how to display image from database?

[mirianCalin]

i am doing a site for appliance center..
i need to display all the products that the company offers, but my problem is that i cant display ALL the images in my database.. the first entry on the database is the only one that displays..
i am using mysql as database

here's the code.. tell me what's my error and pls. kindly edit it..

**imgdata = it is where i store the image.. longblob is the data type

<?php 
    $errmsg = "";
    if (! @mysql_connect("localhost","root",""))
        {
                $errmsg = "Cannot connect to database";
                            }
    @mysql_select_db("upload");
 
    $gotten = @mysql_query("select imgdata from pix");
    header("Content-type: image/jpeg");
    while ($row = mysql_fetch_array($gotten))
    {
        print $row['imgdata'];
 
    }
mysql_free_result($gotten);
?>

[sandeepsandeep]

hi
change in the mysql_fetch_arr ay()

<?php 
    $errmsg = "";
    if (! @mysql_connect("localhost","root",""))
        {
                $errmsg = "Cannot connect to database";
                            }
    @mysql_select_db("upload");
 
    $gotten = @mysql_query("select imgdata from pix");
    header("Content-type: image/jpeg");
 
 while ($row = mysql_fetch_array($gotten,MYSQL_ASSOC))

[harshmaul]

Hi,
the problem here is that you are only able to use the content disposition to send out one file.
As each blob stores a different image you will need another page thats loops through the DB, and displays the content disposition page.

Steps to achieve this....
Firstly change your select query to be based on the ID in the table, and get that ID from the query string ($_GET['ID']) or something like that.

Secondly make another page and in that page loop through the results from the table printing out an image with the content disposition page and an ID in the query string.

I can go in to more detail if you like. let me know

[mirianCalin]

Hi,
the problem here is that you are only able to use the content disposition to send out one file.
As each blob stores a different image you will need another page thats loops through the DB, and displays the content disposition page.

Steps to achieve this....
Firstly change your select query to be based on the ID in the table, and get that ID from the query string ($_GET['ID']) or something like that.

Secondly make another page and in that page loop through the results from the table printing out an image with the content disposition page and an ID in the query string.

I can go in to more detail if you like. let me know

yes, please help me..
and can you show me the code?
im really new with php and im having a hard time..
thanks for the reply..
very much appreciated

[mirianCalin]

hi
change in the mysql_fetch_arr ay()

<?php 
    $errmsg = "";
    if (! @mysql_connect("localhost","root",""))
        {
                $errmsg = "Cannot connect to database";
                            }
    @mysql_select_db("upload");
 
    $gotten = @mysql_query("select imgdata from pix");
    header("Content-type: image/jpeg");
 
 while ($row = mysql_fetch_array($gotten,MYSQL_ASSOC))

hi!! thanks for the reply..
very much appreciated..
i'll try this tomorrow and will inform you about the result..
thanks again!!

[harshmaul]

So you need two pages.

The first page uses content disposition to output an image given an ID in the quesry string just like you have already only modify the query to select an image based on the query string. Like this.....


Call this page pix.php...

<?php 
$errmsg = "";
if (! @mysql_connect("localhost","root",""))
{
$errmsg = "Cannot connect to database";
}
@mysql_select_db("upload");
 
if (IsSet($_GET['pixID'])){
$gotten = @mysql_query("select imgdata from pix where pixID = ".$_GET['pixID']);
header("Content-type: image/jpeg");
while ($row = mysql_fetch_array($gotten))
{
print $row['imgdata'];
 
}
mysql_free_result($gotten);
}
?>

the other page will have a list of img tags with the source pointing to a variation of pix.php and a query string value.


call this page list.php

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root",""))
{
$errmsg = "Cannot connect to database";
}
@mysql_select_db("upload");
 
$strSQL = "select * from pix";
$rsPix = mysql_query($strSQL);
$numRows = mysql_numrows($rsPix);
$i = 0;
 
while($i < $numRows){
?>
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
<?php
$i++;
}
?>

Now put these into the same folder and it should work. i haven't tested you may need to debug slightly

[mirianCalin]

So you need two pages.

The first page uses content disposition to output an image given an ID in the quesry string just like you have already only modify the query to select an image based on the query string. Like this.....


Call this page pix.php...

<?php 
$errmsg = "";
if (! @mysql_connect("localhost","root",""))
{
$errmsg = "Cannot connect to database";
}
@mysql_select_db("upload");
 
if (IsSet($_GET['pixID'])){
$gotten = @mysql_query("select imgdata from pix where pixID = ".$_GET['pixID']);
header("Content-type: image/jpeg");
while ($row = mysql_fetch_array($gotten))
{
print $row['imgdata'];
 
}
mysql_free_result($gotten);
}
?>

the other page will have a list of img tags with the source pointing to a variation of pix.php and a query string value.


call this page list.php

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root",""))
{
$errmsg = "Cannot connect to database";
}
@mysql_select_db("upload");
 
$strSQL = "select * from pix";
$rsPix = mysql_query($strSQL);
$numRows = mysql_numrows($rsPix);
$i = 0;
 
while($i < $numRows){
?>
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
<?php
$i++;
}
?>

Now put these into the same folder and it should work. i haven't tested you may need to debug slightly

hey, thank you very much!! it works!!
i hope i can ask you again for further problems.. :-)
thank you..
thank you..
thank you!!

[harshmaul]

hi,
absolutly no problem. i love to help. But make sure you ask in the forums so every one can learn.

[mirianCalin]

hi,
absolutly no problem. i love to help. But make sure you ask in the forums so every one can learn.

i have another problem..
i also want to display the corresponding title of the image..
"title" is the field that handles the title of the image in my table

[harshmaul]

i have another problem..
i also want to display the corresponding title of the image..
"title" is the field that handles the title of the image in my table

Hi again...

Something like this..... for the page which displays all the images

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root",""))
{
$errmsg = "Cannot connect to database";
}
@mysql_select_db("upload");
 
$strSQL = "select * from pix";
$rsPix = mysql_query($strSQL);
$numRows = mysql_numrows($rsPix);
$i = 0;
 
while($i < $numRows){
?>
<h1><?php echo mysql_result($rsPix,$i,"pixTitle"); ?></h1>
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
<?php
$i++;
}
?>