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Thread: Cant retrieve data from db

  1. #1
    Join Date
    Apr 2009
    Posts
    13

    Cant retrieve data from db

    Hi, I trying to use AJAX to load user's profile through a drop-down list on selected user.

    There are 3 seperate files:
    1) view_info.php
    2) selectuser.js
    3) getuser.php

    I tested and found that the value $q did pass from view_info -> selectuser.js -> getuser.php, and value changes correctly as i select different user from the drop-down list.

    Problem is, it simply can't fetch data from my database using the value $q. I tried debugging it for 3 hours already and is totally out of idea whats wrong with the script. Please help me check what is wrong, i'm new to AJAX, this is my first script..

    view_info.php File:
    Code:
    <?php
    	$con = mysql_connect('yuehyih.hypermartmysql.com', 'username', 'password');
    	if (!$con)
    	 {
    	    die('Could not connect: ' . mysql_error());
    	 } 
    
    	mysql_select_db("yuehyih_ieye", $con); 
    	$sql="SELECT * FROM user";
    	$result = mysql_query($sql); 
    	?>
    	<form> 
    	Select a User:
    	<select name="users" onchange="showUser(this.value)">
    	<?php $num=1;
    	while($row = mysql_fetch_array($result))
    	{  ?>
    	<option value="<? echo $num; ?>"><?php echo $row['username'];?></option>
    	<?php
    	$num=$num+1;
    	} ?> 
    	</select>
    	</form>
    <p>
    <div id="txtHint"><b>User info will be listed here.</b></div>
    </p>
    selectuser.js File
    Code:
    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 || xmlHttp.readyState=="complete")
     { 
     document.getElementById("txtHint").innerHTML=xmlHttp.responseText;
     } 
    }
    
    function GetXmlHttpObject()
    {
    var xmlHttp=null;
    try
     {
     // Firefox, Opera 8.0+, Safari
     xmlHttp=new XMLHttpRequest();
     }
    catch (e)
     {
     //Internet Explorer
     try
      {
      xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
      }
     catch (e)
      {
      xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
     }
    return xmlHttp;
    }
    getuser.php
    Code:
    <?php
    $q=$_GET['q'];
    echo "value of q is ".$q;
    $con = mysql_connect('yuehyih.hypermartmysql.com', 'yuehyih', '000099');
    if (!$con)
     {
     die('Could not connect: ' . mysql_error());
     }
    
    mysql_select_db("yuehyih_ieye", $con);
    
    /*** i have tried with sql="SELECT * FROM user WHERE userID=' ".$q." ' ";  but still didn't work ***/
    $sql="SELECT * FROM user WHERE userID='$q'";
    
    $result = mysql_query($sql);
    
    while($row = mysql_fetch_array($result))
    {
    echo "<table border='0' width=380px> <tr><td width=50px><b>Name</b></td><td>";
    echo ":". $row['Name'] . "</td></tr>";
    echo "<tr><td><b>Email</b></td><td>";
    echo ":". $row['Email'];
    echo "</td></tr>";
    echo "<tr><td><b>Contact</b></td><td>";
    echo ":". $row['Contact'];
    echo "</td></tr>";
    echo "<tr><td><b>Contry</b></td><td>";
    echo ":". $row['Contry'];
    echo "</td></tr>";
    echo "<tr><td><b>State</b></td><td>";
    echo ":". $row['State'];
    echo "</td></tr>";
    echo "<tr><td><b>Address</b></td><td>";
    echo ":". $row['Address'];
    echo "</td></tr>";
    echo "<tr><td><b>Zipcode</b></td><td>";
    echo ":". $row['Zipcode'];
    echo "</td></tr>";
    echo "</table>";
    }
    mysql_close($con);
    ?>
    Thanks in advance!

  2. #2
    Join Date
    Apr 2009
    Posts
    13
    bump

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