Databases
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    <pre><code>$sql="SELECT * FROM customer WHERE name='" . $myusername . "' and cid='" . $mypassword . "'"; // fix this line if ($count === 1){ // and fix this line if (result === 1){ // and this line </code></pre> <p>It looks to me that you're using code from different places. You should be validating that the username / password combination exists, not checking to see if it is available. </p> <p>Nobody is going to do your homework for you, sorry. Here's a good reason why..</p> <pre><code>&lt;?php session_start(); require_once ('db_connect.php'); if(isset($_POST['pswrd']) &amp;&amp; isset($_POST['userid'])) { $sql = "SELECT * FROM users WHERE username='".mysqli_real_escape($dbc, trim($_POST['userid'])."' AND password= '".mysqli_real_escape($dbc, trim($_POST['pswrd'])."'"); $result = mysql_query($dbc, $sql) or die("Could Not make request"); if(mysql_num_rows($result) == 1) { //redirect to the welcome page } else { echo "User Not Found"; } } ?&gt; </code></pre> <p>You're mixing <code>mysql</code> and <code>mysqli</code> here. You should be using JUST <code>mysqli</code>! Your <code>userid</code> and <code>pswrd</code> variables get assigned to <code>username</code> and <code>pass</code> in the AJAX call but the <code>check_login.php</code> script never sees them. It's still looking for <code>userid</code> and <code>pswrd</code>. You've got too many errors and you don't seem to be very responsive about getting them fixed. Besides, we're not a <strong>homework</strong> site.</p>
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    1. POChecking username and password with php ajax
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    1. USDevlshOne
    1. USDevlshOne
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    1. COYes , because I am new to php.
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      1. USBegumm
    2. COThank you , I have changed but again it does not work!
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Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

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