twilitegxa Posted January 29, 2010 Share Posted January 29, 2010 I have the following page: getuser.php <?php $q=$_GET["q"]; $con = mysql_connect('localhost', 'root', ''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ajax_demo", $con); $sql="SELECT * FROM user WHERE id = '".$q."'"; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Age</th> <th>Hometown</th> <th>Job</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName'] . "</td>"; echo "<td>" . $row['Age'] . "</td>"; echo "<td>" . $row['Hometown'] . "</td>"; echo "<td>" . $row['Job'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> But I am getting this error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\getuser.php on line 25 I can't figure out what I am doing wrong. I got it from a tutorial for Ajax and PHP with MySQL. Here are the other two pages: html page: <html> <head> <script type="text/javascript" src="selectuser.js"></script> </head> <body> <form> Select a User: <select name="users" onchange="showUser(this.value)"> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option> <option value="3">Glenn Quagmire</option> <option value="4">Joseph Swanson</option> </select> </form> <br /> <div id="txtHint"><b>Person info will be listed here.</b></div> </body> </html> selectuser.js: var xmlhttp; function showUser(str) { xmlhttp=GetXmlHttpObject(); if (xmlhttp==null) { alert ("Browser does not support HTTP Request"); return; } var url="getuser.php"; url=url+"?q="+str; url=url+"&sid="+Math.random(); xmlhttp.onreadystatechange=stateChanged; xmlhttp.open("GET",url,true); xmlhttp.send(null); } function stateChanged() { if (xmlhttp.readyState==4) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } function GetXmlHttpObject() { if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari return new XMLHttpRequest(); } if (window.ActiveXObject) { // code for IE6, IE5 return new ActiveXObject("Microsoft.XMLHTTP"); } return null; } I got it from this page: http://www.w3schools.com/PHP/php_ajax_database.asp Can anyone help me with why I am getting this error?Here is the table SQL statement: CREATE TABLE IF NOT EXISTS `ajax_demo` ( `id` int(11) NOT NULL, `FirstName` varchar(150) NOT NULL, `LastName` varchar(150) NOT NULL, `Age` int(11) NOT NULL, `Hometown` varchar(150) NOT NULL, `Job` varchar(150) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; INSERT INTO `ajax_demo` (`id`, `FirstName`, `LastName`, `Age`, `Hometown`, `Job`) VALUES (1, 'Peter', 'Griffin', 41, 'Quahog', 'Brewery'), (2, 'Lois', 'Griffin', 40, 'Newport', 'Piano Teacher'), (3, 'Joseph', 'Swanson', 39, 'Quahog', 'Police Officer'), (4, 'Glenn', 'Quagmire', 41, 'Quahog', 'Pilot'); Link to comment https://forums.phpfreaks.com/topic/190284-php-with-ajax-and-mysql-help/ Share on other sites More sharing options...
twilitegxa Posted January 29, 2010 Author Share Posted January 29, 2010 Sorry, guys. I figured out what was wrong. I was connecting to the wrong database and table! Link to comment https://forums.phpfreaks.com/topic/190284-php-with-ajax-and-mysql-help/#findComment-1003903 Share on other sites More sharing options...
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