MySQL/PHP foreach still only displaying first in database

Question

I am querying results from a database, where more than one result should be queried. However, when I tried displaying the result of the query, only one result showed, so I tried to use a foreach function, but it's still not working. I'm beat, no idea what I'm doing wrong. Anyone have a good idea of what's going wrong?

Here's the MySQL query code:

<?php


    //Database Information 
    $dbhost = ""; 
    $dbname = ""; 
    $dbuser = ""; 
    $dbpass = ""; 

    //Connect to database 
    mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); 
    mysql_select_db($dbname) or die(mysql_error()); 

    $filename = $_GET['filename'];

$new_captions = mysql_query("SELECT * from captions where image = 'http://math.stanford.edu/inc/img/PalmDrive.png' ORDER BY idnum DESC LIMIT 5");
while($rows = mysql_fetch_array($new_captions)){
    $caption = $rows;
    }

?>

And here's the foreach:

<?php foreach($caption as $rows) {?>

<div id="set_caption" style="width:<?php echo $caption['width'];?>px; height:<?php echo $caption['height'];?>px; left:<?php echo $caption['posleft'];?>px; top:<?php echo $caption['postop'];?>px;"><?php echo $caption['text'];?></div>

<?php } ?>

It will always store the last value, because you are overwriting the values of $caption each time the loop executes. Instead you follow the answer of Robert which is the right approach. — samir chauhan
– samir chauhan, Commented Nov 23, 2011 at 4:31

robert · Accepted Answer · 2011-11-23 04:46:45Z

3

I think $caption is an array, so your code should be like this

while($rows = mysql_fetch_array($new_captions)){
    $caption[] = $rows;
}

EDIT:

Your foreach loop is also wrong.

Your variable is $rows not $caption.

<div id="set_caption" style="width:<?php echo $rows['width'];?>px; height:<?php echo $rows['height'];?>px; left:<?php echo $rows['posleft'];?>px; top:<?php echo $rows['postop'];?>px;"><?php echo $rows['text'];?></div>

<?php } ?>

edited Nov 23, 2011 at 4:46

answered Nov 23, 2011 at 4:25

robert

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1 Comment

Mark Lyons Over a year ago

I thought so, but when I do that, not even one of the results are displayed. When I leave $caption as a single variable it actually displays one. Maybe I'm missing something.

Dinuka Thilanga · Accepted Answer · 2011-11-23 04:41:10Z

You have following mistakes.

$caption is not declare before.
use array_push or $caption[] = $rows; to make caption array.

Use $row variable in the foreach.

//Database Information 
$dbhost = ""; 
$dbname = ""; 
$dbuser = ""; 
$dbpass = ""; 

//Connect to database 
mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); 
mysql_select_db($dbname) or die(mysql_error()); 

$filename = $_GET['filename'];

$new_captions = mysql_query("SELECT * from captions where image = 'http://math.stanford.edu/inc/img/PalmDrive.png' ORDER BY idnum DESC LIMIT 5");

$caption = array();

while($rows = mysql_fetch_array($new_captions)){
     $caption[] = $rows;
}

foreach($caption as $row) {        
<div id="set_caption" 
     style="width:<?php echo $row['width'];?>px; 
            height:<?php echo $row['height'];?>px; 
            left:<?php echo $row['posleft'];?>px;  
            top:<?php echo $row['postop'];?>px;">
     <?php echo $row['text'];?>
</div>

}

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