Shellscript If statement returns error

Question

I am trying to check if a file is older than 5 minutes and if that is the case I want to call another shell script which sends me a mail.

check_file.sh:

#!/bin/sh

if [$(( (`date +%s` - `stat -L --format %Y /home/ftp/test.txt`) > (5*60) ))] = 1
        then sh ./testmail.sh
fi

Error output: 3: ./check_file.sh: [1]: not found

Add spaces in your if condition: if [ condition ], not if [condition]. — SLePort, Commented Jul 30, 2017 at 14:05

hedgar2017 · Accepted Answer · 2017-07-30 13:51:09Z

1

Try something like:

if find /home/ftp/test.txt -mmin +5 &>/dev/null; then
    <your code>
fi

answered Jul 30, 2017 at 13:51

hedgar2017

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That way the "then" statemant will always be executed... just tried it.
– drifter213
Commented Jul 30, 2017 at 13:56
That's just a hint. Anyway the find program is the best way to check those conditions.
– hedgar2017
Commented Jul 30, 2017 at 13:58

Add a comment |

drifter213 · Accepted Answer · 2017-07-30 14:28:10Z

1

This works:

if test "`find /home/ftp/test.txt -mmin +5`"; then
        echo "file found"
fi

answered Jul 30, 2017 at 14:28

drifter213

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user1934428 · Accepted Answer · 2017-07-31 09:10:15Z

Your script first calculates a number, by virtue of the $((....)) expression. In your case, this number seems to be 1.

This means that you are left with the command

if [1] = 1

This means that bash tries to find a command named [1] and invoke it with two parameters, = and 1.

Since no executable file named [1] is found in your PATH, bash tells you that it can not find the file.

I think

if (( (`date +%s` - `stat -L --format %Y /home/ftp/test.txt`) == 1 ))
then 
    ....

should do the job.

3 Answers 3