---
title: "Capturing Quoted Strings in Sed"
date: 2022-12-18T12:55:32-05:00
draft: false
tags: []
math: false
---
*Disclaimer: This posts assumes some knowledge about regular expressions.*
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:
```
```
Advice you commonly see on the Internet is to use a capture group for anything between the quotes of the href.
In regular expression land, we can represent anything as `.*` and define a capture group of some regular expression `X` as `\(X\)`.
```bash
sed "s/.*href=\"\(.*\)\".*/\1/g"
```
What does this look like for our input?
```bash
echo \\ |\
sed "s/.*href=\"\(.*\)\".*/\1/g"
```
```
https://brandonrozek.com" rel="me
```
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 `[^\"]*`
```bash
sed "s/.*href=\"\([^\"]*\)\".*/\1/g"
```
This then works for our example:
```bash
echo \\ |\
sed "s/.*href=\"\([^\"]*\)\".*/\1/g"
```
```
https://brandonrozek.com
```
Within a bash script, we can make this a little more readable by using multiple variables.
```bash
QUOTED_STR="\"\([^\"]*\)\""
BEFORE_TEXT=".*href=$QUOTED_STR.*"
AFTER_TEXT="\1"
REPLACE_EXPR="s/$BEFORE_TEXT/$AFTER_TEXT/g"
INPUT="\\"
echo "$INPUT" | sed "$REPLACE_EXPR"
```