dropdown php filter problem

Discussion in 'PHP' started by akmal1981, Dec 14, 2011.

0
  1. #1
    Dear prog wizards,

    Please help as I am noob at this. I got source code from other website in which I made small changes to cater my need. I have created a database called 'a_database' and table called 'property' using XAMPP 1.7.7 . The columns are 'id', 'Type', 'Price' and 'Location'.

    My first dropdown box filters Type
    My second dropdown box filters Location

    For example : Type is House and Factory
    Location is Bangkok and Singapore

    When I select House, I will get the desired table.

    Then, I select Bangkok and expect to filter House which are located in Bangkok. Unfortunately, this did not happen. Can someone please help me on this ?

    My ajax code :

    <html> <head> <script type="text/javascript"> 


function showtype(str){
    var ajaxRequest;  // The variable that makes <strong class="highlight">Ajax</strong> possible!
    if (str=="") {   
        document.getElementById("txtHint").innerHTML="";   
        return;   
    }      
    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            var ajaxDisplay = document.getElementById('txtHint');
            ajaxDisplay.innerHTML = ajaxRequest.responseText;
        }
    }
    ajaxRequest.open("GET","gettype.php?q="+str,true); 
    ajaxRequest.send(null);     
}
function showlocation(str){
    var ajaxRequest;  // The variable that makes <strong class="highlight">Ajax</strong> possible!
    if (str=="") {   
        document.getElementById("txtHint").innerHTML="";   
        return;   
    }      
    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            var ajaxDisplay = document.getElementById('txtHint');
            ajaxDisplay.innerHTML = ajaxRequest.responseText;
        }
    }
    ajaxRequest.open("GET","getlocation.php?q="+str,true); 
    ajaxRequest.send(null);     
}
//-->    
</script> 
</head> 
<body>  
<form> 
<select name="type" onchange="showtype(this.value)"> 
<option value="">Select property type:</option> 
<option value="1">All</option> 
<option value="2">factory</option> 
<option value="3">house</option> 
</select><br />
<select name="Location" onchange="showlocation(this.value)"> 
<option value="">Select property location:</option> 
<option value="1">All</option> 
<option value="2">bangkok</option> 
<option value="3">Patpong</option> 
</select> </form> <br /> 
<div id="txtHint"><b>Person info will be listed here.</b></div>  </body> </html>
    Code (markup):
    My getlocation.php

    <?php     $q=$_GET["q"]; 
    // check for your q values, 1 = all, 2 = bangkok, 3 = Patpong
        $con = mysql_connect('localhost', 'root', ''); 
    if (!$con) { 
        die('Could not connect: ' . mysql_error()); 
    } 
    mysql_select_db("a_database", $con); 
    // based on the value passed in we may have to change query, if all run this...
    if ($q == 1) {
        $sql="SELECT * FROM property"; 
    } else if($q == 2){
        $sql="SELECT * FROM property WHERE Location = 'bangkok'"; 
    } else if($q == 3){
        $sql="SELECT * FROM property WHERE Location = 'Patpong'";         
    }
    $result = mysql_query($sql); 
    echo "<table border='1'> <tr> <th>Type</th> <th>Price</th> <th>Location</th> </tr>"; 
    while($row = mysql_fetch_array($result)) { 
        echo "<tr>"; 
        echo "<td>" . $row['Type'] . "</td>"; 
        echo "<td>" . $row['Price'] . "</td>"; 
        echo "<td>" . $row['Location'] . "</td>"; 
        echo "</tr>"; } 
        echo "</table>"; 
        mysql_close($con); 
    ?>
    Code (markup):
    My gettype.php

    <?php     $q=$_GET["q"]; 
    $con = mysql_connect('localhost', 'root', ''); 
    if (!$con) { 
        die('Could not connect: ' . mysql_error()); 
    } 


    mysql_select_db("a_database", $con); 


    


        if ($q == 1) {
        $sql="SELECT * FROM property"; 
    } else if($q == 2){
        $sql="SELECT * FROM property WHERE Type = 'factory'"; 
    } else if($q == 3){
        $sql="SELECT * FROM property WHERE Type = 'house'";         
    }






    $result = mysql_query($sql); 
    echo "<table border='1'> <tr>  <th>Type</th> <th>Price</th> <th>Location</th>  </tr>";
    while($row = mysql_fetch_array($result)) { 
        echo "<tr>"; 
        echo "<td>" . $row['Type'] . "</td>"; 
        echo "<td>" . $row['Price'] . "</td>"; 
        echo "<td>" . $row['Location'] . "</td>"; 
        echo "</tr>"; } 
        echo "</table>"; 
        mysql_close($con); 
    ?>
    Code (markup):
     
    akmal1981, Dec 14, 2011 IP