在javascript中使用正则表达式替换

Question

I have an HTML string in java which has data within certain attributes

One example being

<img class="show-grid" src="http://localhost:4502/content/dam/original" alt="4.7-products.jpg">

Now, the problem that I have is I want to append some string at the end of the string which is within src attribute (string starting with http and ending with original here) which is the url.

Only thing I know:

• String would always start with http .

Somethings I don't know

• What would be the ending sequence of the string.
• There could be another attribute after the target attribute here(target attribute is src in this case).

I tried with following regex:

target.search(/(\bhttp.*\"\b)/gi)

The idea is to find to all strings:

• Which start with http and end with " .

So, I thought this regex will give me the following string:

http://localhost:4502/content/dam/original"

But instead it gives me:

http://localhost:4502/content/dam/original" alt="

Any idea why it goes beyond the current word, Is it because it looks for space between words and then when it finds " in the space seperated new word it stops its search there .

How can I modify target.search(/(\\bhttp.*\\"\\b)/gi) to get the desired result there?

Answer 1

Just split and add back in the character you need with a positive lookaround

String string = "http://yourtestinthstsdfasdfasdfhttp://moreteststrings";
String[] parts = string.split("(?=http)");
String part1 = parts[0]; // is http://yourtestinthstsdfasdfasdf
String part2 = parts[1]; // is http://moreteststrings

Answer 2

Your regex are "greedy" (see : https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Objets_globaux/RegExp https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp ). To change it, modify /(\\bhttp.*\\"\\b)/ to /(\\bhttp.*?\\"\\b)/ .

Edit : This still does not work, see my message below

Answer 3

That regex is using the default greedy modifiers. This means that it is searching for as many * (any characters) as it can find. Instead, what you are looking for is the *? (zero or more, non-greedy) modifier, which changes this behaviour.

Your new regex search would look like:

target.search(/http.*?\"/gi)

Answer 4

/http.*(?=\\"\\s)/gi would match anything from http until the end quote followed by a white space. It uses (?=) which is called Positive Lookahead.

A different approach, if you just want to match anything inside the src attribute without need to speculate if http string exists: src="([^"]*) .

You can play with it https://regex101.com/r/rH0yZ2/3 .