RegEx extract string after asad= and before the next &

Question

How would I match anything after asad= but before the next &?

Test Strings:

1. https://teststore.com/products/mens-sandals-black?asad=253485_e1c6ae3ad&ol_color=13968
2. https://teststore.com/products/womens-sandals-tan-?asad=252485_c1c63c01d&variant=2770725251
3. https://teststore.com/products/mens-shoes-blue?asad=254325_c1c63c01d
4. https://teststore.com/collections/men/products/mens-sneakers?variant=310637539&asad=204207_e1c1756d5

Expected Extraction:

1. 253485_e1c6ae3ad
2. 252485_c1c63c01
3. 254325_c1c63c01d
4. 204207_e1c1756d5

Answer 1

You can use lookahead:

• x(?=y) – positive lookahead (matches 'x' when it's followed by 'y')
• x(?!y) – negative lookahead (matches 'x' when it's not followed by 'y')

and you can also use lookbehinds:

• (?<=y)x – positive lookbehinds (matches 'x' when it's precede by 'y')
• (?<!y)x – negative lookbehinds (matches 'x' when it's not precede by 'y')

These are examples for your need, if you have text with many lines, use this:

(?<=asad=).*?(?=(?:&|\n))

or if you have an array with multiple strings use that

(?<=asad=).*?(?=(?:&|$))

Example on https://regex101.com/r/8LvPtZ/1

Answer 2

You want non-greedy matching, and probably want to match either a & or the end of the line as a terminator:

/asad=(.*?)(?:&|$)/

This tells the regex engine:

1. We're looking for the sequence "asad="
2. Then, capture everything until...
3. We see either a "&" or the end of the line ($).

If you use greedy matching (.*) then it'll capture everything until it can't match terminators anymore.

With the test string:

https://teststore.com/collections/men/products/mens-sneakers?variant=310637539&asad=204207_e1c1756d5&foo=bar

The greedy match /asad=(.*)(?:&|$)/ would capture "asad=204207_e1c1756d5&foo=bar", because it can keep greedily matching up until the end-of-line ($), whereas the non-greedy match /asad=(.*)(?:&|$)/ captures just what you want.