bash/shell 脚本不读取第二个参数 [英] bash/shell script not reading second argument

问题描述

这是一个简单的脚本，我想以某种方式读取第一个和第二个参数，它不读取第二个参数并抛出错误，说明传递值.

this is simple script which i want to read first and second argument somehow its not reading the second argument and throwing error stating pass the value.

这里是脚本//我想克隆git

Here is the script //I want to clone the git

$cat test.sh

#!/usr/bin/env bash

clone () {
  git clone $2
}

case $1
in
   clone) clone ;;

       *) echo "Invalid Argument passed" ;;
esac

运行脚本

$./test.sh clone https://github.com/sameerxxxxx/test.git/
fatal: You must specify a repository to clone.

usage: git clone [<options>] [--] <repo> [<dir>]

    -v, --verbose         be more verbose
    -q, --quiet           be more quiet
    --progress            force progress reporting

推荐答案

当你调用你的函数 clone 时，你必须将参数传递给它.

When you call your function clone, you have to pass the arguments to it.

clone() {
    git clone "$1"
}
...
clone) clone "$2";;

请注意，函数的位置参数与脚本本身是分开编号的.

Note that the function's positional parameters are numbered separately from the script itself.

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