在bash脚本中获取python脚本的退出代码

Question

我是bash的新手,想要抓住我的python脚本退出代码,我script.py看起来像:

#! /usr/bin/python
def foofoo():
    ret = #do logic
    if ret != 0:
        print repr(ret) + 'number of errors'
        sys.ext(1)
    else:
        print 'NO ERRORS!!!!'
        sys.exit(0)
def main(argv):
    #do main stuff
    foofoo()
if __main__ == "__main__":
    main(sys.argv[1:]

我的bash脚本:

#!/bin/bash
python script.py -a a1  -b a2 -c a3
if [ $?!=0 ];
then
    echo "exit 1"
fi
echo "EXIT 0"

我的问题是我总是exit 1在我的bash脚本中打印,如何在我的bash脚本中获取我的python退出代码？

Answer 1

空格很重要因为是参数分隔符

if [ $? != 0 ];
then
    echo "exit 1"
fi
echo "EXIT 0"

或数字测试-ne,看man [的[命令或man bash为内置更多详情

# store in variable otherwise will be overwritten after next command
exit_status=$?
if [ "${exit_status}" -ne 0 ];
then
    echo "exit ${exit_status}"
fi
echo "EXIT 0"