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Shell Scripting: Get string inside quotations

x=1; \
y=2; \
echo x: $x - '{"y": "$y"}'

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I would expect this result:
x: 1 - {"y": "2"}

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but instead I get this results:
x: 1 - {"y": "$y"}

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0
hankknight
Asked:
hankknight
  • 3
1 Solution
 
simon3270Commented:
It's because the double quotes are nested within single quotes.  try:
    echo x: $x - '{"y": '"$y"'}'
(that's a single quote between the space and the double quote before the $y, and one between the double-quote after $y and the curly brace)
0
 
simon3270Commented:
As you can see quotes are hard!

I meant:
    echo x: $x - '{"y": "'$y'"}'
where the extra quotes are between the double-quote and the $y, and between the $y and its trailing double-quote
0
 
simon3270Commented:
Essentially, you put single quotes round text that you don't want changed at all, then double quotes (or no quotes at all) round bits where you want variable substitution to take place.
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