创建一个随机数并检查它是奇数还是偶数

问题描述

我希望创建一个批处理文件，在运行时，创建一个0到100之间的随机整数，然后通过if语句运行它，检查它是奇数还是偶数。
我一直遇到麻烦，所以这就是我目前的情况。

I am looking to create a batch file that, when ran, creates a random integer from 0 to 100 and then runs it through an if statement that checks whether it is odd or even.I've been having some trouble, so here is what I have so far.

@echo off
set /a num=%random% %%100 +1
if ( num % 2 == 0 ) {
     //even
}
else
{
    //odd
}

我得到的错误是 - num此时出乎意料

The error I get is - "num was unexpected at this time"

推荐答案

@ECHO OFF
SETLOCAL
set /a num=%random% %%100 +1
SET /a nummod2=num %% 2
IF %nummod2% == 0 (ECHO %num% is even) ELSE (ECHO %num% is odd)

GOTO :EOF

常规 IF 语法是如果[not] operand1 == operand2 somethingtodo 其中operand1和operand2都是字符串。如果字符串包含分隔符，如 或其他对批处理具有特殊含义的字符，则字符串必须为用引号括起来。

Conventional IF syntax is if [not] operand1==operand2 somethingtodo where operand1 and operand2 are both strings. If the strings contain separators like or other characters that have a special meaning to batch then the string must be "enclosed in quotes".

从根本上说，如果比较字符串。运算符必须是一组固定的运算符之一[== equ neq gtr geq lss leq]因此 cmd 反对 num 预期运营商的位置。

Fundamentally, if compares strings. The operator must be one of a fixed set of operators [== equ neq gtr geq lss leq] hence cmd was objecting to num where it expected an operator.

如果语句的参数，则无法在内执行计算。

A calculation cannot be performed within a if statement's parameters.

%% 是执行 mod 操作所必需的，因为％是一个特殊字符，本身需要通过％进行转义。

%% is required to perform the mod operation, since % is a special character that itself needs to be escaped by a %.

请注意， {和} 是普通字符，对批处理没有特殊意义，必须遵循备注

Note that { and } are ordinary characters with no special meaning to batch and remarks must follow

rem remark string

有一个常用的漏洞利用

::comment

实际上是一个破损的标签，可能会产生无法预料的副作用（比如结束代码块）

actually, a broken label, which can have unforeseen side-effects (like ending a code-block)

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