I know its sounds like repeated but really I didn't find an answer for the problem am having.
am making a form to get specific details which is password this is the form
<form method="post" name="lostpass" action="forgotpass.php">
<ul>
<li>Admin Name:<input name="admin" type="text"></li>
<li> E.mail: <input name="email" type="text"><br></li>
</ul>
<input type="submit" name="get" value="get infos">
<input type="reset" name="reset" value="Reset">
</form>
so from it this data I get I can get the lost password
and this is the php codes
<?php
$con=mysqli_connect("rock","mido","1234","fyp");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT password FROM admin WHERE email = 'email', Admin = 'admin' ");
while($row = mysqli_fetch_array($result))
echo $row['password'];
mysqli_close($con);
?>
its giving me this error (mysql_fetch_array() expects parameter 1 to be mysqli_result).
thanx for your help
Replace this code :
$result = mysqli_query($con,"SELECT password FROM admin WHERE email = 'email', Admin = 'admin' ");
with
$result = mysqli_query($con,"SELECT password FROM admin WHERE email = '$_POST[email]' AND Admin = '$_POST[admin]'");
Thanks.
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