如何在程序中打开％的网站？ [英] How to open a website with % in a program?

问题描述

我只是想创建一个程序来打开特定的网页。

例如打开以下网站：

http：// xxxx / cgi-bin / main- cgi？json =％7b％2cmd％2％3a％2c％2bEnabled％2％3a％7d

xxxx意味着你需要进入的东西



我尝试过的事情：



1.我已经尝试过蝙蝠。

I just want to create a program to open specific webpage.
For example open the website below:
http://xxxx/cgi-bin/main-cgi?json=%7b%2cmd%2%3a%2c%2bEnabled%2%3a%7d
xxxx means something you need to enter

What I have tried:

1.I have tried in bat.

@set /p ip=please enter ip:

@start explorer.exe "http://%ip%/cgi-bin/main-cgi?json=%%7b%22cmd%22%3a654%2c%22bEnabled%22%3a1%7d"

@exit

2.我在C ++中尝试过

2.I have tried in C++

#include<stdio.h>
#include<windows.h>
int main() {
	ShellExecute(NULL, "open", "http://IP/cgi-bin/main-cgi?json=%7b%%22cmd%22%3a654%2c%22bEnabled%22%3a1%7d", NULL, NULL, SW_HIDE);
	return 0;
}

但是如你所知，％不会显示为角色。

我该怎么办？请帮忙。

But as you know, % don't show as a character.
What can I do? Please help.

推荐答案

百分比是url / query字符串中的特殊字符：以下数据作为字符。

加倍你的'％'字符：

Percent is a "special character" in url / query strings: the following data is taken as a character.
Double up on your '%' characters:

explorer.exe "http://codeproject.com/cgi-bin/main-cgi?json=%%7b%%%%22cmd%%22%%3a654%%2c%%22bEnabled%%22%%3a1%%7d"

并且它们以百分比字符传递。

and they wioll pass through as percent characters.

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