bash array - access ellements

Question

I need to extract some information from json strings in bash. The json-strings are like

{"ok":true,"id":"af1aa5cef5dc0c86fd734ff3d42a8188","rev":"1-f28941b049d7d356ae113fa061ddfe1f"}

(outputs from a couchdb insert)

I want to grab just the id, so I tried to

$IFS=;,\" 
json=tail -n1 output
echo $json 
\\ { ok  true  id   af1aa5cef5dc0c86fd734ff3d42a8188   rev   1-f28941b049d7d356ae113fa061ddfe1f }

So, so far, all seems fine and dandy, but, if I try to access an element, ${json[0]} returns the entire string, whereas all other ${json[n]}s are empty.

On the other hand,

for i in $json;
 do echo $i
done

Works with each element of $json...

I must be doing something wrong, but what?

(And yes, I do know this kind of "parsing" of a json string will get me into problems with something just slightly more complicated - but for these strings it should work fine - I thought)

choroba · Accepted Answer · 2013-02-04 10:39:39Z

2

To assing to an array, you have to use parentheses. The line where you capture the output of tail is also invalid, BTW.

json=( $(tail -n1 output) )
echo "${json[2]}"

answered Feb 4, 2013 at 10:39

choroba

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1 Comment

MortenSickel Over a year ago

Ahh, I forgot the backticks when I wrote it.

Suku · Accepted Answer · 2013-02-04 10:38:21Z

1

You can use awk:

$ echo '"ok":true,"id":"af1aa5cef5dc0c86fd734ff3d42a8188","rev":"1-f28941b049d7d356ae113fa061ddfe1f"}' | awk -F, '{ print $2 }'
"id":"af1aa5cef5dc0c86fd734ff3d42a8188"

You can also you two delimiters with awk

 $ awk -F'[,:]' '{ print $4 }' your_json_file.json
"af1aa5cef5dc0c86fd734ff3d42a8188"

answered Feb 4, 2013 at 10:38

Suku

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1 Comment

MortenSickel Over a year ago

Thanks, more manageable than my original idea. Although I ended up with test=tail -n1 output | awk -F'[,:\"]' '{ print $9 }'

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