BetterPHP

friends
i,m making a page that retieve data from database and show them and these data include a picture for each one of them i could upload this pic with the data by a page in which i can insert data and this pic into database but when it is time to retrieve these data to be shown into this page i could show these data but each pic related to it,s specific topic was not retrieved to be shown beside it i can see a red ( X ) instead of the speific pic
so i need you to guide me to a tutorial by which i can follow to show mulitple pics retrieved from database
a snap shot of the page is included to see how do each pic look like
Thank you

could you post the code?

first i made a page called getimage.php and here is the code of it

<?php
 include ("config.php");
 $id=addslashes(@$_REQUEST['id']);
	$image= mysql_query ("SELECT photo FROM tourism_news Where id=$id");
	$row1 = mysql_fetch_assoc($image);
	$myimage= $row1['photo'];
	
	header("content-type:image/jpeg");
	//$type ='content-type:'.$row1['imagetype']; 
	  $type ="content-type:".$row1['imagetype'];  
   echo "<img src='$myimage' width='100' height='200'>";
    //header($type);
	//echo  $row1['photo'];
	echo "$myimage";
	//echo <img src=\"$myimage\">;
?>

second the image where i want to show the image is index.php
here is the code included into this page

		$get=mysql_query ("SELECT * FROM tourism_news  order by id desc  LIMIT $start , $per_page");
		
		 echo "<table border='1' width='100%'  bordercolor='red'> ";	 
			while ($row = mysql_fetch_assoc($get))
			{
			 //$lastid=mysql_insert_id();
			
			echo "<table border='1' width='100%'  bordercolor='red'> ";
			 echo"   <tr align='right'>";
			 echo"
			 <td  align='right'   height='22' width='50%'  ></td> 
			<td   align='right'   height='22' width='60%' background='titlebar.jpg' ><font color='white'><b> "   .$row['topic'].   "  </b></font></td> 
				   ";
				  echo"    </tr>";
			  echo "<table border='1' width='100%'  bordercolor='green'> ";
			  echo"<tr><td align='right' width='70%' ><font color='black'>  ".$row['summary']." <a href='show_this_news.php?id=".$row['id']."'  title='".$row['topic']."'>التفاصيل</a>";
			  			  
echo"</font></td><td align='right' rowspan='2' width='30%' >"; 
			  
			  echo"<img src='getimage.php?id=".$row['id']."' width=200 height=140 />"; 
			 // echo <img src=\"$myimage\">;
			 // echo"</td>";
               echo"    </tr>";
			   echo"<tr><td align='left' width='70%' ><font color='black'>".$row['date']." مضاف بتاريخ</font></td>";
               echo"   </tr>";
			   }
			?>
			</tr>
			</table>

        header("content-type:image/jpeg");
        //$type ='content-type:'.$row1['imagetype'];
          $type ="content-type:".$row1['imagetype'];  
   echo "<img src='$myimage' width='100' height='200'>";
    //header($type);
        //echo  $row1['photo'];
        echo "$myimage";
        //echo <img src=\"$myimage\">;

If $myimage contains the actual data of the image you don't want to output it in an img tag like that, so this section should be

        header("Content-Type: image/jpeg");
        echo $myimage;

However, it would be worth considering storing the image as a file rather than in the database.