CODEWITHSUNDEEP
linuxshellnohup
zjffdu
zjffdu

Reputation: 28924

different behavior when run in nohup mode (linux shell)

I try to do string replacement in linux shell,

str=2011/10/10
echo "$str"
a=${str//\//\_}
echo $a

It can execute when I invoke command : ./test.sh But if I run it in nohup mode, using command : nohup ./test.sh &

It says that ./test.sh: 8: Bad substitution

What's wrong here ?

Thanks

Upvotes: 1

Views: 1236

Answers (1)

EricD
EricD

Reputation: 421

Since you have no #!/bin/bash at the top of your script, the 'nohup' command is using /bin/sh and your system's /bin/sh isn't BASH. Your first and third lines where you assign 'str' and 'a' are not correct Bourne syntax.

Since you likely want to use BASH and not a shell that uses strict Bourne syntax, you should add a #! line at the top of your script like this:

#!/bin/bash
str=2011/10/10
echo "$str"
a=${str//\//\_}
echo $a

Upvotes: 6

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