为什么["$ foo" =="$ bar" ]在bash中始终是正确的? [英] Why is [ "$foo"=="$bar" ] always true in bash?

问题描述

我试图将用户输入与字符串进行比较 这是我的代码

I tried to compare user input between to string Here is my code

Encode="Encode"
Decode="Decode"

printf "Enter name of file: "
read fileName
printf "Encode Or Decode: "
read EncOrDec

if [ "$Encode"=="$EncOrDec" ]; then
    printf "Encode Nice\n"
elif [ "$Decode"=="$EncOrDec" ]; then
    printf "Decode Nice\n"
else
    printf "Nothing\n"
fi

它总是转到Encode语句，为什么? 以及解决方法

Its always go to the Encode statement, Why?. And how to fix it

推荐答案

在bash中，空格是计数.替换:

In bash, spaces count. Replace:

if [ "$Encode"=="$EncOrDec" ]; then

使用:

if [ "$Encode" = "$EncOrDec" ]; then

没有空格，bash只是测试字符串"$Encode"=="$EncOrDec"是否为空.由于从不为空，因此始终执行then子句.

Without spaces, bash is just testing whether the string "$Encode"=="$EncOrDec" is empty or not. Since it is never empty, the then clause is always executed.

此外，作为次要细节，使用[...]时，使用=进行字符串相等是POSIX标准. Bash接受==，但是==不是标准的，并且不会可靠地移植.

Also, as a minor detail, when using [...], the use of = for string equality is POSIX standard. Bash accepts == but == is not standard and won't be reliably portable.

elif行也是如此.替换:

elif [ "$Decode"=="$EncOrDec" ]; then

使用:

elif [ "$Decode" = "$EncOrDec" ]; then

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