如何正确调试bash脚本 [英] How to properly debug a bash script

问题描述

我为了追查code正在使用bash脚本如下code

I am using the following code in a bash script in order to trace the code

#!/bin/bash
function trace()
{
    echo "TRACE" \
         "${BASH_SOURCE[1]}:${BASH_LINENO[0]}:${FUNCNAME[1]}:" \
         "$BASH_COMMAND"
}

set -o functrace
shopt -s extdebug
trap trace DEBUG

# Remainder of code here
exit 0

但是，当我运行它，它吃我的变量。
一些线索？

But when I run it, it eats my variables. Some clue?

推荐答案

最好的办法是使用设置-xv 。在设置-v 将在执行之前回声线。该集 -x 将输出shell脚本插在该行的变量和前pressions后再行。

The best way is to use set -xv. The set -v will echo a line before it is executed. The set -x will output the line after the shell script interpolates the variables and expressions in that line.

作为其中的一部分，还可以创建一个名为 PS4 环境变量这将提示您的每一个shell脚本输出正在执行的线时间打印。大多数人把它设置成类似 PS =\\ $ LINENO：这将打印出的行号正在执行的行

As part of this, you can also create an environment variable called PS4 which will be the prompt printed each time your shell scripts outputs the line being executed. Most people set it to something like PS="\$LINENO: " which will print out the line number for the line being executed.

一旦你完成，你可以通过设置套装+ XV 关闭调试。

Once you're finished, you can turn off debugging by setting set +xv.

#
# Somewhere in this part of my script, I am having problems....
#
export PS4="\$LINENO> "   # Now, I'll see the line numbers while debugging
set -xv                   # Debugging is turned on
....
#
# Done with debugging
#
set +xv                   # Debugging is turned off

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