code for retrieving image in database.

[Paul NIcolai Sunga]

is this applicable?

 
<?php
$query = mysqli_query($Link, "SELECT * from Registration_form where last_insert_id(form_no)" );
 while($row = mysqli_fetch_array($query)){
$picture = $row['picture'];
}
?>
 

thanks...:)

[Dormilich]

what does the manual say?

[Canabeez]

Nope...

You have a few mistakes in your code, (first) the $Link variable should go as a second parameter; (second) last_insert_id() cannot be used in SELECT, only in INSERT; (third) there's no point in while if you want a single row.

Here, if your form_no is auto_increment then this code should work:

$query = mysql_query("SELECT `picture` FROM `Registration_form` ORDER BY `form_no` DESC LIMIT 1;", $Link);
 
if($row = mysql_fetch_row($query))
{
   $picture = $row['picture'];
}

[Dormilich]

@Canabeez
sorry to correct you here, mysqli_* indeed uses the connection instance as first parameter (in contrast to mysql_*)

although if I’d be using MySQLi I’d rather use the object oriented style.

[Canabeez]

@Dormilich
Thanks... you're right... didn't mention the mysqli, didn't even think that anyone really uses that... I think the rest is fine...

[Dormilich]

@Canabeez
although I find MySQLi much more capable than mysql_*. guess most people are already accustomed to mysql_* and don’t want to change for the better…

[Canabeez]

not only accustomed, think of all the code, rewrite all that?! hell no... only I have something like 50,000 lines using mysql_.

[Paul NIcolai Sunga]

thanks for correcting my errors in my codes..thanks a lot.. it is a big help for me..

[Paul NIcolai Sunga]

whats this warning?


Warning: mysql_query() expects parameter 2 to be resource, object given in C:\wamp\www\dfdfdd.php on line 16

Warning: mysql_fetch_row() expects parameter 1 to be resource, null given in C:\wamp\www\dfdfdd.php on line 18

[Paul NIcolai Sunga]

<?Php   
error_reporting (E_ALL ^ E_NOTICE);
 
$Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
            if (!$Link)
              {
              trigger_error("Could not connect", E_USER_ERROR );
              }
$query = mysqli_query($Link, "SELECT picture FROM Registration_form ORDER BY form_no DESC LIMIT 1" );
 
   if($row = mysqli_fetch_row($query))
   {
     $picture1 = $row['picture'];
   }
 
   ?>
 
<table width="200" border="1">
  <tr>
<td><input name="pix" type="image" value="<?Php  error_reporting (E_ALL ^ E_NOTICE);  echo $picture; ?>"/></td>
</tr>
</table>
 

is this correct? the picture is not appearing..

[Dormilich]

@Paul NIcolai Sunga
of course. the image is referenced using the "src" attribute (which btw is a file name). It's completely useless to try inserting the image data (binary type) into html sorce code (text type).

see also Atli's article