO-friendly and therefore a bit more difficult to parse
URL structure: http://www.site.com/item/item_title_that_can_include_1('_etc-32CHARACTERALPHANUMERICGUID
I need to both detect a match of an URL in that format and capture the 32CHARACTERALPHANUMERICGUID which is always placed after the - in the url
something like this:
$ret = preg_replace('#http://www\.site\.com/item/([^-])-([a-zA-Z0-9]+)#','<embed>itemid=$2</embed>', $ret);
For some reason, the above does not find a match for an URL in the specified format. I'm new to regexes, so I think I'm missing something fairly obvious.
You should check out parse_url().
Examine the results - it was made for parsing URLs. You'll be able to extract the data you require from the tokens returned.
If you are regex crazy, try this...
/^http:\/\/www\.site\.com\/item\/[^-]*\-([a-zA-Z0-9]{32})$/
Your example is almost there, but...
When you do the not character range, i.e. [^-], you still need a quantifier. I placed *, or 0 or more.
You don't seem to use the item title, so we won't bother capturing it.
You should use beginning (^) and end ($) anchors if the string is always exactly like that.
You say the GUID is 32 chars, so we may as well explicitly state that with the {32} quantifier.