How do you get php to display more than one image from database?

[twmprys]

I'm trying to display in a web page all the images stored in a database. THis is my code, but it only displays the first picture:

<?php
include "cysylltiad.php";
$result = mysql_query("SELECT * FROM files  ORDER BY fid"); 
while($row = mysql_fetch_array($result)){ 
header("Content-Type: {$row['type']}");
   echo $row["content"]; 
}
?>

where content is the name of the picture field.

Any help would be appreciated.

[rickya100]

hi,

are you storing the images in the database or just the path to the images?

Also I always set the query to die if it fails and output the mysql error when it does IE

$result = mysql_query("SELECT * FROM files ORDER BY fid") or die(mysql_error());

Richard

[twmprys]

Hi, thanks for responding. I'm storing the images actually in the database. I want to be able to allow a user to upload pictures with captions, then output all the pictures with their associated captions on a page.

[almostbob]

I'm trying to display in a web page all the images stored in a database. THis is my code, but it only displays the first picture:

<?php
include "cysylltiad.php";
$result = mysql_query("SELECT * FROM files ORDER BY fid");
while($row = mysql_fetch_array($result)){
header("Content-Type: {$row}");
echo $row["content"];
}
?>

where content is the name of the picture field.

Any help would be appreciated.

in my mind, this should produce a row of images
the content-length header should tell the browser where to break the image,
untested, I do not use blobs
blobs in databases are slow, processor intensive and unneccesarily large, my databases just store the file system pointer to the image

while($row = mysql_fetch_array($result)){ 
header("Content-Type: {$row['type']}");
header("Content-Length: strlen($row['content'])");    
echo $row['content']; 
}

[danielpataki]

Hi there!

I have a nice piece of code I use for queries like this, take a look.

function mysql_fetch_values($result, $numass=MYSQL_BOTH)
{
	$i=0;
	$keys=array_keys(mysql_fetch_array($result, $numass));
	mysql_data_seek($result, 0) ;
	while ($row = mysql_fetch_array($result, $numass)){
		foreach ($keys as $speckey) 
			$got[$i][$speckey]=$row[$speckey];
		$i++;
	}
	return $got;
}

This enables you to fetch results in the form $result[row][column]. So if you have a database holding images and want to display all of them you would include this function and then use:

$result = mysql_fetch_values(mysql_query("SELECT * FROM files ORDER BY fid"));

foreach ($result as $image)
echo $image['content'];

Hope this helps!

[phobia1]

Function  output_array14(){// by default we show first page
$page = 1;
// If current page number, use it 
// if not, set one! 
//echo $_GET['page'];
if(!isset($_GET['page'])){ 
    $page = 1; 
} else { 
    $page = $_GET['page']; 
} 

// Define the number of results per page 
$max_results = 13; 

// Figure out the limit for the query based 
// on the current page number. 
$from = (($page * $max_results) - $max_results);  
//echo $from;
// Perform MySQL query on only the current page number's results 
//echo $from,$max_results;


$sql = mysql_query("SELECT * FROM Garant WHERE Portfolio='existing' LIMIT $from, $max_results") ; 
//$sql = mysql_query("SELECT * FROM Garant"); 
//$query= "SELECT * FROM Garant order by Price ASC"
$result = mysql_query($sql);

while($row = mysql_fetch_array($sql))
{ 
    // Build your formatted results here.  
$count = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM Garant"),0); 
if($count > 0) {  
    $max=4; //set the number of images across here

    echo "<table>";  
    
    for($i=0;$i<$max_results;$i+=$max) /* loop through and build the table */  
    {  
        $headings = $pics = $locations = '<tr>';  
        for($j=0;$j<$max;$j++)  
        { 
            if($row = mysql_fetch_assoc($sql))             
            {  

//$pics .="<TD bgcolor=\"#FFFFFF\" width=\"125\" ><A HREF=\"details.php?detail=".$images["Ref"]." \"><IMG SRC=\"resources/".$images["smallpic"]."\"></A></TD>";
$pics .="<TD bgcolor=\"#B74F8F\"  border=\"2\"><A HREF=\"details.php?detail=".$row["Ref"]." \"><IMG  SRC=\"resources/".$row["smallpic"]."\"></A></TD>";
//$locations .= '<td  align="center" bgcolor=#DDEEFF><B>' .$images['Price']. ' Euro - ref - ' .$images['Ref'].'</td>';  
 $locations .= '<td  align="left" bgcolor=#DDEEFF><B> ' .$row['Ref']. '    Price: ' .$row['Price'].' Euro</td>'; 
            } else {  
               // $headings .= '<td>&nbsp;</td>';  
                $pics .= '<td>&nbsp;</td>';  
                $locations .= '<td>&nbsp;</td>';  
            }              
        }  
        $headings .= "</tr>\n";  
        $pics .= "</tr>\n";  
        $locations .= "</tr>";  
        echo $headings; 
        echo $pics; 
        echo $locations; 
    }  
    echo '</table>'; 
} 
}

// Figure out the total number of results in DB: 
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM Garant"),0); 

// Figure out the total number of pages. Always round up using ceil() 
$total_pages = ceil($total_results / $max_results); 
//echo $total_pages, $total_results, $max_results;
// Build Page Number Hyperlinks 
echo "<center>Select a Page<br />"; 

// Build Previous Link 
if($page > 1){ 
    $prev = ($page - 1); 
    echo "<a href=\"".$_SERVER['PHP_SELF']."?page=$prev\"><<Previous</a> "; 
} 

for($i = 1; $i <= $total_pages; $i++){ 
    if(($page) == $i){ 
        echo "$i "; 
        } else { 
            echo "<a href=\"".$_SERVER['PHP_SELF']."?page=$i\">$i</a> "; 
    } 
} 

// Build Next Link 
if($page < $total_pages){ 
    $next = ($page + 1); 
    echo "<a href=\"".$_SERVER['PHP_SELF']."?page=$next\">Next>></a>"; 
} 
echo "</center>"; 

}
mysql_free_result($result) ;

?>

[twmprys]

No, that doesn't work - nothing shows now. I'm obviously missing something. I think I read somewhere on the net that you need one page to call up the pictures, and another to display them but I can't find an explanation of that anywhere.

[almostbob]

Tutorial:: get multiple images from sql blobs

its tedious
now I'm really glad I dont store blobs in the database

[danielpataki]

I'm not a huge expert on database speed, but I believe it is recommended NOT to use BLOB a lot because file systems are much more quicker at handling files.

Personally I never use blogs, I store the filename in the database. I use the code I posted above to cycle through all the file names and show them on site.

[rickya100]

I've never stored images as BLOB in a DB as every article I've ever read says to not do. I find it much easier (once you learn it) just storing the path in the database and storing the actual files in a folder.

I know that doesn't help. If it's possible I would change to just storing the path, if not I hope someone with experience of BLOBs can help.

Good luck

[phobia1]

you can see an example of image display from a database at http://areabulgaria.net/properties.php let me know if you need other help. This works by storing the filename etc.
best
F

[sourcecode1]

i want to display my image in database to my web page pls i nid the code for dat n also i save it as longbolb in database

SHow us your code so far.