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  1. #1
    SitePoint Member
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    Nov 2011
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    Retrieve data from a php page and display in div with ajax

    I am having a problem displaying data from a php page and outputting on a page in a div tag. can anyone help me. this is what i have done so far

    HTML
    Code:
    include ('scripts/connect_to_db.php');
    	             //Search the Database to check to see if the product in the database column matches the select product and order it by descending order
    				$lftresult = mysqli_query($conn,"SELECT distinct genre FROM video WHERE mediatype='".$cat."' GROUP BY genre");	
    					while($leftrow = mysqli_fetch_array($lftresult))
    					{ 
    						$genre = $leftrow["genre"];
    						?>
    						
    						<div class='left-choicediv'><!--specialoffer-divs-cols-->
    							<input type="button" class="cat" id="catlink" value="<?php echo $genre; ?>" onclick="sendcat(this.value);" />
    							<input type="hidden" name="cattype" id="cattype" value="<?php echo $cat; ?>" />
    						</div><!-- End of left-choicediv-->
    					<?php
    					}
    				?>
    			</div><!-- End of Left Column-->
    			<div class="right-column"><!-- Start of Right Column-->
    		
    			<div id="rightcol"></div>
    			</div><!-- End of Right Column-->
    JAVASCRIPT
    Code:
    function sendcat(elms)
    {
    	if (window.XMLHttpRequest) 
    	{
    		//Firefox, Opera, IE7, and other browsers will use the native object
    		var k = new XMLHttpRequest();
    	} 
    	else 
    	{
    		//IE 5 and 6 will use the ActiveX control
    		var k = new ActiveXObject("Microsoft.XMLHTTP");
    	}
    	
    	var cattype = document.getElementById(cattype).value;
    	k.onreadystatechange = function() 
    	{
    		if (k.readyState == 4 && k.status == 200) 
    		{
    			document.getElementById("rightcol").innerHTML = k.responseText;
    		}
    	}
    	
    	k.open("GET","sendcat.php?catchoice="+elms+"&cattype="+cattype,true);
        // Send the data to PHP now... and wait for response to update the status div
        k.send(null); // Actually execute the request
       // document.getElementsByClassName("right-column").innerHTML = "images/loading/ajax-loader.gif";
    }
    PHP
    Code:
    <?php 
    		include ('scripts/connect_to_db.php');
    		$cat = $_GET['catchoice'];
    		$cattype = $_GET['cattype'];
    		
    		//If A user puts less than 5 exit
    		if (($cat == "all") || ($cat == "")) 
    		{
    			$catchoices = mysqli_query($conn,"SELECT videoID,title,upload_path,thumb FROM video ORDER BY title") 
    			or die(mysqli_error($conn)); 
    		}
    		//If A user puts more than 5 exit
    		else 
    		{
    			$catchoices = mysqli_query($conn,"SELECT videoID,title,upload_path,thumb FROM video WHERE mediatype='".$cattype."' AND genre='".$cat."' ORDER BY title") 
    			or die(mysqli_error($conn)); 
    		}	
    		
    			echo $catchoices;
    			
    			$num_rows = mysqli_num_rows($catchoices);
    			
    			echo "<h2>".$cat."</h2>";	
    			
    			if ($num_rows == 0) 
    			{
    				echo "<p id='notfound'>There are no  available</p>";
    			}
    			else 
    			{
    				while($pdrows = mysqli_fetch_array($catchoices))
    				{ 
    					$vidid = $pdrows["videoID"];
    					$prdtitle = $pdrows["title"];
    					$upload_path = $pdrows["upload_path"]; 
    					$thumb = $pdrows["thumb"];
    
    					echo "<div class='product-divs'><!--product-divs-->";
    					echo  "<h3>".$prdtitle."</h3>";
    					echo  "<a href='product_details.php?vid=".$vidid."'><img src=".$upload_path.$thumb." alt='".$thumb."' border='0' /></a>";
    					echo  "</div><!-- End of product-divs-->";
    				}
    			}	
    ?>
    I Don't know where I'm Making a Mistake

  2. #2
    Keeper of the SFL StarLion's Avatar
    Join Date
    Feb 2006
    Location
    Atlanta, GA, USA
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    So what does it do that its not supposed to do, or not do that it is?
    Never grow up. The instant you do, you lose all ability to imagine great things, for fear of reality crashing in.


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