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Requesting info from a MySQL DB

Posted: Mon Nov 20, 2006 9:05 pm
by Dirge of Cerberus

Code: Select all

<?PHP require('incSession.php'); ?>

<?php

// set your information.
$dbhost='mysql67.secureserver.net';
$dbusername='animestome3';
$dbuserpass='*******';
$dbname='animestome3';

// connect to the mysql database server.
mysql_connect ($dbhost, $dbusername, $dbuserpass);
mysql_select_db($dbname) or die("Cannot select database");

//define a maxim size for the uploaded images in Kb
define ("MAX_SIZE","400");

//This function reads the extension of the file. It is used to determine if the file is an image by checking the extension.
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}

//This variable is used as a flag. The value is initialized with 0 (meaning no error found) and it will be changed to 1 if an errro occures. If the error occures the file will not be uploaded.
$errors=0;
//checks if the form has been submitted
if(isset($_POST['Submit']))
{
//reads the name of the file the user submitted for uploading
$image=$_FILES['image']['name'];
$id = $_POST["id"];
//if it is not empty
if ($image|$id)
{
//get the original name of the file from the clients machine
$filename = stripslashes($_FILES['image']['name']);
//get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
//if it is not a known extension, we will suppose it is an error and will not upload the file, otherwize we will do more tests
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
//print error message
echo '<center><b>Unknown extension!</b></center>';
$errors=1;
}
else
{
//get the size of the image in bytes
//$_FILES['image']['tmp_name'] is the temporary filename of the file in which the uploaded file was stored on the server
$size=filesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
echo '<center><b>You have exceeded the size limit!D=< </b></center>';
$errors=1;
}

$query="SELECT name FROM avatars WHERE id='$id'";
$result=mysql_query($query);
$name=mysql_result($result);

//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="../avatars/".$name."/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
echo '<center><b>Upload Unsuccessful!!! D=</b></center>';
$errors=1;
}}}}

//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors)
{
echo "<center><b>Avatar Uploaded Successfully! =D</b></center>";
}

?>

<!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" -->
<head>
<link rel="shortcut icon" href="../favicon.ico" />
<link rel="stylesheet" href="style_mini.css" type="text/css">
</head>
<form name="newad" method="post" enctype="multipart/form-data" action="">
  <center><table width=250 height="87" border=0 cellpadding=3 cellspacing=0>

    <tr>
      <td class="top">Avatar</td>
    </tr>
    <tr>
      <td class="content"><center><input type="file" name="image"></center>
</td>
    </tr>
    <tr>

      <td height="19" class="bottom">&nbsp;</td>
    </tr>
</table>
<p>
<table width=250 height="87" border=0 cellpadding=3 cellspacing=0>

    <tr>
      <td class="top">Category</td>
    </tr>
    <tr>
      <td class="content"><center>Category: <SELECT name="id">
      <OPTION SELECTED>Choose a category...
      <OPTION>--------------------
      <?php
$query = " SELECT id,name,dir FROM avatars " .
" ORDER BY id";
$result = mysql_query($query) or die(' Error, contact us via forum immediately. Code: List ');
      
while($row = mysql_fetch_array($result))
{
echo "<OPTION value='".$row['id']."'>".$row['name']."";
}
?>
</SELECT> 
</td>
    </tr>
    <tr>

      <td height="19" class="bottom">&nbsp;</td>
    </tr>
</table>
<p>
  <table width=250 border=0 cellpadding=3 cellspacing=0>

    <tr>
      <td class="top">Submit</td>
    </tr>
    <tr>
      <td class="content"><center><input name="Submit" type="submit" value="Submit Avvie"></center>
</td>
    </tr>
    <tr>

      <td height="19" class="bottom">&nbsp;</td>
    </tr>
</table>
</form><p><center>

  <table width=250 height="87" border=0 cellpadding=3 cellspacing=0>

    <tr>
      <td class="top">Copyright</td>
    </tr>
    <tr>
      <td class="content"><center>[DATTEBAYO] Content System ©2006 Anime's Tome</center>
</td>
    </tr>
    <tr>

      <td height="19" class="bottom">&nbsp;</td>
    </tr>
</table><br><br><a href='admin.php'>Back to Admin Area</a></center>
I need help here. Whenever I run it, I get this when using it:

Code: Select all

Warning: Wrong parameter count for mysql_result() in /home/content/a/n/i/animestome/html/admin/avatarreg.php on line 65
Anyone know how to fix it? It uploads an image to a certain directory depending on the Category, whose ID's, names, and Directories are stored in a MySQL DB.

Posted: Mon Nov 20, 2006 9:13 pm
by feyd
Give it the parameters it needs.

Posted: Mon Nov 20, 2006 9:31 pm
by Dirge of Cerberus
??? What do you mean?

Posted: Mon Nov 20, 2006 9:40 pm
by feyd
string mysql_result ( resource result, int row [, mixed field] )
The function requires at least two arguments. A result resource identifier and a row number. You've only given it the former.

Posted: Mon Nov 20, 2006 9:43 pm
by Dirge of Cerberus
Hmm... How do I make it get the directory based on the id, and nothing else?

Code: Select all

mysql_query("SELECT dir FROM avatars WHERE id='$id'");
I know that's how to get the info, but to put in only the directory info?

Posted: Mon Nov 20, 2006 9:49 pm
by feyd
I'm not sure what you're asking about... :?

Posted: Tue Nov 21, 2006 3:50 pm
by Dirge of Cerberus
How do I make it search for a chosen variable inside the mySQL DB and fetch the entry containing it. Then, I want it to take the 'dir' field's value and insert it into this area:

Code: Select all

//the new name will be containing the full path where will be stored (images folder)
$newname="../avatars/" ______HERE!!!!______ "/".$image_name;

Posted: Tue Nov 21, 2006 4:41 pm
by RobertGonzalez
Ok, no code. Just plain simple English. What are you trying to accomplish, step by step?

Posted: Tue Nov 21, 2006 5:33 pm
by impulse()
Hello.

Try:

Code: Select all

$query="SELECT name FROM avatars WHERE id='$id'"; 

while ($res = mysql_fetch_array($query)) {
  echo $res['<field name goes here>'];
}
Hope that helps,

Posted: Tue Nov 21, 2006 6:26 pm
by volka
impulse() wrote:

Code: Select all

$query="SELECT name FROM avatars WHERE id='$id'"; 

while ($res = mysql_fetch_array($query)) {
  echo $res['<field name goes here>'];
}
needs mysql_query

Posted: Tue Nov 21, 2006 8:22 pm
by Dirge of Cerberus
Here's what I want to do:
  • 1. Connect to DB 'avatars' (Done)
    2. Search for a chosen ID in 'Avatars'
    3. Retrieve entry with the chosen ID
    4. Retrieve text from the 'DIR' field in the entry
    5. Set that text to be the variable '$dir'

Posted: Tue Nov 21, 2006 8:53 pm
by Dirge of Cerberus
Nevermind. I found out how.

Code: Select all

$sql="SELECT dir FROM avatars WHERE id='$id'";

$result = mysql_query( $sql );

while( $dir = mysql_fetch_array( $result ) ) { 
<Uploader Code>
}