display image from mysql database

mase

i have the following code and would like to display all the image in the table... the follwoing code will only display one image. can someone plase help

thanks.
<?php
// Connect to the database server
$dbcnx = @mysql_connect(xxxxxx, 'xxxxxx', 'xxxxxx');
if (!$dbcnx) {
exit('Unable to connect to the ' .
'database server at this time.');
}

// Select the jokes database
if (!@mysql_select_db('dbnation')) {
exit('Unable to locate the joke ' .
'database at this time.');
}

// Request the text of all the image
$result = @('SELECT * FROM ae_gallery');
if (!$result) {
exit('Error performing query: ' . mysql_error() . '');
}

// Display the text of each image in a paragraph
$row =mysql_fetch_array($result);
$content=$row["id"];
$name=$row['data'];

header('content-type: image/jpeg');
print $content

mase

Please I have one more request...

Trying to create a web page where users can upload their image and organize the image by date and by user..... How is this possible...?

Thanks.

slick101

The reason is because your only grabbing 1 row from the table.
You need to replace this code:

$row =mysql_fetch_array($result);
$content=$row["id"];
$name=$row['data'];

$images = array();
while($r = mysql_fetch_row($result)
array_push($images, $r);

now if you do
print_r($images);
you will notice all the stored information from your table.

Are you storing the the data as binary? or as a file name?

slick101

mase;10882415 wrote:
Please I have one more request...

Trying to create a web page where users can upload their image and organize the image by date and by user..... How is this possible...?

Thanks.

How is your table setup? You can setup a date/user field for the table so taht you can sort images by date and user, or display images by only this user. etc.

mase

thanksk slick101 for your help... i replace the code with your's but i get the following error...
Parse error: syntax error, unexpected T_STRING in /viewme.php on line 24

as for the sorting, i will do it by user name and date...
i will try that later...

slick101

mase;10882453 wrote:
thanksk slick101 for your help... i replace the code with your's but i get the following error...
Parse error: syntax error, unexpected T_STRING in /viewme.php on line 24

as for the sorting, i will do it by user name and date...
i will try that later...

i noticed i forgot a ) at the end of the first line i showed you.

while($r = mysql_fetch_assoc($result))

mase

i don't get any errors, the image have a little box whti an x in side... will not display it...

slick101

mase;10882490 wrote:
i don't get any errors, the image have a little box whti an x in side... will not display it...

Well youll need to change more of your code around in order for it to work.

you wont need
// Display the text of each image in a paragraph
$row =mysql_fetch_array($result);
$content=$row["id"];
$name=$row['data'];

basically $images has the array now.
so you can do a foreach loop if you wanted

header('Content Type: image/jpeg'); # NOT TOTALLY SURE IF THIS WILL WORK FOR ABSOLUTELY EVERY IMAGE i havent tested doing this by binary data before. normally i just store the file in a directory for albums and do echo '<img src="{$im['id']}" />';
foreach($images as $im)
   echo $im['id'];

mase

i remove that this is the code now...thanks for you help

<?php
// Connect to the database server
$dbcnx = @mysql_connect(xxxxxx, 'xxxxxx', 'xxxxxx');
if (!$dbcnx) {
exit('Unable to connect to the ' .
'database server at this time.');
}

// Select the image database
if (!@mysql_select_db('dbnation')) {
exit('Unable to locate the image' .
'database at this time.');
}

// Request the text of all the image
$result = @('SELECT * FROM ae_gallery');
if (!$result) {
exit('Error performing query: ' . mysql_error() . '');
}

// Display the text of each image in a paragraph
$images = array();
while($r = mysql_fetch_assoc($result))
array_push($images, $r);

header('Content-type: image/jpeg');
print_r($images, $r);
?>

slick101

mase;10882497 wrote:
i remove that this is the code now...thanks for you help

<?php
// Connect to the database server
$dbcnx = @mysql_connect(xxxxxx, 'xxxxxx', 'xxxxxx');
if (!$dbcnx) {
exit('Unable to connect to the ' .
'database server at this time.');
}

// Select the image database
if (!@mysql_select_db('dbnation')) {
exit('Unable to locate the image' .
'database at this time.');
}

// Request the text of all the image
$result = @('SELECT * FROM ae_gallery');
if (!$result) {
exit('Error performing query: ' . mysql_error() . '');
}

// Display the text of each image in a paragraph
$images = array();
while($r = mysql_fetch_assoc($result))
array_push($images, $r);

header('Content-type: image/jpeg');
print_r($images, $r);
?>

change:
header('Content-type: image/jpeg');
print_r($images, $r);

to what i posted before. try that, because right there you are just printing out the array print_r($images) and $r will be the last row within the array.

mase

ok i am lost the code is as follwo..

<?php
// Connect to the database server
$dbcnx = @mysql_connect(asdfasdfasdfasdfasdf);
if (!$dbcnx) {
exit('Unable to connect to the ' .
'database server at this time.');
}

// Select the image database
if (!@mysql_select_db('dbnation')) {
exit('Unable to locate the image' .
'database at this time.');
}

// Request the text of all the image
$result = @('SELECT * FROM ae_gallery');
if (!$result) {
exit('Error performing query: ' . mysql_error() . '');
}

// Display the text of each image in a paragraph
$images = array();
while($r = mysql_fetch_assoc($result))
array_push($images, $r);

header('Content Type: image/jpeg'); # NOT TOTALLY SURE IF THIS WILL WORK FOR ABSOLUTELY EVERY IMAGE i havent tested doing this by binary data before. normally i just store the file in a directory for albums and do echo '<img src="{$im['id']}" />';
foreach($images as $im)
echo $im['data'];
?>

when i echo im['data']; i get one image any thing else, it will not display no image.

mase

can someone help...

cant get more that one image to dispaly...i want to display only the first 10 new images that was uploaded...
pluse the comments that was placed.
thanks.