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How do I extract the last part of a url?

asked 2019-02-18 16:07:10 +0200

rickn gravatar image

I have a column that looks like this: I only need this information which is the last par of the file name: 987611480.jpg Thank you so much if you can help me solve this problem.

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answered 2019-02-18 16:23:57 +0200

Grantler gravatar image

You may use the FIND&REPLACE tool. Activate Regular Expressions.

Search: ^.*\/([:digit:]+\.jpg)$

Replace (all): $1

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Asked: 2019-02-18 16:07:10 +0200

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Last updated: Feb 18