匹配JSON文本中的键值模式并获取另一个键的值

Question

Suppose I have a file with this JSON:

[
   {
      "label" : "deploy",
      "pk" : 2388175,
      "key" : "gsfd45"
   },
   {
      "label" : "jenkins",
      "key" : "eQtIAwP",
      "pk" : 2388165
   }
]

I want to get the value for key "pk" if it is in the hash that has label = "deploy".

How can I do this? Do I need to write a script?

Answer 1

To parse JSON in Bash, use jq !

$ jq '.[] | select(.label=="deploy").pk' file
2388175

If you want to store deploy in a variable, use --arg . From jq manual → Invoking jq :

--arg name value

This option passes a value to the jq program as a predefined variable. If you run jq with --arg foo bar, then $foo is available in the program and has the value "bar". Note that value will be treated as a string, so --arg foo 123 will bind $foo to "123".

$ v="deploy"
$ jq --arg var "$v" '.[] | select(.label==$var).pk' file
2388175

$ v="blabla"
$ jq --arg var "$v" '.[] | select(.label==$var).pk' file
            # empty!

$ v="jenkins"
$ jq --arg var "$v" '.[] | select(.label==$var).pk' file
2388165

By pieces:

Print everything:

$ jq '.[]' file
{
  "key": "gsfd45",
  "pk": 2388175,
  "label": "deploy"
}
{
  "pk": 2388165,
  "key": "eQtIAwP",
  "label": "jenkins"
}

Print those records where label equals "deploy":

$ jq '.[] | select(.label=="deploy")' file
{
  "key": "gsfd45",
  "pk": 2388175,
  "label": "deploy"
}

Print just the field pk in such case:

$ jq '.[] | select(.label=="deploy").pk' file
2388175

Answer 2

If jq was not availale on your server, python should be there, right? ^_*

#!/bin/python
import json
with open('data.json') as data_file:    
    data = json.load(data_file)
    for d in data:
        if d['label'] == 'deploy':
            print(d["pk"])

assume your file named as data.json save it as id.py , and run with:

python id.py

It needs python3 installed on your system.

change the line print (d["pk"]) into print d["pk"] if you only have python2 installed.

The output would be:

2388175

Edit

added the if check, didn't notice OP wanted to check the label .

Answer 3

In awk. It's a bit incomplete but as you didn't have anything to show, you can work on this one:

$ awk -F: '$1~/"label"/{l=$2} l~/deploy/ && $1 ~ /pk/ {sub(/,/,"",$2);print $2}' file
 2388175

When awk meets a record with "label" on it, it stores the $2 . Once the pk is found and flag l has deploy in it, remove comma and print.

Answer 4

If the elegant solution provided by James Brown does not work (eg different ordering of the key/value pairs) here is something that tries to get at least the string between the braces into one record (by setting RS ), then the string is splittet at the key value pair with key "pk" (by setting FS ).

After that setup the pattern looks for the label/deploy key/value pair in $0 and then, only if there are two fields (eg the pk was present and a field split took place) the string after the comma in $2 is deleted and the value of key pk remains and is printed:

script.awk

BEGIN {
    RS="[{}[\\]]"
    FS="\"pk\"[^:]*:"
    }

/"label"[^:]*:[^\"]*"deploy"/ {
        if( NF == 2 ) { 
            # "pk" is present in $0, remove everything after comma
            sub(/,.*/, "", $2)
            print $2
        }
    }

You use this script with awk like this: awk -f script.awk yourfile .

I have only tried it with GNU awk, but RS and FS should also work with awk, too.