请教同一个文件中php如何获取html的变量值?
代码如下:
<html> <head> <title>配置器</title> </head> <body> <br> 请设置您要修改的属性:<br> <form> <table width="220" height="10" border="0" cellpadding="1" cellspacing="1"> <tr> <td align="center">金币: </td> <td align="center"><input name=jb type=TEXT value="555"></td> </tr><tr> <td align="center">经验:</td> <td align="center"><input name=jy type="TEXT" value="444" /></td> </tr><tr> <td align="center">道具:</td> <td align="center"><input name=dj type="TEXT" value="333" /></td> </tr> <input type=SUBMIT value="测试"> </table> </form> </body> <?php echo "</br>test connect.</br>"; echo ($jb); echo ($jb); echo ($jy); ?></html>
在浏览器打开这个index.php页面后 显示错误
Notice: Undefined variable: jb in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\command.php on line 32
Notice: Undefined variable: jb in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\command.php on line 33
Notice: Undefined variable: jy in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\command.php on line 34
我该如何在php代码中获取html的变量值了?
回复讨论(解决方案)
这个恐怕要用 js 来获取吧,你这个值是动态的,要么就是ajax传到后台php页面
这个恐怕要用 js 来获取吧,你这个值是动态的,要么就是ajax传到后台php页面
我是在《php高级编程》第50页看到的,用php获取html变量的值是可以的!但是为什么会提示未定义的变量了?
你的这个代码早就过时了,是 php4 的早期代码
为了在 php 5.3 以前正确执行,需要有以下设置
register_globals = on
error_reporting = E_ALL & ~E_NOTICE
php5.4 开始取消了 register_globals 开关,你的代码不修改是不能正确运行的
如果你不想修改 php.ini 文件,也不想改动原有的代码
那么至少需要在程序的开始处,加上以下代码
error_reporting(E_ALL ^ E_NOTICE);extract($_GET);
php5.4 开始取消了 register_globals 开关,你的代码不修改是不能正确运行的
如果你不想修改 php.ini 文件,也不想改动原有的代码
那么至少需要在程序的开始处,加上以下代码
error_reporting(E_ALL ^ E_NOTICE);extract($_GET);
加上这两行代码后,不报错了,但是 echo ($jb);为什么什么都没有输出了?有其他方法获取html的变量吗?
你没有提交(点击 测试 按钮)是不会有数据的

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