i want to extract the category id from the response message. the regex i had used is categoryId=(.*?)>
I am doing this on the following response messages. can you please correct me like what is going wrong here ?
<img border="0" src="../images/sm_fish.gif" />
Try this regex:
categoryId=(.*?)"
This uses the non greedy operator to make sure that it only matches the content between the categoryId label and the ending quotation.
Try this: categoryId=([^"]+)"
[^"] matches any character, which is not in the list. So, in this case everything, but "
Related
I am trying to create a regex to match all a href links that contain my domain and I will end up removing the links. It is working fine until I run into an a href link that has another HTML tag within the tag.
Regex Statement:
(<a[^<]*coreyjansen\.com[^<]*>)([^"]*?)(<\/a>)
It matches the a href links in this statement with no problem
Need a lawyer? Contact <span style="color: #000000">Random text is great Corey is awesome</span>
It is unable to match both of the a href links this statement:
<strong><a href="http://coreyjansen.com/"><img class="alignright size-full
wp-image-12" src="http://50h0.com/wp-content/uploads/2014/06/lawyers.jpg"
alt="lawyers" width="250" height="250" /></a>
I have been trying to play with the neglected character set with no luck. If I remove the neglected character set what ends up happening is it will match two links that are right after each other such as example 2 as one match.
The issue here is that [^<]*> matches everything up until last >. That's the greedy behaviour of * asterisk. You can make it non-greedy by appending ? after asterisk(which you already do in other part of your query). It will then match everything until first occurrence of >. Then you have to change the middle part of your regex too ie. to catch everything until first tag </a> like this:
(<a[^<]*coreyjansen\.com[^<]*?>)(.*?)(<\/a>)
Use below regex which matches only a tag
(<a[^>]*coreyjansen\.com[^>]*>)
Example data
<strong><a href="http://coreyjansen.com/"><img class="alignright size-full
wp-image-12" src="http://50h0.com/wp-content/uploads/2014/06/lawyers.jpg"
alt="lawyers" width="250" height="250" /><a href="http://coreyjansen.com/"/>
Above regex will match all three a tag with your required domain.
Try above on regex
I'm playing with the following regex and it seems to be working:
<a.*coreyjansen\.com.*</a>
it captures anything between anchor tags that contain your site name. I am using javascript pattern matching from www.regexpal.com, depending on the language it could be slightly different
You need to match start of tag <a then match address before > char. You are matching wrong char. When you match that, then everithing between <a> and </a> is displayed link. I don't know why you compare to not contain quotes, every tag attribute (in HTML5) has value inside quotes, so you need to match everything except link ending tag </a>. It's done by ((?!string to not match).)* and after that should follow </a>. The result regex is:
(<a[^>]*coreyjansen\.com[^>]*>)((?!<\/a>).)*(<\/a>)
I need some help with Regular expression to Search and Replace in Sublime to do the following.
I have HTML-code with links like
href="http://www.example.com/test=123"
href="http://www.example.com/test=6546"
href="http://www.example.com/test=3214"
I want to replace them with empty links:
href=""
href=""
href=""
Please help me to create a Reg. ex. filter to match my case. I guess it would sound like "starts with Quote, following with http:// .... ends with Quote and has digitals and '=' sign", but I'm not very confident of how to write this in Reg. ex. way.
(?<=href=")[^"]*
Try this.Replace by empty string.
See demo.
https://regex101.com/r/sH8aR8/40
Via a Regex, I'm trying to match the word one, only when it's within an HTML <p> tag.
<p>zero one two three</p>
zero one two<p>three</p>
<p>zero one <b>two</b></p><p>three</p>
<p>two</p>three one
#1 and #3 above should be matches. It feels like I need a lookahead that makes sure there is a closing </p> tag without an opening <p> tag that comes before it (or a lookbehind that does the opposite). But I can't seem to come up with the right expression. Any ideas are appreciated.
<p>(?:(?!<\/p>).)*(\bone\b)(?:(?!<\/p>).)*<\/p>
You can try this.Just grab the capture.See demo.
http://regex101.com/r/xT7yD8/12
You could try the below regex to match the string one which is inside the <p> tag.
\bone\b(?=(?:(?!<\/?p>).)*<\/p>)
DEMO
How can I tell a character that comes after a wildcard to use the first occurrence of it?
I did the following to find any tag with the word "title" in it:
<(.*?)(title)(.*?)>
but clearly what happens is I end up with the entire tag to the end of
</title>
So that in
<Bla bla ="nametitle">Yada yada</title>
I want
<Bla bla ="nametitle">
but end up with the whole tag.
Please if anyone is offended by the use of parsing html with regex simply move on and accept my apologies for the transgression. I am simply trying to find out how to use the wildcard which I have not used before correctly and apply as I see fit. Thank you.
You can use this regex:
<title.+?>
The above matches <title and goes till it encounters a >
Stop parsing at the first >. Using your example, you could do this with: <(.*?)(title)([^>]*?)>
<(?![\/]).*?title.*?>
This will find title inside any set of < > tags except for closing tags beginning with </
Example:
https://regex101.com/r/QFs4ny/1
In my HTML I have below tags:
<img src="../images/img.jpg" alt="sometext"/>
Using regex expression I want to remove alt=""
How would I write this?
Update
Its on movable type. I have to write it a like so:(textA is replaced by textB)
regex_replace="textA","textB"
Why don't you just find 'alt=""' and replace it with ' ' ?
On Movable Type try this:
regex_replace="/alt=""/",""
http://www.movabletype.org/documentation/developer/passing-multiple-parameters-into-a-tag-modifier.html
What regex you are asking for ? Straight away remove ..
$ sed 's/alt=""//'
<img src="../images/img.jpg" alt=""/>
<img src="../images/img.jpg" />
This does not requires a regex.
The following expression matches alt="sometext"
alt=".*?"
Note that if you used alt=".*" instead, and you had <img alt="sometext src="../images/img.jpg"> then you would match the whole string alt="sometext src="../images/img.jpg" (from alt=" to the last ").
The .* means: Match as much as you can.
The .*? means: Match as little as you can.
s/ alt="[^"]*"//
This regex_replace modifier should match any IMG tag with an alt attribute and capture everything preceding the alt attribute in group #1. The matched text is then replaced with the contents of group #1, effectively stripping off the alt attribute.
regex_replace='/(<img(?:\s+(?!alt\b)\w+="[^"]*")*)\s+alt="[^"]*"/g','$1'
Is that what you're looking for?