将整数作为字符串添加到变量bash中 [英] Add integers as strings to a variable bash

问题描述

我想通过向变量添加随机整数来输出字符串来创建字符串。然而，Bash只是将数字加在一起。

I want to output a string by adding random integer to a variable to create the string. Bash however, just adds the numbers together.

#!/bin/bash
b=""
for ((x=1; x<=3; x++))
do
 number=$RANDOM
 let number%=9
 let b+=$number
done
echo ${b}

假设每个随机数为1，脚本将输出3，而不是111.
如何达到111的预期结果？

Say every random number is 1, the script will output 3 instead of 111. How do I achieve the desired result of 111?

推荐答案

有几种可能性来实现你理想的行为。让我们先来看看你做了什么：

There are several possibilities to achieve your desired behavior. Let's first examine what you've done:

let b+=$number

运行帮助让：

let: let ARGUMENT...
    Evaluate arithmetic expressions.

这解释了为什么让b + = $ number 执行整数加法（ 1 ， 2 ， 3 ） $ number 到 b 而不是字符串连接。

That explains why let b+=$number performs an integer addition (1, 2, 3) of $number to b instead of string concatenation.

只需删除让和所需行为 1 ， 11 ， 111 将会发生。

Simply remove let and the desired behavior 1, 11, 111 will occur.

执行字符串连接的另一种方法：

The other method to perform string concatenation:

b="$b$number"

是的，只需让 b 成为连接 b 和数字。

Yes, simply "let b become the result of concatenating b and number.

作为附注， b =相当于 b = as 扩展为空字符串。对变量的模块操作可以通过算术扩展完成： number = $（（RANDOM％9））。

As a side note, b="" is equivalent to b= as "" is expanded to an empty string. Module operation on a variable can be done with arithmetic expansion: number=$((RANDOM%9)).

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