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May 27, 2015 at 19:07 vote accept Suzana
May 27, 2015 at 18:56 comment added John Because at that point, you're trying to execute the variable's value, not echo it. You have to have an echo command in front of the variable if you're not assigning it (the modified value) to another variable.
May 27, 2015 at 18:54 comment added Suzana Using a second variable name resolves the problem. It's weird that you can't do this in bash. But why do I get the same problem with just the line ${input//\n/,} without any assignment?
May 27, 2015 at 18:47 history answered John CC BY-SA 3.0