Inverse Match in Regex

Question

I have a string

<img width="24" src="https://someurl.com" height="24" alt="FirstName LastName" id="ember44" class="global-nav__me-photo ember-view"> id="ember44" class="global-nav__me-photo ember-view">

In the RegEx I need to select everything except

alt="FirstName LastName"

Tried some kind of expression

alt.+(?!alt)

but still not in place. Thank you in advance!

Answer 1

Instead of trying to match everything that isn't your "anti-search" string, how about replacing that string with nothing?

s = """<img width="24" src="https://someurl.com" height="24" alt="FirstName LastName" id="ember44" class="global-nav__me-photo ember-view"> id="ember44" class="global-nav__me-photo ember-view">"""

s_new = re.sub(r'alt=\"[^\"]+\"\s+', '', s)
# '<img width="24" src="https://someurl.com" height="24" id="ember44" class="global-nav__me-photo ember-view"> id="ember44" class="global-nav__me-photo ember-view">'

Explanation ( Try online ):

alt=\"[^\"]+\"\s+
-----------------

alt=\"       \"     : Literally alt=, followed by quotes
      [^\"]+        : One or more non-quote characters
               \s+  : One or more whitespace

Answer 2

How about this then

<(.*)(?=alt=\"[^\"]+\")(?:alt=\"[^\"]+\")([^>]+)>

You can test this here https://regex101.com/r/P0DLw5/1

This is basically get me everything until alt="...", then match but ignore alt="..." then get me everything after it.
This is not perfect by any means but i'm going by your current example.

Answer 3

Well, in order to invert a regex, you could use re.sub() which takes 3 required arguments. a pattern, replacement, and a original string.

So, you could invert like this

import re

s = '<img width="24" src="https://someurl.com" height="24" alt="FirstName LastName" id="ember44" class="global-nav__me-photo ember-view"> id="ember44" class="global-nav__me-photo ember-view">'

pattern = r'alt=".*?"'
without_alt = re.sub(pattern, '', s)
print(without_alt)