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I am trying to display 4 images using PHP and MySQL database. I have to display the 4 images as rows.

I use the table cars with fields (`id_car`, `car_image1`, `car_image2`, `car_image3`, `car_image4`), with all the images being blob datatype.


$id = $_GET['id'];

$link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");
$sql = "SELECT car_image1, car_image2, car_image3, car_image4 FROM cars WHERE id_car='$id'";
$result = mysql_query("$sql");

mysql_close($link);
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
echo $row['car_image1'];
echo $row['car_image2'];
echo $row['car_image3'];
echo $row['car_image4'];
}


I can only display 1 image and not the other images. Since I am a newbie to this technology I need help.
Posted

i believe your problem is there are not correct data at your database,

İf your uploading code is something like that;

$handle = fopen ($_FILES['avatar']['tmp_name'], "rb");
$image = fread ($handle, filesize($_FILES['avatar']['tmp_name']));
$image = mysql_real_escape_string($image);

and if you forgot to add this code: (which is for take the reading pointer back to top, since fread takes this pointer to the end you need to use it before fread use)
rewind( $handle);
adding this before reuse of fread, may help.
   
PHP
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
$img1 = $row['car_image1'];
$img2 = $row['car_image2'];
$img3 = $row['car_image3'];
$img4 = $row['car_image4'];
}
echo "<img src="/filepath/".$img1."" hold=" />echo "><img src="/filepath/".$img2."" hold=" />echo "><img src="/filepath/".$img3."" hold=" /></pre></xml>"></img></img></img>
   

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