Python 2.7 string

Question

I need to print [something] only one time, even if exactly the same string occurs next time. I tried this:

msg = "[something]" + random.randint(1,5)*str(random.randint(1,1000000000)) + "[something]"
i = msg[msg.index('['):msg.rindex(']')+1]
print i

but it works wrong. Message is prints up to last "]", I would like it to be up to the first "]". Between two "[something]" is randomly amount of strings. Is it posible with this code?

This might help: i = msg[msg.index('['):msg.index(']')+1]

dotdeb
– dotdeb

2015-08-23 13:25:05 +00:00
Commented Aug 23, 2015 at 13:25 — dotdeb
– dotdeb, Commented Aug 23, 2015 at 13:25

Martin Evans · Accepted Answer · 2015-08-23 14:13:25Z

2

Here are two ways you could do it:

import re

msg = "[something]" + random.randint(1,5)*str(random.randint(1,1000000000)) + "[something]"

print msg[msg.index('['):msg.index(']')+1]
print re.search("(\[.*?\])", msg).group(1)

Which will display:

[something]
[something]

answered Aug 23, 2015 at 14:13

Martin Evans

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jorispilot · Accepted Answer · 2015-08-23 13:25:54Z

1

Use msg.index(']') instead of msg.rindex(']') as the last method searches "]" from the end of the string. For more details, see the doc.

answered Aug 23, 2015 at 13:25

jorispilot

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