How to pass the variable value in echo statement?
testvar="actualvalue"
echo 'testing "${testvar}", "testing", "testing" ;'
Expected output:
testing "actualvalue", "testing", "testing" ;
But, I am getting the below output:
testing "${testvar}", "testing", "testing" ;
Can someone help me with this?
Remove the single quotes:
$ testvar="actualvalue"
$ echo testing "${testvar}", "testing", "testing" ;
testing actualvalue, testing, testing
The single-quote inhibits variable expansion.
Without double-quotes gives you the same output:
$ echo testing ${testvar}, testing, testing ;
testing actualvalue, testing, testing
But if you really want double-quotes in the output, escape them:
$ echo "testing \"${testvar}\", \"testing\", \"testing\" ;"
testing "actualvalue", "testing", "testing" ;
${testvar} outside double quotes is invoking the split+glob operator, so it would only give the same output in the special cases where ${testvar} contains no wildcard characters nor characters of $IFS. Also note that echo can generally not be used to output arbitrary data. You'd use printf for that instead.
echocommand remove the special meaning of $. Replace the single quotes with double quotes, and put backslashes before the other double quotes:echo "testing \"${testvar}\", \"testing\", \"testing\" ;".