如何在 Bash 的“if"语句中比较两个字符串变量? [英] How do I compare two string variables in an 'if' statement in Bash?

问题描述

我正在尝试让 if 语句在 Bash(使用 Ubuntu):

I'm trying to get an if statement to work in Bash (using Ubuntu):

#!/bin/bash

s1="hi"
s2="hi"

if ["$s1" == "$s2"]
then
  echo match
fi

我已经尝试了各种形式的 if 语句，使用 [["$s1" == "$s2"]]，带和不带引号，使用=、== 和 -eq，但我仍然收到以下错误:

I've tried various forms of the if statement, using [["$s1" == "$s2"]], with and without quotes, using =, == and -eq, but I still get the following error:

[嗨:找不到命令

我查看了各种网站和教程并复制了它们，但它不起作用 - 我做错了什么?

I've looked at various sites and tutorials and copied those, but it doesn't work - what am I doing wrong?

最后，我想说如果 $s1 包含 $s2，那我该怎么做?

Eventually, I want to say if $s1 contains $s2, so how can I do that?

我只是算出了空格...:/我怎么说包含?

I did just work out the spaces bit... :/ How do I say contains?

我试过了

if [[ "$s1" == "*$s2*" ]]

但是没有用.

推荐答案

对于字符串相等比较，使用:

For string equality comparison, use:

if [[ "$s1" == "$s2" ]]

对于字符串不等于比较，使用:

For string does NOT equal comparison, use:

if [[ "$s1" != "$s2" ]]

对于a包含b，使用:

if [[ $s1 == *"$s2"* ]]

(并确保在符号之间添加空格):

(and make sure to add spaces between the symbols):

不好:

if [["$s1" == "$s2"]]

好:

if [[ "$s1" == "$s2" ]]

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