/表单设计/
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<form name="form1" method="post" action=""><label><center>Register Form</center></label>
<p><center></center><label>Name:</label>
<input type="text" name="name"></center>
</p>
<p><label>Rollno:</label>
<input type="text" name="rno">
</p>
<p><label>Address:</label>
<input type="text" name="add">
</p>
<p>
<input type="submit" name="Submit" value="Submit">
</p>
</form>
/************PHP********************/ /表单提交/
<?php
$con=mysql_connect("localhost","root","");
mysql_select_db("alpha") or die( "Unable select database");
// Check connection
// escape variables for security
$name = mysql_real_escape_string($_POST['name'],$con);
$rno = mysql_real_escape_string( $_POST['rno'],$con);
$add = mysql_real_escape_string($_POST['add'],$con);
$sql=mysql_query("INSERT INTO test (name, rno, add)
VALUES ('$name','$rno','$add')",$con);
$result=mysql_query($sql) or die(mysql_error());
if (!mysql_query($result,$con)) {
die(mysql_error($con));
}
echo "1 record added";
mysql_close($con);
?>
</body>
</html>
请有人帮我解决这个错误。 我收到消息查询为空。 我怎样才能解决这个问题。 我是 php 和 sql 的新手。 我不知道如何解决这个问题。
最佳答案
mysql_query 应该执行一个查询:
<?php
$con=mysql_connect("localhost","root","");
mysql_select_db("alpha") or die( "Unable select database");
// Check connection
// escape variables for security
$name = mysql_real_escape_string($_POST['name'],$con);
$rno = mysql_real_escape_string( $_POST['rno'],$con);
$add = mysql_real_escape_string($_POST['add'],$con);
$query = sprintf("INSERT INTO test (name, rno, add)
VALUES ('%s','%s','%s')", $name, $rno, $add);
$result = mysql_query($query, $con);
if (!$result) {
$message = 'Invalid request : ' . mysql_error() . "\n";
$message .= 'Your query : ' . $query;
die($message);
}
// @see http://php.net/manual/en/function.mysql-affected-rows.php
echo mysql_affected_rows() + " record added";
mysql_close($con);
?>关于php - 提交表单时查询为空，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26483877/