<html><head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>交互测试页</title>
</head>
<body>
<form method="POST" action="show.php">
<p><input type="submit" value="show" name="B3"></p>
</form>
<?php
if (count($_POST) > 0)
{
$link = mysql_pconnect("localhost", "root", "3033")
or die("Could not connect: " . mysql_error());
mysql_select_db('123') or die ('can not select dtb');
$query='select * from ben';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
echo"<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
?></body>
</html>
不知道可行不?改用SMARTY模版是最好的.
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>交互测试页</title>
</head>
<body>
<form method="POST" action="show.php">
<p><input type="submit" value="show" name="B3"></p>
</form>
<?php
if (count($_POST) > 0)
{
$link = mysql_pconnect("localhost", "root", "3033")
or die("Could not connect: " . mysql_error());
mysql_select_db('123') or die ('can not select dtb');
$query='select * from ben';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
echo"<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
?></body>
</html>
不知道可行不?改用SMARTY模版是最好的.
<form method="POST" action="show.php">
<p><input type="submit" value="show" name="B3"></p>
显然 action="show.php" 是说如果 被按了 激发 这个action="show.php"
那么如果我不是想打开 这个 网页,我想做些什么.
就是 说我想在 action="show.php">的这个地方添加代码 应该怎么弄呀?
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>交互测试页</title>
</head>
<body>
<form method="POST" action="show.php">
<p><input type="submit" value="show" name="B3"></p>
</form><?php
if($_SERVER['REQUEST_METHOD'] == "POST"):
$link = mysql_pconnect("localhost", "root", "3033")
or die("Could not connect: " . mysql_error());
mysql_select_db('123') or die ('can not select dtb');
$query='select * from ben';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
echo"<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
endif;
?>
</body>
</html>
我想改成 当" 按键被按下时"运行我添加的代码,这个在具体格式上怎么写呀?
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>交互测试页</title>
</head>
<body>
<form method="POST" action=<?php $_SERVER['PHP_SELF'] ?> >
<p><input type="submit" value="show" name="B3" onClick="javascript:document.getElementById('button_name').value = "show" >
<input type="hidden" name="button_name" id="button_name" /></p>
</form>
<?php
if (count($_POST) > 0)
{
$link = mysql_pconnect("localhost", "root", "3033")
or die("Could not connect: " . mysql_error());
mysql_select_db('123') or die ('can not select dtb');
$query='select * from ben';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
echo"<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
?></body>
</html>
麻烦你再看下,我改了点东西.早上比较闲,现在有点忙,加我MSN吧.CSDN名@hotmail.com
下班了...明天再研究了。..