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In Topic: About Myself

Posted 4 Jul 2012

hello friends. i m saravanan. now i am doing computer science in an institution. and i am learning php. so please help me to learn well. thanks..,
In Topic: Color change to php commands

Posted 4 Jul 2012
demoloution, on 03 July 2012 - 10:50 AM, said:
also in my website i want to allow user to see some of my images
i use this code:
```
<?php
$con = mysql_connect("localhost","root","");

if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("photoshare", $con);

$Title = $_POST['Title'];
$Creator = $_POST['Creator'];


$result = mysql_query("SELECT Title, Creator FROM images WHERE Title = '$Title' OR  Title LIKE '%$Title%' 
OR Creator = '$Creator' OR  Creator LIKE '%$Creator'");


// escape from PHP ... ?>

<img src="<? echo $result; ?>">

<? // starting up PHP again...


//some code
mysql_close($con);
?>
```
but it doeasnt display me any image.It displays me the icon you have when the image dont load correctly or is broken.
Also i noticed that in my database, the images array doesnt have any elements in it.is this wrong?
My upload code works fine i think.
Any suggestions?
I THINK, THE VARIABLE NAME AND FIELD NAME OF YOUR DATABASE SHOULD NOT BE THE SAME. JUST TRY TO CHANGE THEIR NAME PLEASE. FOR EXAMPLE REPLACE $CREATOR BY $CREATORS AND ALSO $TITLE BY $TITLES.