jdgower Posted March 12, 2010 Share Posted March 12, 2010 Please help First of all, my actual question is much more complicated but if i can find the answer to the below it will probably answer my deeper question... http://www.w3schools.com/php/php_ajax_database.asp At the above URL, it connects to a database to grab the information when you select a person. Let's say that I wanted to modify this so that it actually has a master list of people in the drop down, but not all the people are in one database, they are in 2 different databases. Theres a lot of script here but basically all i did was add 2 lines on the index page (script and div), then copy the selectuser.js and getuser.php and put 2 after everything that [i THINK] is relevant. I would think that i could just change the script at the top of index to: <script type="text/javascript" src="selectuser.js"></script> <script type="text/javascript" src="selectuser2.js"></script> at the div on the index, change it to: <div id="txtHint"><b>Person info will be listed here.</b></div> <div id="txtHint2"><b>Person info will be listed here.</b></div> the script in selectuser2.js (the one i make): var xmlhttp; function showUser(str) { xmlhttp=GetXmlHttpObject(); if (xmlhttp==null) { alert ("Browser does not support HTTP Request"); return; } var url="getuser2.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("txtHint2").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; } the script for getuser2.php (the one i make): <?php $q=$_GET["q"]; $con = mysql_connect('localhost', 'peter', 'abc123'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ajax_demo2", $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); ?> The issue I'm having is that it only shows information from the second database, not both. There is probably something wrong with the JS, but i dont know what it is. Thanks in advance for your help! -Jeremy Quote Link to comment Share on other sites More sharing options...
jdgower Posted March 12, 2010 Author Share Posted March 12, 2010 nevermind, figured it out, thanks Quote Link to comment Share on other sites More sharing options...
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