How to get a substring of this string Python3

Question

string:

ecsdcsdcsdfvdfv":"https://scdsscdcsdent-mxp1-1.cdninstdscsdcagdssdcsdam.com/v/t51.283485-19/s320x320/79000872_1455436197941341_7513464347075543040_n.pnk?_nc_ht=scontent-mxp1-1.cdninadcdcdm.codcsdcm&_nc_ohc=0fehqjedb48AX8r72Hi&oh=eb1f6a78a2dcd67e443aa7f74eee91b4&oe=5E7F0A0C","vsvdfvsfvcfvfdcvfd

substring that i want to get:

https://scdsscdcsdent-mxp1-1.cdninstdscsdcagdssdcsdam.com/v/t51.283485-19/s320x320/79000872_1455436197941341_7513464347075543040_n.pnk?_nc_ht=scontent-mxp1-1.cdninadcdcdm.codcsdcm&_nc_ohc=0fehqjedb48AX8r72Hi&oh=eb1f6a78a2dcd67e443aa7f74eee91b4&oe=5E7F0A0C

i tried this but doesn't work

print (log.split("ecsdcsdcsdfvdfv",1)[1])

@Sayse if you find a simple solution using json.parse i'll follow it — the Pythoner
– the Pythoner, Commented Dec 20, 2019 at 13:05

Guimoute · Accepted Answer · 2019-12-20 13:03:22Z

2

This yield the expected result.

The way you used .split() was wrong, you did not include the " character.

string = 'ecsdcsdcsdfvdfv":"https://scdsscdcsdent-mxp1-1.cdninstdscsdcagdssdcsdam.com/v/t51.283485-19/s320x320/79000872_1455436197941341_7513464347075543040_n.pnk?_nc_ht=scontent-mxp1-1.cdninadcdcdm.codcsdcm&_nc_ohc=0fehqjedb48AX8r72Hi&oh=eb1f6a78a2dcd67e443aa7f74eee91b4&oe=5E7F0A0C","vsvdfvsfvcfvfdcvfd'
substring = string.split('ecsdcsdcsdfvdfv":"')[1].split('","vsvdfvsfvcfvfdcvfd')[0]

answered Dec 20, 2019 at 13:03

Guimoute

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2 Comments

the Pythoner Over a year ago

This is what i was looking for! Thanks

the Pythoner Over a year ago

you forgot a " on the start

Henry Harutyunyan · Accepted Answer · 2019-12-20 12:55:29Z

2

You can try this

log.split('":"')[1].split('","')[0]

But this is not the best way to do what you are trying to achieve. Better parse it and get what you want.

answered Dec 20, 2019 at 12:55

Henry Harutyunyan

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7 Comments

kuro Over a year ago

That won't give the expected output though

Henry Harutyunyan Over a year ago

I know :) Edited

the Pythoner Over a year ago

I don't like this solution because i i got a string with more ":" it doesn't work

Henry Harutyunyan Over a year ago

More ":"? What do you mean? :D

the Pythoner Over a year ago

int the start and in the end of the string more ":"

|

Henry Harutyunyan · Accepted Answer · 2019-12-20 13:09:26Z

If you want to get the string between the ":" and ",", then you can use regular expression to do it.

re.match('(.*\":\")([^\",\"]*)(\",\".*)', log).group(2)

Given your input

'ecsdcsdcsdfvdfv":"https://scdsscdcsdent-mxp1-1.cdninstdscsdcagdssdcsdam.com/v/t51.283485-19/s320x320/79000872_1455436197941341_7513464347075543040_n.pnk?_nc_ht=scontent-mxp1-1.cdninadcdcdm.codcsdcm&_nc_ohc=0fehqjedb48AX8r72Hi&oh=eb1f6a78a2dcd67e443aa7f74eee91b4&oe=5E7F0A0C","vsvdfvsfvcfvfdcvfd'

you will get

'https://scdsscdcsdent-mxp1-1.cdninstdscsdcagdssdcsdam.com/v/t51.283485-19/s320x320/79000872_1455436197941341_7513464347075543040_n.pnk?_nc_ht=scontent-mxp1-1.cdninadcdcdm.codcsdcm&_nc_ohc=0fehqjedb48AX8r72Hi&oh=eb1f6a78a2dcd67e443aa7f74eee91b4&oe=5E7F0A0C'

And if you input something like

'ecsdcsdcs":"dfvdfv":"https://s...F0A0C","vsvdfvsfvcfvfdc","vfd'

you will get

'https://s...F0A0C'

Dont forget to import re.

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