Bash脚本变量范围问题

Question

username="hello"
password="3333"

function login {
    # 1 - Username
    # 2 - Password
    match=0
    cat LoginsMaintMenu.txt | while read line; do
            x=`echo $line | awk '{print $1}'`
            y=`echo $line | awk '{print $2}'`
            if [ "${x}" == "${1}" ] && [ "${y}" == "${2}" ]; then
                    echo "match"
                    match=1
                    echo $match
                    break
            fi
    done
    echo $match
    return $match
}

echo $username $password
login ${username} ${password}
if [ $? -eq 0 ]; then
echo "FAIL"
else
echo "success"
fi

输出:

hello 3333
match
1
0
FAIL

问题:我不明白为什么它会回应"失败"."匹配"变量在while循环中设置为1,但由于某种原因,一旦我离开while循环,它仍然认为它是声明的初始零.

我尝试过做很多不同的事情,所以如果有人能给我一些具体的东西,那就太棒了!

谢谢

Answer 1

这不起作用的原因实际上是UUOC.在bash中,管道的右侧在子shell内部运行.在子shell中设置的任何变量都不会在父shell中设置.要解决此问题,请使用重定向而不是管道:

username="hello"
password="3333"

function login {
    # 1 - Username
    # 2 - Password
    match=0
    while read x y _; do
        if [ "${x}" == "${1}" ] && [ "${y}" == "${2}" ]; then
            echo "match"
            match=1
            echo $match
            break
        fi
    done < LoginsMaintMenu.txt
    echo $match
    return $match
}

echo $username $password
if login "${username}" "${password}"; then
    echo "FAIL"
else
    echo "success"
fi