Question

我目前正在学习bash编程，并且真的不明白为什么传递给我的论点不起作用。

我有一个像这样的剧本

#!/bin/bash
# the following environment variables must be set before running this script
# SIM_DIR name of directory containing armsim
# TEST_DIR name of the directory containing this script and the expected outputs
# LOG_DIR name of the directory that your output is written to by the run_test2 script
# ARMSIM_VERBOSE set to "-v" for verbose logging or leave unset

# First check the environment variables are set
giveup=0
if [[ ${#SIM_DIR} -eq 0 || ${#TEST_DIR} -eq 0 || ${#LOG_DIR} -eq 0 ]] ; then
    echo One or more of the following environment variables must be set:
    echo SIM_DIR, TEST_DIR, LOG_DIR
    giveup=1
fi

# Now check the verbose flag
if [[ ${#ARMSIM_VERBOSE} != 0 && "x${ARMSIM_VERBOSE}" != "x-v" ]] ; then
    echo ARMSIM_VERBOSE must be unset, empty or set to -v
    giveup=1
fi

# Stop if environment is not set up
if [ ${giveup} -eq 1 ] ; then
    exit 0
fi

cd ${TEST_DIR}
for i in test2-*.sh; do
  echo "**** Running test ${i%.sh} *****"
  ./$i > ${LOG_DIR}/${i%.sh}.log
done

当我运行.sh文件并传入3个示例参数时，如下所示： -

$ ./run_test2 SIM_DIR TEST_DIR LOG_DIR

它仍显示：One or more of the following environment variables must be set: SIM_DIR, TEST_DIR, LOG_DIR

任何人都可以指导我吗？谢谢。

Answer 1

这不是它的工作方式。必须事先在脚本或终端中设置环境变量，如

export SIM_DIR=/home/someone/simulations
export TEST_DIR=/home/someone/tests
export LOG_DIR=/home/someone/logs

./run_test2

如果您经常使用这些变量，则可能需要在export中~/.bashrc使用这些变量。语法与上例中的export相同。

Answer 2

从我的问题/示例中我理解的环境变量不是真正的参数。听起来我觉得你想给一个函数/脚本提供参数，如果你这样做，你可以在$ 1-9中找到你的参数（我认为bash支持更多，不确定），参数的数量存储在$＃< / p>

需要两个参数的示例函数：

my_func() {
    if [ $# -ne 2 ]; then
        printf "You need to give 2 arguments\n"
        return
    fi

    printf "Your first argument: %s\n" "$1"
    printf "Your second argument: $s\n" "$2"
}

# Call the functionl like this
my_func arg1 arg2

Bash编程传递参数

2 个答案: