For each loop from mysql

[pgrevents]

I dont understand the problem as it is a loop. I have two sets of image information so thats two images. I have set up a for each loop so that I can display these images as a thumbnail

 

here is the loop statement

 

<?php foreach($gallery as  $image) : ?>                      
<li>

<a class="thumb" href="<?=$imageloc?>" title="Title #0">
<img src="<?=$imagethumb?>" alt="Title #0" /></a>

						<div id="clear"></div>
                                <div class="caption">

							<div class="image-title"><?=$imagename?></div>

							<div class="image-desc">see larger</div>
						    </div>
					</li>

<?php  endforeach ; ?>
?>

 

now it does display but as I said I have only two images to show but it returns 8 loops of the same first image.

 

Here is the code php mysql code that i include

 

<?php
$dbhost = 'localhost';
$dbuser = 'dbuser';
$dbpass = 'dbpass';
$dbname = 'dbname';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);
$query = mysql_query("SELECT * FROM gallery");
$gallery = mysql_fetch_array($query);
$imageloc = $gallery['image_loc'];
$imagename = $gallery['image_name'];
$imagethumb = $gallery['thumnail_loc'];

?>

There is no errors outputting so I am at a loss can any of you help?

[Julian]

Try this:

<?php foreach ($gallery as $image) { ?>                      
<li>

<a class="thumb" href="<?php echo $imageloc; ?>" title="Title #0">
<img src="<?php echo $imagethumb; ?>" alt="Title #0" /></a>

						<div id="clear"></div>
                                <div class="caption">

							<div class="image-title"><?php echo $imagename; ?></div>

							<div class="image-desc">see larger</div>
						    </div>
					</li>

<?php } ?>

 

[pgrevents]

tried it it didnt work same result as before I am seeing 8 copies of 1 image when I am supposed to be seeing 2 different images

 

thanks

[PFMaBiSmAd]

The following line of code -

 

$gallery = mysql_fetch_array($query);

 

Fetches ONE row from the result set. From the php manual -

Returns an array of strings that corresponds to the fetched row, or FALSE if there are no more rows

 

Looping over $gallery is just looping over all the columns in that row. You must use the mysql_fetch_array in a while() loop to fetch all the rows.

[Julian]

What about using while

<?php  while ($gallery = mysql_fetch_assoc($query)) { ?>                      
<li>

<a class="thumb" href="<?php echo $imageloc; ?>" title="Title #0">
<img src="<?php echo $imagethumb; ?>" alt="Title #0" /></a>

						<div id="clear"></div>
                                <div class="caption">

							<div class="image-title"><?php echo $imagename; ?></div>

							<div class="image-desc">see larger</div>
						    </div>
					</li>

<?php } ?>

 

[pgrevents]

a big DOOH to me I used while in something yesterday but now its only showing one result i will keep working on it

 

thanks

 

peeps

[ted_chou12]

$dbhost = 'localhost';
$dbuser = 'dbuser';
$dbpass = 'dbpass';
$dbname = 'dbname';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);
$query = mysql_query("SELECT * FROM gallery");
while ($gallery = mysql_fetch_array($query)) {
$imageloc = $gallery['image_loc'];
$imagename = $gallery['image_name'];
$imagethumb = $gallery['thumnail_loc'];?>                      
<li>

<a class="thumb" href="<?=$imageloc?>" title="Title #0">
<img src="<?=$imagethumb?>" alt="Title #0" /></a>

						<div id="clear"></div>
                                <div class="caption">

							<div class="image-title"><?=$imagename?></div>

							<div class="image-desc">see larger</div>
						    </div>
					</li>
					<?php } ?>

Ted

[pgrevents]

hi thanks,

 

there wasnt a problem with the vars it does display correctly but instead of the 2 test images there is only one image being shown in the look. Saying that I did try your code but unfortunately it has the same result as before only showing one image

 

thanks

[ted_chou12]

Hi, sorry didn't solve ur problem, but remind u to check ur mysql db and make sure it does have two images. just to make sure

Ted

[pgrevents]

yeah i thought it the second might not have gone through but it is set in stone there thes is confudling me ???

[pgrevents]

This works it displays the content BUT it will not show the last entry

here is the DB code

<?php
$internal = $_GET['more'];
$dbhost = 'localhost';
$dbuser = 'bates';
$dbpass = 'pass';
$dbname = 'bates';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);
$query = mysql_query("SELECT * FROM contact WHERE internal= '$internal'");
$more = mysql_fetch_array($query);
?>

 

here is my page display code

<?php 
while($row=mysql_fetch_array($query, MYSQL_ASSOC)){ ;
foreach ($row as $onbox)?>        
<div class="id_box"><?=$row['internal']?></div>
<div class="from_box"><?=$row['name']?></div>
<div class="from_email_box"><?=$row['email']?></div>
<div class="msg_status_box"><?=$row['status']?></div>
<div class="reply_status_box">
<a href="?more=<?=$row['internal']?>">more</a>
    </div>
<div class="updated_box">
<? if(empty($row['reply_time'])){
print "".$row['time']." on ".$row['date']."";
}
else{
print "";
}
?>

 

what could be the reason that it wont show the latest entry?

[PFMaBiSmAd]

Because you are fetching a row into $more before your while loop and then not using it -

$more = mysql_fetch_array($query);
while($row=mysql_fetch_array($query, MYSQL_ASSOC)){ ;

 

Why do you have this - $more = mysql_fetch_array($query); in your code at all?

[pgrevents]

just remnants of something else i was doing i removed it since you saying that and still same outcome

[pgrevents]

Does any one have any ideas?

 

The latest entry is not showing ?????????? I cant find an answer an since its a messaging system it seems a little silly not showing the latest mail

[pgrevents]

http://www.barrybatemanphotography.co.uk/?id=155 now Im not the php programmer for the group im just a designer lol he is on holiday and this project has to be handed in like tommorow.

 

as you will see from the above it displays 2 images but they are the same there are two images in the database and they are different but its looping with the while statement at the top of the page

[haku]

$images = mysql_query("SELECT stuff FROM gallery");

while ($image = mysql_fetch_array($images))
{
  echo $image['column1']; // column 1 in the database
  echo $image['column2']; // column 2 in the database
  //... and so on
}

[pgrevents]

so the while statement that i have doesnt pull everything from the database?

[Ken2k7]

while($row=mysql_fetch_array($query, MYSQL_ASSOC)){ ;

Take out the semi-colon at the end.

 

haku's example is a good one. The SQL selects one column, but the while loop assumes more than one is returned.

[Julian]

I don't see any image, only the background image...  Please post the page code completely.

[J.Daniels]

Working from your display code, you have a semi-colon after the while and you do not need the foreach:

 

<?php 
while($row=mysql_fetch_array($query, MYSQL_ASSOC)) {
?>
<div class="id_box"><?=$row['internal']?></div>
<div class="from_box"><?=$row['name']?></div>
<div class="from_email_box"><?=$row['email']?></div>
<div class="msg_status_box"><?=$row['status']?></div>
<div class="reply_status_box">
<a href="?more=<?=$row['internal']?>">more</a>
</div>
<div class="updated_box">
<? if(empty($row['reply_time'])){
        print "".$row['time']." on ".$row['date']."";
    }
    else{
        print "";
    }
}
?>

 

Also, before the while loop, you can echo out the number of rows to make sure there are two rows returned:

 

echo mysql_num_rows($query);