这是我的表单页面。按下登录时,会将其重定向到checkLogin页面,但它是空的。没有回声是有效的。
<html>
<title>User Login Form</title>
<head></head>
<body>
<form name="form1" method="post" action="checkLogin.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1"
bgcolor="#FFFFFF">
<tr>
<td width="80">Username</td>
<td width="6">:</td>
<td width="300"><input name="myusername" type="text" id="myusername"></td>
</tr>
<tr>
<td>Password</td>
<td width="6">:</td>
<td><input name="mypassword" type="text" id="mypassword"></td>
</tr>
<tr>
<td> </td>
<td> </td>
<td><input type="submit" name="Submit" value="Login"></td>
</tr>
</table>
</td>
</form>
</td>
</tr>
</body>
</html>
这是checkLogin页面 这没有显示任何东西。我试过mysql_error但仍然没有显示。请帮忙
<?php
include("config.php");
echo "Check Login";
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$myusername=stripslashes($myusername);
$mypassword=stripslashes($mypassword);
$myusername=mysql_real_escape_string($myusername);
$mypassword=mysql_real_escape_string($mypassword);
$sql="select j.jobid from jobs j,rcuser r where r.userName='$myusername'and r.password='$mypassword'";
echo "Hey".$myusername." ";
$result=mysql_query($sql);
while($col=mysql_fetch_array($result)){
echo "<tr><th>Job Id</th>"</tr>;
echo "<tr><td>" . $col['jobid'] . "</td><tr>";
}
$count=mysql_num_rows($result);
echo $count;
if($count==1){
session_register("myusername");
session_register("mypassword");
session_start();
if(!session_is_registered($myusername){
echo "Session expired";
}
echo "is it???";
}
else{
echo "Invalid Username or Password";
}
?>
答案 0 :(得分:1)
这是问题所在,请注意:Id</th>"</tr>;
您没有构建HTML并以正确的方式回显。检查下面,你误认的是。
echo "<tr><th>Job Id</th>"</tr>;
echo "<tr><td>" . $col['jobid'] . "</td><tr>";
答案 1 :(得分:0)
请先检查一下这个脚本,然后告诉我结果,因为你的编码方式不够明确。
`
$myusername=stripslashes(mysql_real_escape_string($_POST['myusername']));
$mypassword=stripslashes(mysql_real_escape_string($_POST['mypassword']));
$sql=" SELECT j.jobid
FROM jobs j,rcuser r
WHERE r.userName='$myusername' AND r.password='$mypassword' ";
$result=mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($result) === 1) {
echo "Username and Password is found";
}else {
echo "Error! No results found.";
}
&GT;`