issue encountered when attempting to use string.replace with regex

Question

issue encountered when attempting to use string.replace with regex

When using a regex like this:

I am attempting to replace "subdir" with a custom string using the string.replace method.

However, when I use

myStr.replace(/^.*\/\/.*\.net\/.*\/(.*)\/.*\z/, otherStr)

The result is not as expected, you can see it here: http://jsfiddle.net/nLmbV/

The Rublar tool seems to capture what I want correctly, but the Fiddle fails to replace it.

I would like to understand why this discrepancy occurs and identify any mistakes in my approach. While corrections in the regex or replace call are welcome, understanding where I went wrong is most important for future reference.

EDIT

I have updated the fiddle by changing my regex from:

/^.*\/\/.*\.net\/.*\/(.*)\/.*\z/

To:

/^.*\/\/.*\.net\/.*\/(.*)\/.*$/

Yet, the fiddle now returns only hello instead of

https://xxxxxxxxxxx.cloudfront.net/dir/hello/Slide1_v2.PNG

javascript regex

Answer 1

Answer №1

That tiny \z in your regular expression is causing the issue.

You may have overlooked replacing it with a $ symbol. In JavaScript, ^ and $ are used as anchors, whereas Ruby uses \A and \z.

In response to your edit:

The entire match is always replaced at once. To correctly handle the replacement, you need to group both sides of the segment to be replaced and then include them in the replacement string:

url.replace(/^(.*\/\/.*\.net\/.*\/).*(\/.*)$/,"$1hello$2")

Answer 2

That tiny \z in your regular expression is causing the issue.

You may have overlooked replacing it with a $ symbol. In JavaScript, ^ and $ are used as anchors, whereas Ruby uses \A and \z.

In response to your edit:

The entire match is always replaced at once. To correctly handle the replacement, you need to group both sides of the segment to be replaced and then include them in the replacement string:

url.replace(/^(.*\/\/.*\.net\/.*\/).*(\/.*)$/,"$1hello$2")

Answer 3

Answer №2

Instead of using complex regular expressions to process URLs reliably, consider a different approach. Using an A tag in your script can simplify the task significantly. The A tag is capable of parsing URLs into pieces and allows you to modify those pieces independently. This way, you only need to focus on dealing with the pathname:

//create a temporary a tag
var a = document.createElement('a');
//assign the URL you want to process to the href property
a.href = "scheme://host.domain/path/to/the/file?querystring"
//split the path part of the URL into an array of directories
var dirs = a.pathname.split('/');
//modify the 2nd directory name - array is ['','path','to','file']
dirs[2]='hello';
//reconstruct the path
a.pathname = dirs.join('/');

Now, a.href contains the desired URL. Although it may require more lines of code, this method will save you from experiencing a headache when you revisit the code later on.

Answer 4

Instead of using complex regular expressions to process URLs reliably, consider a different approach. Using an A tag in your script can simplify the task significantly. The A tag is capable of parsing URLs into pieces and allows you to modify those pieces independently. This way, you only need to focus on dealing with the pathname:

//create a temporary a tag
var a = document.createElement('a');
//assign the URL you want to process to the href property
a.href = "scheme://host.domain/path/to/the/file?querystring"
//split the path part of the URL into an array of directories
var dirs = a.pathname.split('/');
//modify the 2nd directory name - array is ['','path','to','file']
dirs[2]='hello';
//reconstruct the path
a.pathname = dirs.join('/');

Now, a.href contains the desired URL. Although it may require more lines of code, this method will save you from experiencing a headache when you revisit the code later on.

issue encountered when attempting to use string.replace with regex

EDIT

Answer №1

Answer №2

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