passing command line arguments to a shell script doesn't work

Question

I want to write a script that will change to different directories depending on my input. something like this:

test.sh:
#!/bin/bash
ssh machine001 '(chdir ~/dev$1; pwd)'

But as I run ./test.sh 2 it still goes to ~/dev. It seems that my argument gets ignored. Am I doing anything very stupid here?

Necto · Accepted Answer · 2013-02-11 19:01:45Z

5

Bash ignores any variable syntax inside the single-quoted(') strings. You need double quotes(") in order to make a substitution:

#!/bin/bash
ssh machine001 "(chdir ~/dev$1; pwd)"

answered Feb 11, 2013 at 19:01

Necto

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Comments

chepner · Accepted Answer · 2013-02-11 19:01:26Z

2

The parameter is enclosed in single quotes, so it isn't expanded on the local side. Use double-quotes instead.

#!/bin/bash

ssh machine001 "chdir ~/dev$1; pwd"

There's no need for the (...), since you are only running the pair of commands then exiting.

answered Feb 11, 2013 at 19:01

chepner

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