从另一个脚本呼叫一个Bash脚本传递给它的参数加上引号和空格 [英] Calling one Bash script from another Script passing it arguments with quotes and spaces

问题描述

    echo "TestScript1 Arguments:"
    echo "$1"
    echo "$2"
    echo "$#"
    ./testscript2 $1 $2

的 TestScript2如下：的

    echo "TestScript2 Arguments received from TestScript1:"
    echo "$1"
    echo "$2"
    echo "$#"

的当我通过以下方式执行testscript1：的

    ./testscript1 "Firstname Lastname" [e-mail]  

的所需的输出应该是：的

    TestScript1 Arguments:  
    Firstname Lastname  
    [e-mail]  
    2
    TestScript2 Arguments received from TestScript1:  
    Firstname Lastname  
    [e-mail]  
    2  

的但是实际的输出是：的

    TestScript1 Arguments:  
    Firstname Lastname  
    [e-mail]  
    2
    TestScript2 Arguments received from TestScript1:  
    Firstname
    Lastname      
    3  

我要如何解决这个问题？我想获得所需的输出，而不是实际的输出。

How do i solve this problem? I want to get the desired output instead of the actual output.

推荐答案

引用您的ARGS在Testscript 1：

Quote your args in Testscript 1:

echo "TestScript1 Arguments:"
echo "$1"
echo "$2"
echo "$#"
./testscript2 "$1" "$2"

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