while loop multiple conditions Python

while ((j <= len(input)) |  (j != ' ') | (input[j+1] != "'")):

0) You should be using or.

1) You should not use input as a variable name; it hides a built-in function.

2) j is an integer, so it can never be equal to ' ', so that test is useless.

3) j <= len(input) passes when j == len(input). The length of a string is not a valid index into the string; indices into a string of length N range from 0 to (N - 1) (you can also use negative numbers from -1 to -N, to count from the end). Certainly j+1 doesn't work either.

4) I can't tell what the heck you are actually trying to do. Could you explain it in words? As stated, this isn't a very good question; making the code stop throwing exceptions doesn't mean it's any closer to working correctly, and certainly doesn't mean it's any closer to being good code.

It looks like j+1 is a number greater than or equal to the length of the string you have (input). Make sure you structure your while loop so that j < (len(input) - 1) is always true and you won't end up with the string index out of range error.

if j >= len(input) - 1 then j+1 will most certainly be out of bounds. Also, use or and not |.

You get an error IndexError: string index out of range. The only index reference is in part input[j+1]. Situation when j = len(input) will cause an error, as the following code demonstrates:

>>> input = "test string"
>>> len(input)
11
>>> input[11]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: string index out of range
>>> input[10]
'g'

If you try to reference element number j+1, then condition j < ( len(input) - 1 ) needs to be satisfied.

When using != in if statements, make sure that or is actually what you need. Here's an example:

import random
a = random.randint(1, 10)
b = random.randint(1, 10)
c = random.randint(1, 10)
if a != 1 or b != 1 or c != 1:
    print "None of the values should equal 1"
    # The interpreter sees `a != 1`.
    # If `a` is not equal to 1 the condition is true, and this code gets excecuted.
    # This if statement will return true if ANY of the values are not equal to 1.
if a != 1 and b != 1 and c != 1:
    print "None of the values are equal to 1" # True
    # This if statement will return true if ALL of the values are not equal to 1.

This is a hard thing to understand at first (I made this mistake all the time), but if you practise it a bit, it will make perfect sense.

So, to get your code working, replace those |s with and, and it will work (and stick with the keywords or and and unless you specifically need boolean or or and (|/&):

while ((j <= len(input)) and  (j != ' ') and (input[j+1] != "'")):

and the output is:

100px

Not the solution to your problem. Code that probably does what you are aiming for.

Just use a single regular expression:

import re

def width(input, attr):
    """
    >>> width("a='100px'", 'a')
    '100px'
    """
    regex = re.compile(attr + r""".*?'(?P<quoted_string>[^']+)'""")
    m = regex.match(input.strip())
    return m.group("quoted_string") if m else ''

if __name__ == '__main__':
    import doctest
    doctest.testmod()

This code skips attr and searches for a quoted string that follows. (?P<quoted_string>[^']+) captures the quoted string. m.group("quoted_string") recovers the quoted string.