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Printing double quotes in echo command

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# 1  
Old 09-03-2017
Printing double quotes in echo command
Please help me to use echo or printf type of command to print some value from variable within double quotes - I want to print the double quote ( " ") also.

I tried
HTML Code:
#!/bin/bash
VALUE=some_value
echo '{"value" : "$VALUE"}'
I was expecting the above script would produce ..
HTML Code:
{"value" : "some_value"}
But I am not getting the desired result. Rather the above script is producing
HTML Code:
{"value" : "$VALUE"}
Thank you very much
# 2  
Old 09-03-2017
The single quotes are preventing variable expansion. Try:
Code:
#!/bin/bash
VALUE=some_value
echo "{\"value\" : \"$VALUE\"}"

or
Code:
#!/bin/bash
VALUE=some_value
printf '{"value" : "%s"}\n' "$VALUE"

# 3  
Old 09-03-2017
Got the answer from other threads..

HTML Code:
echo "{\"javapath\" : \"$JAVAPATH\"}"
Smilie
Thanks
# 4  
Old 09-03-2017
Or
Code:
echo '{"value" : "'"$VALUE"'"}'

a concatanation of 'string' "string" 'string'.
$VALUE is evaluated (but not otherwise expanded) because it's inside "string".
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