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66 lines
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1.6 KiB
Markdown
66 lines
No EOL
1.6 KiB
Markdown
---
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date: 2022-12-18 12:55:32-05:00
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draft: false
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math: false
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medium_enabled: true
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medium_post_id: 9801b2556737
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tags: []
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title: Capturing Quoted Strings in Sed
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---
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*Disclaimer: This posts assumes some knowledge about regular expressions.*
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Recently I was trying to capture an HTML attribute in `sed`. For example, let's say I want to extract the `href` attribute in the following example:
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```
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<a href="https://brandonrozek.com" rel="me"></a>
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```
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Advice you commonly see on the Internet is to use a capture group for anything between the quotes of the href.
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In regular expression land, we can represent anything as `.*` and define a capture group of some regular expression `X` as `\(X\)`.
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```bash
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sed "s/.*href=\"\(.*\)\".*/\1/g"
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```
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What does this look like for our input?
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```bash
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echo \<a href=\"https://brandonrozek.com\" rel=\"me\"\>\</a\> |\
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sed "s/.*href=\"\(.*\)\".*/\1/g"
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```
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```
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https://brandonrozek.com" rel="me
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```
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It matches all the way until the second `"`! What we want, is to not match *any* character within the quotations, but match any character that is not the quotation itself `[^\"]*`
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```bash
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sed "s/.*href=\"\([^\"]*\)\".*/\1/g"
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```
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This then works for our example:
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```bash
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echo \<a href=\"https://brandonrozek.com\" rel=\"me\"\>\</a\> |\
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sed "s/.*href=\"\([^\"]*\)\".*/\1/g"
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```
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```
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https://brandonrozek.com
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```
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Within a bash script, we can make this a little more readable by using multiple variables.
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```bash
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QUOTED_STR="\"\([^\"]*\)\""
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BEFORE_TEXT=".*href=$QUOTED_STR.*"
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AFTER_TEXT="\1"
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REPLACE_EXPR="s/$BEFORE_TEXT/$AFTER_TEXT/g"
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INPUT="\<a href=\"https://brandonrozek.com\" rel=\"me\"\>\</a\>"
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echo "$INPUT" | sed "$REPLACE_EXPR"
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``` |