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New posts
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58
content/blog/blogroll-from-subscriptions.md
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58
content/blog/blogroll-from-subscriptions.md
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---
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title: "Blogroll From Subscriptions"
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date: 2022-12-18T13:05:49-05:00
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draft: false
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tags: []
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math: false
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---
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While I was browsing around personal websites, I found a fun little piece of code from Jake Bauer's [links page](https://www.paritybit.ca/links).
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```bash
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grep "xmlUrl" static/subscriptions.opml |\
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sed 's/.*text=\"\(.*\)\" xmlUrl=\"\(https\?:\/\/[^\/]*\/\)\(.*\)\" .*/<li><a href=\"\2\">\1<\/a> (<a href=\"\2\3\">feed<\/a>)<\/li>/g'
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```
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This takes the subscriptions exported from [yarr](https://github.com/nkanaev/yarr) and generates a HTML list which you can then include in a blogroll page.
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Running this script on my export from [Feedbin](https://feedbin.com/) yielded some extra metadata being shown in the HTML. For example: `Joke Bauer type="rss" (Feed)`. So let's edit the code snippet above so that it works for my subscription export.
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From my `subscriptions.xml` here's an example entry:
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```xml
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<outline text="Jake Bauer" title="Jake Bauer" type="rss" xmlUrl="https://www.paritybit.ca/feed.xml" htmlUrl="https://www.paritybit.ca/"/>
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```
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It looks like I need to extract the `title`, `xmlUrl`, and `htmlUrl` attributes in that specific order. I'll use the same technique from a previous post on [capturing quoated strings](/capturing-quoted-string-sed).
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```bash
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grep "xmlUrl" subscriptions.xml |\
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sed 's/.*title=\"\([^\"]*\)\".*xmlUrl=\"\([^\"]*\)\".*htmlUrl=\"\([^\"]*\)\".*/<li><a href=\"\3\">\1<\/a> (<a href=\"\2\">feed<\/a>)<\/li>/g'
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```
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We can then clean this up into a `create_blogroll` script saved within [`~/.local/bin`](/blog/customexec/).
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```bash
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#!/bin/sh
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set -o errexit
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set -o nounset
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show_usage() {
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echo "Usage: create_blogroll [subscriptions.xml]"
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exit 1
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}
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# Check argument count
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if [ "$#" -ne 1 ]; then
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show_usage
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fi
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QUOTED_STR="\"\([^\"]*\)\""
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XML_EXPR=".*title=$QUOTED_STR.*xmlUrl=$QUOTED_STR.*htmlUrl=$QUOTED_STR.*"
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HTML_EXPR="<li><a href=\"\3\">\1<\/a> (<a href=\"\2\">feed<\/a>)<\/li>"
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REPLACE_EXPR="s/$XML_EXPR/$HTML_EXPR/g"
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grep "xmlUrl" "$1" | sed "$REPLACE_EXPR"
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```
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65
content/blog/capturing-quoted-string-sed.md
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content/blog/capturing-quoted-string-sed.md
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---
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title: "Capturing Quoted Strings in Sed"
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date: 2022-12-18T12:55:32-05:00
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draft: false
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tags: []
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math: false
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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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```
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