bash 比较shell脚本中的两个数字

Compare two numbers in shell script

提问by MO.

Write a Bash shell script “order.sh” that takes two integer arguments “a” and “b”, and prints out the appropriate relationship “a < b”, “a == b” or “a > b” (with “a” and “b” replaced by their values).

编写一个 Bash shell 脚本“order.sh”，它接受两个整数参数“a”和“b”，并打印出适当的关系“a < b”、“a == b”或“a > b”（带有“ a”和“b”替换为它们的值）。

code:

代码：

#!/bin/bash
echo -n "enter the first number:"; read x
echo -n " enter the second number:"; read y

if ["$x " -lt "$y"]
then
echo "$x < $y"
else
echo"$y < $x"

if [ "$x" -eq "$y"]
then
echo " $x == $y "

fi 

i am not being able to compile his code as it is failing and says "/bin/sh: make command not found" can someone tell me what this means? i am new to shell scripting and i have no idea what is the problem...

我无法编译他的代码，因为它失败并说“/bin/sh: make command not found”有人能告诉我这是什么意思吗？我是 shell 脚本的新手，我不知道有什么问题...

回答by janos

i am not being able to compile his code as it is failing and says "/bin/sh: make command not found" can someone tell me what this means? i am new to shell scripting and i have no idea what is the problem...

我无法编译他的代码，因为它失败并说“/bin/sh: make command not found”有人能告诉我这是什么意思吗？我是 shell 脚本的新手，我不知道有什么问题...

Several problems in that statement:

该声明中有几个问题：

• "compile this code" ... a Bash script doesn't need to be compiled. Bash is an interpreted language
• "/bin/sh: make command not found" means exactly what it looks like: the makecommand is not found. You don't have a makecommand on your PATH. But it doesn't matter, because you don't need makehere

• “编译此代码”……不需要编译 Bash 脚本。Bash 是一种解释型语言
• “/bin/sh: make command not found”的意思就是它的样子：make找不到命令。您make的PATH. 不过没关系，因为你不需要make这里

---

---

Your script has syntax errors, for example:

您的脚本有语法错误，例如：

if ["$x " -lt "$y"]

if ["$x " -lt "$y"]

You need to put a space after [and before ], like this:

您需要在[前后放置一个空格]，如下所示：

if [ "$x " -lt "$y" ]

Other problems:

其他问题：

• Not using if-elif-elsefor the 3 cases
• Broken conditions: there are 2 ifbut only 1 closing fi

• 不if-elif-else用于 3 种情况
• 破坏条件：有 2if但只有 1 个关闭fi

---

---

A few other tips:

其他一些提示：

• For doing arithmetic in Bash, use ((...))instead of [...].
• Instead of echo -n; read, use read -p: it's one command instead of two, and the flags of echoare not portable, so it's better to avoid using them
• Indent the content of if-elif-elseto make the script easier to read

• 要在 Bash 中进行算术运算，请使用((...))代替[...]。
• 而不是echo -n; read，使用read -p：它是一个命令而不是两个命令，并且 的标志echo不可移植，因此最好避免使用它们
• 缩进 的内容if-elif-else以使脚本更易于阅读

With the corrections and improvements applied:

应用更正和改进后：

#!/usr/bin/env bash

read -p "enter the first number: "
read -p "enter the second number: "

if ((x < y)); then
    echo "$x < $y"
elif ((x > y)); then
    echo "$y < $x"
else
    echo "$x == $y"
fi

回答by Elliott Frisch

You should probably use /usr/bin/envto find bash. I think you also wanted an elif(and an else- your current one is missing a fiand won't test for greater than), and you should be using [[and ]](also you missed a space with echo"$y < $x"). Something like,

您可能应该使用/usr/bin/envfind bash。我认为您还想要一个elif（并且一个else- 您当前的一个缺少一个fi并且不会测试大于），并且您应该使用[[and ]]（您也错过了一个带有 的空格echo"$y < $x"）。就像是，

#!/usr/bin/env bash

echo -n "enter the first number:"; read x
echo -n "enter the second number:"; read y
if [[ "$x" -lt "$y" ]]; then
        echo "$x < $y"
elif [[ "$x" -gt "$y" ]]; then
        echo "$y < $x"
else
        echo "$x == $y"
fi

Which I tested, and does as you would expect. I suggest you see 7.02. More advanced if usage - Bash Beginner's Guide

我测试过，并且按您的预期执行。我建议你看7.02。更高级的 if 用法 - Bash 初学者指南