Can´t read out pictures out of database, because using foreign key

Question

My Database has a table (a) where i have a id as PK and a table for images (output_images), which contents id from (a) as foreign key.

output_images table:

image_id tinyint(3) NOT NULL AUTO_INCREMENT,

image_type varchar(25) NOT NULL,

imageData   mediumblob BINARY

a_ID int(11)

image.php to get every image out of database:

<?php
mysql_connect("... .com","user","pw");
mysql_select_db("database");
if(isset($_GET['image_id'])) {
$sql = "SELECT imageType,imageData FROM output_images, a WHERE 
output_images.a_ID=a.a_ID" . $_GET['image_id'];
$result = mysql_query("$sql") or die("<b>Error:</b> Problem on Retrieving Image BLOB<br/>" . mysql_error());
$row = mysql_fetch_array($result);
header("Content-type: " . $row["imageType"]);
echo $row["imageData"];
}
mysql_close();
?>

listimages.php calls image.php to build a list of images:

        <?php
        mysql_connect("... .com","user","pw");
        mysql_select_db("database");
        $sql = "SELECT imageId FROM output_images,a Where output_images.a_ID=a.a_ID ORDER BY imageId DESC"; 
        $result = mysql_query($sql);
        ?>
        <HTML>
        <HEAD>
        <TITLE>List BLOB Images</TITLE>
        <link href="imageStyles.css" rel="stylesheet" type="text/css" />
        </HEAD>
        <BODY>
        <?php
        while($row = mysql_fetch_array($result) or die (mysql_error()) 
        {
        ?>
        <img src="image.php?image_id=<?php echo $row["imageId"]; ?>" /><br/>
        <?php       
        }
        mysql_close();
        ?>
        </BODY>
        </HTML>

actual error message:

Parse error: syntax error, unexpected '{' in home/a9996478/public_html/listImages.php on line 15

if i change the fifth line of listImages.php to:

$sql = "SELECT imageId FROM output_images ORDER BY imageId DESC";

i get the same problems, even i also get before for this line 12 broken pics

i don´t know whats the problem....

score 1 · Answer 1 · answered Jun 10 '15 at 11:25

1

missing ) in while loop

 while($row = mysql_fetch_array($result) or die (mysql_error()) )// missing

answered Jun 10 '15 at 11:25

You´re right , but pictures are still broken....you know why? – Lama Dingo Jun 10 '15 at 11:29
i think `$sql = "SELECT imageType,imageData FROM output_images, a WHERE output_images.a_ID=a.a_ID" . $_GET['image_id'];` not correct – Jun 10 '15 at 11:30
which line could be false? cause i don´t have any idea – Lama Dingo Jun 15 '15 at 19:43

Can´t read out pictures out of database, because using foreign key

1 Answers1