如何使用curl传递curl的JSON输出以使用python查询JSON值

Question

I am writing a bash script where the curl command returns a JSON array of objects and then I need to filter the required values out of those objects. So I need to iterate through the objects to check then and parse them and finally get the result.

But inside my bash script if I do something like,

for i in 1 2 3 4
    do
        curl -XGET 'https://gitlab.com/user/api/v4/projects/1/pipelines/1/jobs' | python -c 'import sys, json; print(json.load(sys.stdin)[$i]["stage"])'
    done

I get the following error:

File "<string>", line 1
    import sys, json; print($i)
                            ^
SyntaxError: invalid syntax
100  3016  100  3016    0     0  18748      0 --:--:-- --:--:-- --:--:-- 19210
(23) Failed writing body

Everything Ok if I remove the python part after the pipe, but why can't I use python in combination with curl ? what am I doing wrong here ?

Answer 1

In bash (and other shells), variables are not expanded when put inside single quotes.

Better approach than my original (as @chepner pointed out), pass the shell variable as command line argument to python command itself:

curl ... | python -c \
   'import sys, json; print(json.load(sys.stdin)[sys.argv[1]]["stage"])' "$i"

Original:

You can get around by using double quotes like:

curl ... | python -c \
            'import sys, json; print(json.load(sys.stdin)['"$i"']["stage"])'

ie terminating the single quote just before variable reference, and continuing afterwards.