How to solve the following error in python: "IndexError: string index out of range"

Question

I have written a python code which gives the following error "IndexError: string index out of range"

Please tell me how to fix this error.

actuallink = 'http://www.exxonmobilperspectives.com'
slashcounter = 0
indexslash = 0
while slashcounter < 3:
    if(actuallink[indexslash] == '/'):
        slashcounter = slashcounter + 1
    indexslash = indexslash + 1
    PLink = actuallink[:indexslash - 1]

PS. When I change the link to anything else, it works perfectly

The link you provided doesn't have 3 slashes, so the while loop will not stop incrementing indexslash so it will eventually become too high for the string. — burgerhex
– burgerhex, Commented Feb 24, 2019 at 19:06
@AvivShai , shouldn't the while loop automatically exit after it has indexed to the last character? — Prashant Kumar
– Prashant Kumar, Commented Feb 24, 2019 at 19:11
No. It continues until 3 is hit for slashcounter. You need an additional guard that accounts for len(actuallink) — QHarr
– QHarr, Commented Feb 24, 2019 at 19:12
@QHarr I want to extract the "http://" part from the link. I have an array of links and this link is one of that — Prashant Kumar
– Prashant Kumar, Commented Feb 24, 2019 at 19:13

QHarr · Accepted Answer · 2019-02-24 19:14:55Z

1

Something like

actuallink = 'http://www.exxonmobilperspectives.com'
endPoint = len(actuallink.split('/')) - 1
if endPoint > 0:
    slashcounter = 0
    indexslash = 0
    while slashcounter < endPoint:
        if(actuallink[indexslash] == '/'):
            slashcounter = slashcounter + 1
        indexslash = indexslash + 1
        PLink = actuallink[:indexslash]

answered Feb 24, 2019 at 19:14

QHarr

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Mahrez BenHamad · Accepted Answer · 2019-02-24 19:19:27Z

1

Try this:

actuallink = 'http://www.exxonmobilperspectives.com'
slashcounter = 0
indexslash = 0
while indexslash < len(actuallink):
    if(actuallink[indexslash] == '/'):
        slashcounter = slashcounter + 1
        print("Slash number {},Index ={}".format(slashcounter,indexslash))
    indexslash = indexslash + 1
    PLink = actuallink[:indexslash - 1]
print("Slashcounter = {}".format(slashcounter))

The result :

Slash number 1,Index =5
Slash number 2,Index =6
Slashcounter = 2

answered Feb 24, 2019 at 19:19

Mahrez BenHamad

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