PHP - Issues Displaying One Image From 2 Tables

Below is a page which is supposed to output the name, blog contribution and picture of contributing members of a website.
<div id="blog_content" class="" style="height:90%; width:97%; border:5px solid #c0c0c0; background-color: #FFFFFF;"> <!--opens blog content--> 



                             
<?php


//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);



//include the config file

require_once("config.php");



//Define the query. Select all rows from firstname column in members table, title column in blogs table,and entry column in blogs table, sorting in ascneding order by the title entry, knowing that the id column in mebers table is the same as the id column in blogs table.


$sql = "SELECT


    blogs.title,blogs.entry,members.firstname,images.image
FROM
    blogs
LEFT JOIN
    members
ON
    blogs.member_id = members.member_id
LEFT JOIN
    images
ON
    blogs.member_id = images.member_id

ORDER BY
     blogs.title ASC

"; 

$query = mysql_query($sql);

                 if($query !== false && mysql_num_rows($query) > 0)

                 {

         while(($row = mysql_fetch_assoc($query)) !== false)

         {

 echo

                
                              '<div id="blog_content1" style="float:left; position:relative;bottom:18px;left:13px; background-color: #FFFFFF; height:16.7%; width:100%; border:0px none none;"                              
                              <!--opens blog_content1  same as main center top 1 and 2 from index page everything scaled down by a factor of 3, heightwise--> 
        
 
                                   <div class="red_bar" style="height:3%; width:100%; border:1px solid #959595;"> <!--a--> 
 
                                       <div class="shade1" style="height:5px; width:100%; border:0px none none;"> </div>                                      
                                       <div class="shade2" style="height:5px; width:100%; border:0px none none"> </div> 
                                       <div class="shade3" style="height:5px%; width:100%; border:0px none none"> </div>

 
                                   </div> <!-- closes red bar--> 
 
 
 
                                   <div class="content" style="height:28.3%; width:100%; border:0px none none;"> <!----> 
 
                                       
                                           <div class="slideshow" id="keylin" style="float:left; width:20%;  border:0px none none;"> <!--a-->
 
                                                  
                                                  
 
                                                 <div><img header("Content-type: image/jpeg"); name="" alt="" id="" height="105" width="105" src="$row[image]" /></div>                              
 
 
                                       
                                           </div> <!-- closes pic--> 
 
 
 
                                           <div class="content_text"  style="float:right;  position:relative;top:7px;left:0px; max-height:150px; width:78.5%; border-width:4.5px; border-bottom-style:solid; border-right-style:solid; border-color:#c0c0c0; "> <!--a-->'; 
 
                                           

                                               
                                                echo "<h3>".$row['title']."</h3>";

echo "<p>" .$row['entry']."<br />".$row['firstname']."</p>";

         
                                             
                                                                                           
                                           echo                                         
                                           '</div> <!-- closes content text--> 
 
  
                                   </div> <!-- closes content--> 
 
                                   
                                                                                                                                                                                                                            
                            </div> <!-- closes blog_content1-->';



                                                 }

                 }

                 else if($query == false)

                 {

               

 echo "<p>Query was not successful because:<strong>".mysql_error()."</strong></p>";

         echo "<p>The query being run was \"".$sql."\"</p>";

                 }

                 else if($query !== false && mysql_num_rows($query) == 0)

                 {

         echo "<p>The query returned 0 results.</p>";

                 }


                                                 mysql_close(); //Close the database connection.





                                                 ?> 

                      
                        </div> <!-- closes blog content-->


The select query is designed to retrieve all the blog contributions(represented by the fields blogs.title and blogs.entry) from the database, alongside the contributing member (member.firstname) and the member's picture(images.image), using the member_id column to join the 3 tables involved, and outputs them on the webpage.   The title, entry and firstname values are successfully displayed on the resulting page.  However, I can't seem to figure out how to get the picture to be displayed.  Note that the picture was successfully stored in the database and I was able to view it on a separate page using a simple select query.  It is now just a question of how to get it to display on this particularly crowded page.  Anyone knows how I can output the picture in the img tag?  I tried placing the header("Content-type: image/jpeg"); statement at the top of the php segment, then just right below the select query and finally just right above the img tag, but in every case, I just got a big white blank page starring at me.  How and where should I place the header statement? And what else am I to do to get this picture displayed?  Any help is appreciated.

So I have 3 tables, student, student_course and course. I'm only using student and course atm, probably should look into student_course haha well those are the tables provided to us for the project I'm working on.

I'm trying to list all the courses from the course table and then when clicking on a course I want to display all the students who registered for this course( which would probably be the student table)

Here is my code for displaying the courses:
Code: [Select]
<?php
$display_sql ="SELECT cname FROM course ORDER BY cid DESC";
$display_query = mysql_query($display_sql) or die(mysql_error());
$rsDisplay = mysql_fetch_assoc($display_query);
?>


<html>

<head>

</head>


<body>
<p> Click course to display all students registered for that course</p>
<?php do { ?>
<p><a href="test.php?cname=<?php echo $rsDisplay['cid']; ?>">


Course:
<?php echo $rsDisplay['cname']; ?>

</a>
</p>


<?php } while ($rsDisplay = mysql_fetch_assoc($display_query)) ?>
And here is my code for displaying the students from the student table registered for the specific course you click on:
Code: [Select]
<?php
$q = "SELECT * FROM student WHERE cname = $_GET[cname]";
$confirm_query = mysql_query($q);
$rsconfirm = mysql_fetch_assoc($confirm_query);
?>
<html>
<head>
</head>
<body>
<p> Displaying all student </p>
<p>
First Name
<?php echo $rsconfirm['sname']; ?>
</p>
<p>
Surname:
<?php echo $rsconfirm['fname']; ?>
</p>
I'm very new to php. Im using "test.php?cname=" to transfer the cname (course name) info from my course table to the next page where I use $_GET[cname]";. cname isn't a primary key in my table, rather just a row. Can it be done like that? Or should you just use primary keys?

I managed to display the courses but when clicking it, it wont display the registered users and gives me an error:
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\Program Files\EasyPHP-5.3.3\www\Project\test.php on line 28

What does that mean?

Line 28 is this part :
Code: [Select]
$q = "SELECT * FROM student WHERE cname = $_GET[cname]"; //26
$confirm_query = mysql_query($q);                                                       //27
$rsconfirm = mysql_fetch_assoc($confirm_query);                     //28

Any criticism welcome. When coding I'm still trying to figure out what it is I'm doing nevermind getting it to work haha. Thanks in advance

I want to display a bunch of records from table 1.  Table 1 has a field called "user_id".
In table 2 i have id and name.

Hoiw do I query table 1 and then display the name of the user?

I have
$query = "SELECT * from table1 where status = '$status' ORDER by ????";
$results = mysql_query($query);
while $row = $mysql_fetch_array($results){
echo "status is "$row['status']." and work type is ".$row['work_type']. "for the user" $row['name']."<br>"";
}

Obviosuly thius wouldnt work but basically I want to pull everything fromt able 1, but sort by the name which is located in table 2.  The only thing I need from table two is the name associated with user_id.
Oh I ALSO want to keep all records from table1 with the same user_id on 1 line (users in table2 are unique)!

Hey all,

First off, if this is in the wrong section I apologize. I wasn't sure if it should be here or the mySQL section.

What's going on is, I'm in the process of learning the Prepared Statement way of doing things and am changing / updating my code to reflect the changes. Everything was going fine until I attempted to do what I could do using old MySQL methods and that is display the queried results on the same page.


I can place a query and display the results as they should be displayed if I only use one block of code. However, if I try to do any additional queries on the same page, they get killed and do not display anything even though I know the query is fine because I can test the exact same syntax below one a different page and it works.

Here's a code snippet for an example:
Code: [Select]

Code:
<table>
<tr>
<td>
// The below code will display a selection box containing various strings such as "hello world", "great to be here", "Wowserz", "this is mind blowing" etc. that are stored in the database.

<?php

   echo "<select = \"SpecialConditions\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
   
       echo "<option value=\"$specialId\">$specialcondition</option>";
   }
}
   echo "</select>";
?>
</td>
<td>
// If I place another fetch query below the above fetch() query, this one will not show up. This one is supposed to display values 1 - 20 that have been stored in the DB.

<?php
echo "<select = \"NumberSets\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
      
          echo "<option value=\"".$numbers."\">".$numbers."</option>";
      }
   }
echo "</select>";

?>

</td>
</tr>
</table>


What am I doing wrong with this? When I use regular SQL queries I can display multiple results on the same page.

The results are being pulled from two separate joined tables but I don't think that's the issue.

Here's the code, this is the PHP in my html contacts page (that IS in fact saved with a PHP extension): 
  <div class = "centercontainer">