这个 Zenity 代码有什么问题？

Question

这段代码有什么问题？

#!/bin/bash
ARCH=$(uname -m)
if ["$ARCH" = "i686"]; then
 zenity --info --title="Architechture Checker" --text="Your Architechture is 32-Bit"
if ["$ARCH" = "x86_64"];then
 zenity --info --title="Architechture Checker" --text= "Your Architechture is 64-Bit"

Answer 1

1. 没有与“if”匹配的“fi”

2. 您需要在“[”和“]”周围添加空格

3. “--text=”后的空格会导致参数丢失。

工作版本：

#!/bin/bash
ARCH=$(uname -m)
if [ "$ARCH" = "i686" ]; then
 zenity --info --title="Architechture Checker" --text="Your Architechture is 32-Bit"
fi
if [ "$ARCH" = "x86_64" ]; then
 zenity --info --title="Architechture Checker" --text="Your Architechture is 64-Bit"
fi

Answer 2

或者，使用 case 代替（也可以使用函数来稍微缩短它）。

#!/bin/bash

zinfo() { zenity --info --title="Architecture Checker" --text="$1"; }

case $(uname -m) in
  i686) zinfo "Your architecture is 32-bit" ;;
  x86_64) zinfo "Your architecture is 64-bit" ;;
  *) zinfo "Your architecture is unknown to me" ;;
esac