Using a for loop inside a ssh statement replaces variable with empty string

Question

I have the following statement in a bash script:

ssh $host "cd /directory; for i in *$date.gz; do echo $i; done; exit"

I expect it to print the name of each file in the directory that ends with the date, and is a zip file. By ssh-ing to the host on the command line, and searching the directory, I find that there should be 5 such files. However, this script returns 5 blank lines. I checked if the $date variable was properly defined inside the quotes (it was). When I replaced $i with 'adf', the script printed

adf
adf
adf
adf
adf

So it is correctly filtering out those 5 files, but it is just not printing their names, and is replacing the $i in the statement with nothing (so that that line is just echo). Why is it doing this, and how can I make it print the filenames? The same thing happens when I run this line on the command line.

vdavid · Accepted Answer · 2020-03-12 17:42:48Z

1

By double-quoting your command, the variable expansion occurs before the ssh call.

So when you call this command line:

ssh $host "cd /directory; for i in *$date.gz; do echo $i; done; exit"

It calls the ssh command with two arguments: $host and "cd /directory; for i in *$date.gz; do echo $i; done; exit"

The second argument picks the content of the date variable and the i variable when the string is built. But at this time, you do NOT have the correct value for i yet.

I think that escaping $i into \$i should solve your issue:

ssh $host "cd /directory; for i in *$date.gz; do echo \$i; done; exit"

answered Mar 12, 2020 at 17:42

vdavid

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