CODEWITHSUNDEEP
regexapacheregex-groupssirequest-uri
SteveL
SteveL

Reputation: 77

How to use Regex on Apache REQUEST_URI?

Currently using Apache 2.4 with SSI Legacy Parser On and trying to use Regex code to obtain the results I am looking for.

<!--#set var="test" value=$REQUEST_URI -->
<!--#echo var="test" -->

The example result I get from the above code is:

/path1/path2/path3/filename.html

/path1/path2/path3/filename.html?id=2019

The example result I would like to get is:

/path1/path2/path3/

Anything after the last forward slash removed regardless of the number of paths there may be.

Is there a regex code or something I could use to do this?

Upvotes: 1

Views: 2055

Answers (1)

Emma
Emma

Reputation: 27743

Sure! Here we can simply add a slash boundary and swipe everything from beginning to the last slash:

(.*)\/

The expression could stop right there, and it can be simply called using $1. However, we can also add more boundaries to it, if you wish, such as start and end chars:

^(.*)\/.*$

enter image description here

RegEx

If this wasn't your desired expression, you can modify/change your expressions in regex101.com.

RegEx Circuit

You can also visualize your expressions in jex.im:

enter image description here

JavaScript Demo

const regex = /^(.*)\/.*$/gm;
const str = `/path1/path2/path3/filename.html
/path1/path2/path3/filename.html?id=2019
/path1/path2/path3/path4/filename.html?id=2019
/path1/path2/path3/path4/path5/filename.html?id=2019

`;
const subst = `$1`;

// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);

console.log('Substitution result: ', result);

Upvotes: 1

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