PHP script

[matheussousuke]

May someone help me correct this script? there a few ' and " and ; in the wrong places:

<?php
include "./comm.inc"; 
connectdb(); 
$sql = "SELECT imgid,imgtype FROM tblimage ORDER BY imgid";  
 
$result = @mysql_query($sql) or die(mysql_error());  
 
echo '<table border=1>n';  
echo '<tr><th>imgid</th><th>imgtype</th><th>imgdata</th></tr>n';  
while ($rs=mysql_fetch_array($result)) {  
  echo '<tr><td>.$rs[0].</td>';  
  echo '<td>.$rs[1].</td>';  
  echo '<td><img src="post.php?imgid=.$rs[0]."" width=100 height=100></td></tr>n';  
echo '</table>n';  
 
?>
 
 
 
 
 
 

[Markus]

@matheussousuke
Wherever you are outputting strings (and they are not showing correctly).

[matheussousuke]

Well, true is, I found that script on forum, dont remember witch, had to correct a lot of things, it was worst than that.


Let's go to the point, all I wanted was a simple script that could upload a image to mysql and show it on page. And this script "does" all of this.


So if someone could give a script with that simple task, I would be really greatful, I'm not being rude, I'm tellgin that cause I spent more than 4 hours looking on google for a script that could do what I need.

The goal is put a option on my site where the user can be able to manage the index logo diretly from the admin area by using image upload function.

[code green]

What Markus is referring to is lines such as this

echo '<td><img src="post.php?imgid=.$rs[0]." 
width=100 height=100></td></tr>n'; 

You have wrapped the whole line in single quotes so the variable inside will not be parsed.
In fact the variable is concatenated but the quotes are not opened or closed
This is better

echo '<td><img src="post.php?imgid='.$rs[0].'" 
width=100 height=100></td></tr>n'; 

You need to correct all lines such as this

[matheussousuke]

I'm using a image upload script without mysql, it works perfectly, all I want is to show its image on my site index.

The script works like this:

It simple upload a image with the name logo. with extension in front (eg. logo.png).

What I need is a way to show it on index with a code such as img src


eg.:

 
<?
echo '<img src="admin/fotos/logo.   I need a way to put a extension variable here, so, undepeding on extension format, it will always show image file, rembering that the name is aways "logo.".
 
">';
?>
 
 
 

[code green]

It has been explained to you what you are doing wrong in your thread http://bytes.com/topic/php/answers/876650-php-script

[maheswaran]

Have u checked that logo extension got coorectly for example logo.png. If you echo any values using single code, for example

<?
 echo '<img src="admin/fotos/logo.".$ext."">
?>
 
then you must add ". ." in variables.
 

[Markus]

@maheswaran
That won't work - and will definitely throw an error.

The correct PHP would look like:

<?
 echo "<img src='admin/fotos/logo" . $ext . "'>";
?>
 

Furthermore, please do not double post your questions.

[maheswaran]

i think your code is wrong

[Markus]

@maheswaran
Would you care to clarify?

[Dormilich]

the dot between file name and file extension is missing. there is also the required alt attribute missing.