PHP - Image Select Query Distorts Page Display

Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance.

BTW, is using php only doable? Is there a simpler or more elegant way to do this?

Thanks all!

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.

Hello all. I'm using this code to go through the database and output certain user-defined numbers

Code: [Select]
<?php
mysql_connect ("pdb1.awardspace.com", "anastasov_db","moscow1945")  or die (mysql_error());
mysql_select_db ("anastasov_db");
 
$term = $_POST['term'];
 
$sql = mysql_query("select * FROM countries WHERE cocode = '$term'");
 
while ($row = mysql_fetch_array($sql)){
    echo '<br/> Code: '.$row['cocode'];
    echo '<br/> Country: '.$row['coname'];
    echo '<br/><br/>';
    }
 
?>
Now how would I go about making a page that appears if the script can't find any results in the database?

Thank you in advance

Good day and Merry Christmas to all,

I just spent a good time of my christmas eve trying to figure out this problem. I hope one of you santas would be so kind as to help me with it.

First off I have two tables; employee and employee_works both connected via employee_id key.

Basically I have a parent window we'll call parent.php.
inside the parent page is a search button that once clicked will open a child window we'll call child.php

inside the child page is a list, lets say employees with name, employee_id, etc.

My main concern is this:
How do I populate parent.php based off the employee selection I made in the child window.

Example:
-Access parent.php
-Click on search
-Click on [ID: 004] [NAME: JOHN SMITH] [PHONE: 1233456] [DATE HIRED: JULY 16, 1992] <---format of a row in child.php
-child.php automatically closes and parent.php now shows all data from employee_works with the employee_id = 004

Is this even possible? I know this is vary vague and would be willing to explain more if needed. My website is built mostly on javascript and php.

Hello all,   Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).   The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).   Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:   In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).   I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.   Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.   So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:   1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red. 2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red. 3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.   Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.   Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.   Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.   Cheers guys!

case "update-delete":
	if(isset($_POST['highlight-purchased'])) 
		{
			// ****** Setup customized query to obtain only items that are checked ******
			$sql = "SELECT * FROM shoplist WHERE";
					for($i=0; $i < count($_POST['checkboxes']); $i++)
						{
							$sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or";
						}
					$sql= rtrim($sql, "or");
											
			$statement = $conn->prepare($sql);
			$statement->execute();
																						
			// *** fetch results for all checked items (1st query) *** //
			$result = $statement->fetchAll();

			$statement->closeCursor();
											
			// Setup query that will change the purchased flag to "N", if it's currently set to "Y"
			$sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'";
											
			// Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL
			$sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL";

			$statementSetToN = $conn->prepare($sqlSetToN);
			$statementSetToY = $conn->prepare($sqlSetToY);
											
			/*** loop through the results/collection of checked items ***/
			foreach($result as $row)
				{
					if ($row["purchased"] != "Y")
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToY = $statementSetToY->fetch();
															
							foreach($resultSetToY as $row)
								{															
									$statementSetToY->execute();																
								}
						}
					else
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToN = $statementSetToN->fetch();
															
							foreach($resultSetToN as $row)
								{															
									$statementSetToN->execute();																
								}
						}
				}
				break;
		}

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