如何写HTML，让按钮触发事件运行服务器端的一个.php文件

小弟我正在做毕设，现在想要实现在一个页面上点击一个按钮，让服务器端运行一个.php的文件（这个php文件在服务器本地生成一个txt文件）。
up.php文件如下：
<?php
$filename = "Test.txt";
$file = fopen($filename, "w+");
fwrite($file, "UP\n");
fclose($file);?>
不知道这个html文件怎么写，如何写按钮调用php文件？
看了论坛上一些东西，初步了解以下两种，不知哪种是对的：
1. <FORM action="up.php" method="post"><input type=submit name="test" value="Up"></FORM>
2. <FORM><INPUT NAME="UpButton" TYPE="BUTTON" VALUE="Up" onclick="up.php" ></FORM>请各位大侠指教。小弟先谢过。

解决方案 »

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<FORM><INPUT NAME="UpButton" TYPE="BUTTON" VALUE="Up" onclick="up.php" ></FORM>
这个是错的吧  没见过  触发一个js来submit还差不多
你第一条就可以
up.php写点东西吧，否则是显示一片空白
那就是点击完第一条的按钮，网页就会转向到up.php？如果不让他转向该怎么做？
很明显用 AJAX 最简单了前台不刷新，跳转而执行后台功能。AJAX
1. <FORM action="up.php" method="post"><input type=submit name="test" value="Up"></FORM>
這種就OK了啊...
不会ajax，能否原网页调用？
index.php：
<?phpif($dongzuo == "UP")   {$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "UP\n");fclose($file); print("up");}else{     if($dongzuo == "DOWN"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "DOWN\n");fclose($file);print("down"); }     else{          if($dongzuo == "LEFT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "LEFT\n");fclose($file);print("left"); }          elseif($dongzuo == "RIGHT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "RIGHT\n");fclose($file); print("right");}         }    }?>
<html><head><title>Test</title></head><body><FORM action="test1.php" method="post"><INPUT type="submit" name="dongzuo" value="UP"> <INPUT type="submit" name="dongzuo" value="DOWN"><INPUT type="submit" name="dongzuo" value="LEFT"><INPUT type="submit" name="dongzuo" value="RIGHT">
</FORM></body></html>
能否原网页调用？
index.php：
<?phpif($dongzuo == "UP")   {$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "UP\n");fclose($file); print("up");}else{     if($dongzuo == "DOWN"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "DOWN\n");fclose($file);print("down"); }     else{          if($dongzuo == "LEFT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "LEFT\n");fclose($file);print("left"); }          elseif($dongzuo == "RIGHT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "RIGHT\n");fclose($file); print("right");}         }    }?>
<html><head><title>Test</title></head><body><FORM action="test1.php" method="post"><INPUT type="submit" name="dongzuo" value="UP"> <INPUT type="submit" name="dongzuo" value="DOWN"><INPUT type="submit" name="dongzuo" value="LEFT"><INPUT type="submit" name="dongzuo" value="RIGHT">
</FORM></body></html>
可是没有反应………
写在一个页面也可以  不过是会刷新的
你所谓的没反应  是不是压根就没搭建web服务器
有搭建服务器，环境是ubuntu8.04 apache2 php5,通过进入http://localhost:8081/info.php，得到php的信息表格知道已经搭建了服务器，可以解析php。是不是代码里有错误？
我说的没反应意思有两个：
1.我在每个按钮动作后都加了print("down"); 这样的语句，期望在刷新后的页面显示出来，以说明前面的写文件已经运行过。但是却没有这种文字。
2.本地文件test.txt的内容也没有改变。
是不是代码错误了？可无从调试
补充一下，info.php内容就是<?php phpinfo();?>。另外，把PHP代码放到<BODY>里，也没用。
<html><head><title>Test2</title></head><body>
<?phpif($dongzuo == "UP")   {$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "UP\n");fclose($file);print("up");}else{     if($dongzuo == "DOWN"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "DOWN\n");fclose($file);print("down"); }     else{          if($dongzuo == "LEFT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "LEFT\n");fclose($file);print("left"); }          elseif($dongzuo == "RIGHT"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "RIGHT\n");fclose($file); print("right");}         }    }?><FORM action="test2.php" method="post"><INPUT type="submit" name="dongzuo" value="UP"> <INPUT type="submit" name="dongzuo" value="DOWN"><INPUT type="submit" name="dongzuo" value="LEFT"><INPUT type="submit" name="dongzuo" value="RIGHT">
</FORM></body></html>
<html><head><title>Test2</title></head><body>
<?php
$dongzuo = $_POST['dongzuo'];
if($dongzuo == "UP")  {$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "UP\n");fclose($file);print("up");}else{  if($dongzuo == "DOWN"){$filename = "Test.txt";$file = fopen($filename, "w+");fwrite($file, "DOWN\n");fclose($file);print("down"); }
.....
OH yeah～！Thank you very much~!!!!!!!!!!!
借助jquery用ajax吧
<?php
if (!empty($_POST['dongzuo'])) {
switch (strtoupper($_POST['dongzuo'])) {
case 'UP':
case 'DOWN':
case 'LEFT':
case 'RIGHT':
$filename = "Test.txt";
$file = fopen($filename, "w+");
fwrite($file, strtoupper($_POST['dongzuo'])."\n");
fclose($file);
print(strtolower($_POST['dongzuo']));
break;
default:
print('error');
break;
}
exit;
}
?>
<html>
<head>
<title>Test</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script>
</head>
<body>
<FORM action="?" method="post">
<INPUT type="submit" name="dongzuo" value="UP">
<INPUT type="submit" name="dongzuo" value="DOWN">
<INPUT type="submit" name="dongzuo" value="LEFT">
<INPUT type="submit" name="dongzuo" value="RIGHT">
</FORM>
<script type="text/javascript">
$('input[name=dongzuo]').click(function(){
$.post('?',{dongzuo:$(this).val()},function(data){alert(data);});
return false;
});
</script>
</body>
</html>