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POHow to selectively replace spaces in a string with Bash only
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Id
12819795
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AcceptedAnswerId
12820006
AnswerCount
3
ClosedDate
2012-10-11T21:44:11.807
CommentCount
1
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CreationDate
2012-10-10T12:51:26.777
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2012-10-10T13:36:19.360
LastEditDate
2012-10-10T12:59:40.370
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1
Score
1
ViewCount
136
LastEditorDisplayName
text
Body
I have this string: <pre><code>str='1 2 3 var="foo bar" 4 5 6' </code></pre> Also it could be like this: <pre><code>str="1 2 3 var='foo bar' 4 5 6" </code></pre> How would I replace spaces with <code>+</code> only inside quotes(double or single)? The result should be like: <pre><code>1 2 3 var="foo+bar" 4 5 6 </code></pre> or <pre><code>1 2 3 var='foo+bar' 4 5 6 </code></pre> In case the var is single quoted. <code>var</code> can be anything, it is not strong match. Also it could be missing at all, like this: <pre><code>str='1 2 3 "foo bar" 4 5 6' </code></pre> I don't want to use awk, sed, or perl to do this.
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<bash>
Title
How to selectively replace spaces in a string with Bash only
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1. USPeter O.
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1. USBill Parker
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1. COsee my answer here - http://stackoverflow.com/questions/12821302/split-a-string-only-by-spaces-that-are-outside-quotes - it does exactly what you asking
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