jq-hopkok | jq snippets to manipulate json files | JSON Processing library
kandi X-RAY | jq-hopkok Summary
kandi X-RAY | jq-hopkok Summary
A bunch of jq snippets and shell scripts to manipulate JSON files. See also jqnpm for jq modules/packages. jq is a lightweight and flexible command-line JSON processor. ️ This project has been archived No future updates are planned. Feel free to continue using it, but expect no support.
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QUESTION
Requirement: I have a Json file which is in .gz format. So, when it is compressed it is around ~500 MB in size. When I extract it, the json file becomes nearly around ~10 GB. The extracted JSON file contains individual JSON objects line by line.What I want is to sort the file based on a field ps
using either any bash script or python programs.
Because the file is too large, its not advisable to load it into memory. So, I used gzcat and cat bash command to stream the JSON data and then pipe them to jq for sorting purpose. But either the system doesn't respond during the process or I get empty file in the output.json
...ANSWER
Answered 2019-Apr-16 at 20:21Here is one solution based on the suggestion in one of the comments:
If you can e.g. prefix the lines with the sort key so that they can be sorted as text rather than JSON, then GNU sort can easily sort 10GB+ files without loading them into memory. – that other guy
You can use jq to do this along the following lines:
QUESTION
I have a question on how to provide a variable wrapped within '"..."'
, the expected format for a value read by shell script using read -d '' var <<- EOF
.
I wish to use this jq
utility library: jq-hopkok. Specifically, bash shell script to-components.sh, which uses jq
for parsing URL into its components.
This script is great, except, it is not clear to me how to use this script by passing it a variable $URL
. It appears I can only pass this script with a static URL string that must be wrapped within double-quotes then wrapped again in single-quotes, e.g. '"..."'
.
'"https://..."'
From README.md:
...ANSWER
Answered 2019-Oct-18 at 00:15You need to echo literal double quotes before and after $URL
. A way to write this clearly is to create a variable that contains double quotes.
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