警告：mysql_num_rows（）期望参数1是资源，布线在第18行的C：\ xampp \ htdocs \ LOGIN TEST \ login.php中给出[复制]

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mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

Hi im trying to do a very simple login system using php and mysql, it can be nonsecure

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\LOGIN TEST\login.php on line 18}

I followed the same exact guide from the internet and i still get this error ^

This is my following codes

<?php
$username = "root";
$password = "";
$hostname = "localhost";

$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");

$selected = mysql_select_db("login", $dbhandle);

$myusername = $_POST['user'];
$mypassword = $_POST['pass'];

$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);

$query = "SELECT * FROM admin WHERE Username='$myusername' and Password='$mypassword'";
$result = mysql_query($query);
$count = mysql_num_rows($result);

  if($count==1) {
      echo 'It worked';
  }
  ?>

btw mysql is nonsecure

i also tried to input this code from this link

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xamppserver\htdocs\www\login.php on line 18

and i still got the same error....

this is my index.php

<body>
    <form action="login.php" method="POST">
        <p>Username:</p><input type="text" name="user" />
        <p>Password:</p><input type="password" name="pass" />
        <br />
        <input type="submit" value="Login" />
    </form>


</body>

this is the video which i followed

https://www.youtube.com/watch?v=NuiTzSdGmKM&feature=iv&src_vid=9SdqRKhZwfY&annotation_id=annotation_232683

</div>

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1条回答默认最新

dsfo22654 2015-08-29 05:43

关注

try this code....using mysqli

<?php
$mysqli = new mysqli("localhost","root","","login");
if(mysqli_connect_errno())
{
  trigger_error('Connection failed: '.$mysqli->error);
}
$myusername = $_POST['user'];
$mypassword = $_POST['pass'];

$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);

$query = "SELECT * FROM admin WHERE Username='".$myusername."' and    Password='".$mypassword."' ";
$result = $mysqli->query($query);

if($result->num_rows>0)
   {
     echo 'It worked';
   }
?>