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Old 02-29-2016, 04:49 PM   #1
Paltieldav
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Bulk Metadata Edit search replace Question

Hey All,

I have a number of books that have (XXXX) before the title. x being a number.

I want to get rid of the parentheses and numbers all in one shot. The numbers vary from title to title.

Whats the best way of doing that?

Thanks
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Old 02-29-2016, 05:45 PM   #2
eschwartz
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Using a regular expression.
(You need to switch from "character match" mode to "regular expression" mode. See HERE.)

Find:
Code:
\(\d+\)(.*)
Replace:
Code:
\1

Last edited by eschwartz; 02-29-2016 at 05:48 PM.
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Old 02-29-2016, 05:47 PM   #3
Quadrivium
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Delete (123...) from title

Use Edit metadata in bulk
Select [I]Search and replace[/u] tab
Load search/replace: blank
Search field: title
Search for: a regex
Replace with: blank
Destination field: title

Use a regular expression (regex) to find the thing you want to replace
Given a titles of the form
(123) …
(6789)
The regex is
‘\(\d+\) ‘ [without the ‘s]
The regex: find left paren followed by 1 or more digits followed by right paren folloed by space.
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Old 02-29-2016, 06:47 PM   #4
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Quote:
Originally Posted by Paltieldav View Post
Hey All,

I have a number of books that have (XXXX) before the title. x being a number.

I want to get rid of the parentheses and numbers all in one shot. The numbers vary from title to title.

Whats the best way of doing that?

Thanks
Pick answer in #2
REGEX (mode) is king
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Old 02-29-2016, 07:58 PM   #5
Paltieldav
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Quote:
Originally Posted by eschwartz View Post
Using a regular expression.
(You need to switch from "character match" mode to "regular expression" mode. See HERE.)

Find:
Code:
\(\d+\)(.*)
Replace:
Code:
\1


Thank you!
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Old 03-01-2016, 06:12 AM   #6
Paltieldav
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Quote:
Originally Posted by Quadrivium View Post
Use Edit metadata in bulk
Select [I]Search and replace[/u] tab
Load search/replace: blank
Search field: title
Search for: a regex
Replace with: blank
Destination field: title

Use a regular expression (regex) to find the thing you want to replace
Given a titles of the form
(123) …
(6789)
The regex is
‘\(\d+\) ‘ [without the ‘s]
The regex: find left paren followed by 1 or more digits followed by right paren folloed by space.

Thank you!
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