Bash command substitution executing before earlier code

Question

ssh -i "path/to/my/key.key" ip.ip.ip.ip "mkdir a/a; sql 'query;' > a/a/file.txt; if [ -f a/a/file.txt ]; then if [ $(wc -l a/a/file.txt) -ge 2 ]; then echo 'done'; fi; fi; exit"

After connecting via ssh, I'm basically making a directory, writing a sql query to a file. If that file exists, then I check if its word count is greater than 1. If word count greater than 1, I echo done, then exit.

However, I get this error:

wc: a/a/file.txt open: No such file or directory

It seems as if the command substitution $(wc -l a/a/file.txt) is executing first (at least before the sql query), so the file hasn't been created yet. Is there some way to force this to execute in the order its written, or is there some better way to do this?

Yes, you have the string in double quotes so variables/command substitutions/etc. all happen on the calling side. You need single quotes or to escape the expansions to avoid that. — Etan Reisner
– Etan Reisner, Commented Jul 1, 2015 at 16:26
Also the output from wc -l file isn't directly usable in a -gt test it has the filename in it. Use wc -l <file to avoid that. — Etan Reisner
– Etan Reisner, Commented Jul 1, 2015 at 16:27
The -f test is not necessary; once sql returns, that file is guaranteed to exist because the shell creates it before it executes sql. — chepner
– chepner, Commented Jul 1, 2015 at 17:12

chepner · Accepted Answer · 2015-07-01 18:15:41Z

2

You need to protect the command substitution from being expanded before ssh is called; you can wrap the entire command in single quotes, or simply escape the $ within the double quotes (shown here). (Newlines inserted for clarity; they aren't necessary, but can be used if you want.)

ssh -i "path/to/my/key.key" ip.ip.ip.ip "mkdir -p a/a
 sql 'query;' > a/a/file.txt
 [[ \$(wc -l < a/a/file.txt) -ge 2 ]] && echo 'done'"

The -f test is unnecessary; the shell creates the file before running sql, and the file will exist whether or not the command produces any output. The exit is unnecessary as well; the shell will exit as soon as the last command is executed anyway.

One (hacky?) way to avoid the command substitution altogether is to simply try to read 2 lines from the file with read:

ssh ... "mkdir -p a/a
 sql 'query;' > a/a/file.txt
 { read && read; } < a/a/file.txt && echo 'done'"

edited Jul 1, 2015 at 18:15

answered Jul 1, 2015 at 17:14

chepner

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2 Comments

Alex Forsyth Over a year ago

I'm dealing with a buggy version of sql which is why I have to do these checks in the first place. The query seems to have 3 types of errors. 2 types write a one line error message to the file, which is why I check if it's at least two lines. The third type exits without writing to file, which is why I check if the file exists.

chepner Over a year ago

If the file doesn't exist at all, that's not due to sql; the shell creates the target of a redirection before the associated command is ever run. The one-line error message does require the line count; I'll amend my answer.

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