Bash脚本未检测到失败的退出代码

Question

I can't get my bash script (a logging file) to detect any other exit code other than 0, so the count for failed commands isn't being incremented, but the successes is incremented regardless of whether the command failed or succeeded.

Here is the code:

#!/bin/bash
#Script for Homework 8
#Created by Greg Kendall on 5/10/2016
file=$$.cmd
signal() {
    rm -f $file
    echo
    echo "User Aborted by Control-C"
    exit
}
trap signal 2
i=0
success=0
fail=0
commands=0
read  -p "$(pwd)$" "command"
while [ "$command" != 'exit' ]
do
    $command
    ((i++))
    echo $i: "$command" >> $file
    if [ "$?" -eq 0 ]
        then
            ((success++))
            ((commands++))
        else
            ((fail++))
            ((commands++))
    fi
    read -p "$(pwd)" "command"
done
if [ "$command" == 'exit' ]
    then
    rm -f $file
    echo commands:$commands "(successes:$success, failures:$fail)"
fi

Any help would be greatly appreciated. Thanks!

Answer 1

That's because echo $i: "$command" is succeeding always.

The exit status $? in if [ "$?" -eq 0 ] if [ "$?" -eq 0 ] is actually the exit status of echo , the command that is run immediately before the checking.

So do the test immediate after the command:

$command
if [ "$?" -eq 0 ]

and use echo elsewhere

Or if you prefer you don't need the $? check at all, you can run the command and check status within if alone:

if $command; then .....; else ....; fi

If you do not want to get the STDOUT and STDERR:

if $command &>/dev/null; then .....; else ....; fi

** Note that, as @Charles Duffy mentioned in the comment, you should not run command(s) from variables .

Answer 2

Your code is correctly counting the number of times that the echo $i: "$command" command fails. I presume that you would prefer to count the number of times that $command fails. In that case, replace:

$command
((i++))
echo $i: "$command" >> $file
if [ "$?" -eq 0 ]

With:

$command
code=$?
((i++))
echo $i: "$command" >> $file
if [ "$code" -eq 0 ]

Since $? captures the exit code of the previous command, it should be placed immediately after the command whose code we want to capture.

Improvement

To make sure that the value of $? is captured before any other command is run, Charles Duffy suggests placing the assignment on the same line as the command like so:

$command; code=$?
((i++))
echo $i: "$command" >> $file
if [ "$code" -eq 0 ]

This should make it less likely that any future changes to the code would separate the command from the capture of the value of $? .