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DeanWhitehouse

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Posts posted by DeanWhitehouse

    [SOLVED] wierd mysql count result.

    in PHP Coding Help

    Posted

    i am using this code

    $totalCount = "SELECT image_id, COUNT(image_id) FROM hayleyimages GROUP BY image_id ";
$rsCOUNT = mysql_query($totalCount, $conn) or die(mysql_error());

$rowCount = mysql_fetch_object($rsCOUNT);

echo "$rowCount";

     

    and the echo is the word

    Object

    anyone no why??

    also this is the code i was told to use

     

    "SELECT COUNT (*) as 'Total' FROM mysql.helpkeyword";

     

    mysql is there database name

    and

    helpkeyword is there table name

     

    this didn't work when i tryed it . is it outdated?

     

    pagination help

    in PHP Coding Help

    Posted

    i found this code,

    <?php
$max = 2; //amount of articles per page. change to what to want

$p = $_GET['p'];

if(empty($p))

{

$p = 1;

}

$limits = ($p - 1) * $max; 

//view the news article!

if(isset($_GET['act']) && $_GET['act'] == "view")

{

$id = $_GET['id'];

$sql = mysql_query("SELECT * FROM hayleyimages WHERE image_id = '$id'");

while($r = mysql_fetch_array($sql))

{

$title = $r['image_name'];

$story = $r['image_caption'];

$author = $r['image_link'];

echo "<div><p>$title</p><p>$author</p><p>$story</p></div>";

}



}else{



//view all the news articles in rows

$sql = mysql_query("SELECT * FROM hayleyimages LIMIT ".$limits.",$max") or die(mysql_error());

//the total rows in the table

$totalres = mysql_result(mysql_query("SELECT COUNT(image_id) AS tot FROM hayleyimages"),0);	

//the total number of pages (calculated result), math stuff...

$totalpages = ceil($totalres / $max); 

//the table

echo "<table><tr><td>Title</td><td>Author</td></tr><tr>";

while($r = mysql_fetch_array($sql))

{

$id = $r['id'];

$title = $r['title'];

$author = $r['author'];

echo "<td><a href='news.php?act=view&id=$id'>$title</a></td><td>$author</td>";

}

//close up the table

echo "</tr></table>";

for($i = 1; $i <= $totalpages; $i++){ 

//this is the pagination link

echo "<a href='news.php?p=$i'>$i</a>|";

}

}
?>

     

    now can some explain how to intergrate it into this code,

     

    <?php
require_once 'nav_bar.php';
if (isset($_GET['image_id']))
{
if ((int) $_GET['image_id'] > 0) 
{

$imageid = $_GET['image_id'];
$sql = "SELECT * FROM hayleyimages WHERE image_id=$imageid";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$imageprofname = $row['image_name'];
$imageprofcaption = $row['image_caption'];
$imageproflink = $row['image_link'];
if(mysql_num_rows($result) > 0)
{
?>
<div id="prof_img">
	<table id="portfolio_image">
<tr><th><?php echo "$imageprofname"; ?></th></tr><tr><td><?php echo '<a href="'.$row['image_link'].'" rel="lightbox" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="200px" /></a>'; ?></td><td width="200px"><?php echo 'Image ID :'.$row['image_id'].''; ?></td></tr><tr><td><?php echo ' '.$row['image_caption']. ' </a>'; ?></td></tr><tr><td><a href="portfolio.php">Return To Gallery</a></td></tr></table></div>
<?php
}
	if (mysql_num_rows($result) < 1) {
		 echo 'This ID does not exist in the database<br>';
		 echo "<a href=\"index.php\">Return to Image Page</a>";
		 exit();
	}
exit();
}
	else {
	 echo "Unknown Image ID! <br />";
	 echo "<a href=\"index.php\">Return to Image Page</a>";
	 exit();
	}
}
//No ID passed to page, display user list:
$query = "SELECT image_id, image_name, image_link, image_caption FROM `hayleyimages`";
$result = mysql_query($query) or die("Error:" . mysql_error());

if (mysql_num_rows($result) > 0) {
?>
<table class="welcome_port"><tr><td>Welcome this is my portfolio.</td></tr></table>
<table id="portfolio">
<?php
               $loop = 3;
		   $loop1 = 3;
	while($row=mysql_fetch_array($result)){ // Start looping table row

	if ($loop == 3) {
                        echo "<tr>";
                }
	$toecho1 .= '<td><a href="?image_id='.$row['image_id'].'">Details</a></td><td>    </td>';
	$toecho .= '<td><a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="150px" /></a></td><td>    </td>';

                
                if ($loop == 3) 
			{
                        echo "</tr>";
                        $loop=0;
                }   


	}
               if (substr($toecho, -5) != "</tr>") {
                  $toecho .= "</tr>";
               }
               echo $toecho;
               echo $toecho1;
}

?>		
</table>
<?php
		if (mysql_num_rows($result) < 1) {
		echo "No Images To Display";
		}

mysql_close();
?>
</div>
</body>
</html>

     

    this bit in particular

    <?php
               $loop = 3;
		   $loop1 = 3;
	while($row=mysql_fetch_array($result)){ // Start looping table row

	if ($loop == 3) {
                        echo "<tr>";
                }
	$toecho1 .= '<td><a href="?image_id='.$row['image_id'].'">Details</a></td><td>    </td>';
	$toecho .= '<td><a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="150px" /></a></td><td>    </td>';

                
                if ($loop == 3) 
			{
                        echo "</tr>";
                        $loop=0;
                }   


	}
               if (substr($toecho, -5) != "</tr>") {
                  $toecho .= "</tr>";
               }
               echo $toecho;
               echo $toecho1;
}

?>

    pagination help

    in PHP Coding Help

    Posted

    how can i paginate this 

    require_once 'nav_bar.php';
if (isset($_GET['image_id']))
{
if ((int) $_GET['image_id'] > 0) 
{

$imageid = $_GET['image_id'];
$sql = "SELECT * FROM hayleyimages WHERE image_id=$imageid";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$imageprofname = $row['image_name'];
$imageprofcaption = $row['image_caption'];
$imageproflink = $row['image_link'];
if(mysql_num_rows($result) > 0)
{
?>
	<table id="portfolio_image">
<tr><th><?php echo "$imageprofname"; ?></th><td><?php echo '<a href="'.$row['image_link'].'" rel="lightbox" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="200px" /></a>'; ?></td><td width="200px"><?php echo 'Image ID :'.$row['image_id'].''; ?></td></tr><tr><td><?php echo ' '.$row['image_caption']. ' </a>'; ?></td></tr></table>
<?php
}
	if (mysql_num_rows($result) < 1) {
		 echo 'This ID does not exist in the database<br>';
		 echo "<a href=\"index.php\">Return to Image Page</a>";
		 exit();
	}
exit();
}
	else {
	 echo "Unknown Image ID! <br />";
	 echo "<a href=\"index.php\">Return to Image Page</a>";
	 exit();
	}
}
//No ID passed to page, display user list:
$query = "SELECT image_id, image_name, image_link, image_caption FROM `hayleyimages`";
$result = mysql_query($query) or die("Error:" . mysql_error());

if (mysql_num_rows($result) > 0) {
?>
<table class="welcome_port"><tr><td>Welcome this is my portfolio.</td></tr></table>
<table id="portfolio">
<?php
               $loop = 3;
		   $loop1 = 3;
	while($row=mysql_fetch_array($result)){ // Start looping table row

	if ($loop == 3) {
                        echo "<tr>";
                }
	$toecho1 .= '<td><a href="?image_id='.$row['image_id'].'">Details</a></td><td>    </td>';
	$toecho .= '<td><a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="150px" /></a></td><td>    </td>';

                
                if ($loop == 3) 
			{
                        echo "</tr>";
                        $loop=0;
                }   


	}
               if (substr($toecho, -5) != "</tr>") {
                  $toecho .= "</tr>";
               }
               echo $toecho;
               echo $toecho1;
}

?>		
</table>
<?php
		if (mysql_num_rows($result) < 1) {
		echo "No Images To Display";
		}

mysql_close();
?>
</div>
</body>
</html>

     

    this is the exact piece that needs paginating

     

    <?php
               $loop = 3;
		   $loop1 = 3;
	while($row=mysql_fetch_array($result)){ // Start looping table row

	if ($loop == 3) {
                        echo "<tr>";
                }
	$toecho1 .= '<td><a href="?image_id='.$row['image_id'].'">Details</a></td><td>    </td>';
	$toecho .= '<td><a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="150px" /></a></td><td>    </td>';

                
                if ($loop == 3) 
			{
                        echo "</tr>";
                        $loop=0;
                }   


	}
               if (substr($toecho, -5) != "</tr>") {
                  $toecho .= "</tr>";
               }
               echo $toecho;
               echo $toecho1;
}

?>

     

    i have tried pagination tutorials, but i can't ever get it to work

    [SOLVED] limit mysql_query amount

    in PHP Coding Help

    Posted

    how can i start a new line after every three database entry's ?

     

    here is my code

    <div id="body"> 
<?php
if (isset($_GET['image_id']))
{
if ((int) $_GET['image_id'] > 0) 
{

$imageid = $_GET['image_id'];
$sql = "SELECT * FROM hayleyimages WHERE image_id=$imageid";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$imageprofname = $row['image_name'];
$imageprofcaption = $row['image_caption'];
$imageproflink = $row['image_link'];
if(mysql_num_rows($result) > 0)
{
?>
	<table align="center" border="1"><th><?php echo "$imageprofname"; ?></th>
<tr><td><?php echo '<a href="'.$row['image_link'].'" rel="lightbox" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="200px" /></a>'; ?></td><td width="200px"><?php echo 'Image ID :'.$row['image_id'].''; ?></td></tr><tr><td><?php echo ' '.$row['image_caption']. ' </a>'; ?></td></tr></table>
<?php
}
	if (mysql_num_rows($result) < 1) {
		 echo 'This ID does not exist in the database<br>';
		 echo "<a href=\"index.php\">Return to Image Page</a>";
		 exit();
	}
exit();
}
	else {
	 echo "Unknown Image ID! <br />";
	 echo "<a href=\"index.php\">Return to Image Page</a>";
	 exit();
	}
}
//No ID passed to page, display user list:
$query = "SELECT image_id, image_name, image_link, image_caption FROM `hayleyimages`";
$result = mysql_query($query) or die("Error:" . mysql_error());

if (mysql_num_rows($result) > 0) {
?>
<table border="1"><th>Images</th>
<?php
	while($row=mysql_fetch_array($result)){ // Start looping table row
	?>

	<tr><td><?php echo ' <a href="?image_id='.$row['image_id'].'">'.$row['image_name']. ' Details</a>'; ?></td><td><?php echo ' <a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" >'.$row['image_name']. ' </a>'; ?></td><td><?php echo ' <img src="'.$row['image_link'].'"width="150px" /></a>'; ?></td></tr>
	<?php

	}
	?>
</table>
<?php
		if (mysql_num_rows($result) < 1) {
		echo "No Images To Display";
		}
}
mysql_close();
?>
</div>
</body>
</html>

    [SOLVED] textarea problem

    in PHP Coding Help

    Posted

    what are the HTML tags in the database?

     

    also try this

    ?>

<div class="editor" style="text-align: center">
<p class="editortitle" style="font-size: 175%; padding: 4px 0px 4px 0px; margin: 0px;">Welcome Letter</p>
<textarea cols="80" rows="25" name="welcome"><?php echo"" . $row['welcome'] . ""; ?></textarea>
</div>
<?php

    pagination and word wrap(sort of thing) on tables??

    in PHP Coding Help

    Posted

    with this code i need to have pagination for it, and i need it so only 3 images a row on the page.

    so like ,

    |image1|image2|image3|

    instead of

    image1

    image2

    image3

     

    this is my code

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Hayley - Home</title>
<script type="text/javascript" src="js/prototype.js"></script>
<script type="text/javascript" src="js/scriptaculous.js?load=effects,builder"></script>
<script type="text/javascript" src="js/lightbox.js"></script>
<link rel="stylesheet" href="css/lightbox.css" type="text/css" media="screen" />
<link rel="stylesheet" href="css/style.css" type="text/css" media="screen" />
</head>
<body>
<div id="body"> 
<?php
if (isset($_GET['image_id']))
{
if ((int) $_GET['image_id'] > 0) 
{

$imageid = $_GET['image_id'];
$sql = "SELECT * FROM hayleyimages WHERE image_id=$imageid";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$imageprofname = $row['image_name'];
$imageprofcaption = $row['image_caption'];
$imageproflink = $row['image_link'];
if(mysql_num_rows($result) > 0)
{
?>
	<table align="center" border="1"><th><?php echo "$imageprofname"; ?></th>
<tr><td><?php echo '<a href="'.$row['image_link'].'" rel="lightbox" title="'.$row['image_caption']. '" ><img src="'.$row['image_link'].'"width="200px" /></a>'; ?></td><td width="200px"><?php echo 'Image ID :'.$row['image_id'].''; ?></td></tr><tr><td><?php echo ' '.$row['image_caption']. ' </a>'; ?></td></tr></table>
<?php
}
	if (mysql_num_rows($result) < 1) {
		 echo 'This ID does not exist in the database<br>';
		 echo "<a href=\"index.php\">Return to Image Page</a>";
		 exit();
	}
exit();
}
	else {
	 echo "Unknown Image ID! <br />";
	 echo "<a href=\"index.php\">Return to Image Page</a>";
	 exit();
	}
}
//No ID passed to page, display user list:
$query = "SELECT image_id, image_name, image_link, image_caption FROM `hayleyimages`";
$result = mysql_query($query) or die("Error:" . mysql_error());

if (mysql_num_rows($result) > 0) {
	while($row=mysql_fetch_array($result)){ // Start looping table row
	?>
	<table border="1"><th>Images</th>
	<tr><td><?php echo ' <a href="?image_id='.$row['image_id'].'">'.$row['image_name']. ' Details</a>'; ?></td><td><?php echo ' <a href="'.$row['image_link'].'" rel="lightbox [main]" title="'.$row['image_caption']. '" >'.$row['image_name']. ' </a>'; ?></td><td><?php echo ' <img src="'.$row['image_link'].'"width="200px" /></a>'; ?><br /></td></tr>
	</table>
	<?php
	}
		if (mysql_num_rows($result) < 1) {
		echo "No Images To Display";
		}
}
mysql_close();
?>
</div>
</body>
</html>

    Check Screen Resoultion in PHP

    in PHP Coding Help

    Posted

    i found this on the net.

    <?
if(isset($HTTP_COOKIE_VARS["users_resolution"]))
$screen_res = $HTTP_COOKIE_VARS["users_resolution"];
else //means cookie is not found set it using Javascript
{
?>
<script language="javascript">
<!--
writeCookie();

function writeCookie()
{
var today = new Date();
var the_date = new Date("December 31, 2023");
var the_cookie_date = the_date.toGMTString();
var the_cookie = "users_resolution="+ screen.width +"x"+ screen.height;
var the_cookie = the_cookie + ";expires=" + the_cookie_date;
document.cookie=the_cookie

location = 'get_resolution.php';
}
//-->
</script>
<?
}
?>

     

    or this

     Put this on the page:

<script language="javascript">
window.location.href = "screen.php?width=" + screen.width + "&height=" + screen.height;
</script>

Then create a file called screen.php and use this:

<?php
$width = $_GET['width'];
$height = $_GET['height'];

echo "You are using a $width x $height screen resolution";
?>

    [SOLVED] mysql loop or while or something related, not working

    in PHP Coding Help

    Posted

    this is my code

    <?php
if (isset($_GET['image_id']))
{
if ((int) $_GET['image_id'] > 0) 
{

$imageid = $_GET['image_id'];
$sql = "SELECT * FROM `hayleyimages` WHERE `image_id`='{$imageid}' LIMIT 0,1;";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$imageprofname = $row['image_name'];
$imageprofcaption = $row['image_caption'];
$imageproflink = $row['image_link'];
echo "$imageprofname<br>$imageprofcaption<br>";
?>
<table align="center" border="1"><th>Image<?php echo "$imageprofname"; ?></th>
<tr><td><?php echo ' <img src="'.$rows['image_link'].'"width="200px" /></a>'; ?></td><td><?php echo 'ID'.$rows['image_id'].''; ?></td></tr><td><?php echo ' <a href="'.$rows['image_link'].'" rel="lightbox" title="'.$rows['image_caption']. '" >'.$rows['image_name']. ' </a>'; ?></td><td><br /></td></tr>
</table>
<?php
	if (mysql_num_rows($result) < 1) {
		 echo 'This ID does not exist in the database<br>';
		 echo "<a href=\"index.php\">Return to Image Page</a>";
		 exit();
	}
exit();
}
	else {
	 echo "Unknown Image ID! <br />";
	 echo "<a href=\"index.php\">Return to Image Page</a>";
	 exit();
	}
}
//No ID passed to page, display user list:
$query = "SELECT image_id, image_name, image_link, image_caption FROM `hayleyimages`";
$result = mysql_query($query) or die("Error:" . mysql_error());

if (mysql_num_rows($result) > 0) {
	while($rows=mysql_fetch_array($result)){ // Start looping table row
	?>
	<table align="center" border="1"><th>Images</th>
	<tr><td><?php echo ' <a href="?image_id'.$rows['image_id'].'">'.$rows['image_name']. ' Details</a>'; ?></td><td><?php echo ' <a href="'.$rows['image_link'].'" rel="lightbox" title="'.$rows['image_caption']. '" >'.$rows['image_name']. ' </a>'; ?></td><td><?php echo ' <img src="'.$rows['image_link'].'"width="200px" /></a>'; ?><br /></td></tr>
	</table>
	<?php
	}
		if (mysql_num_rows($result) < 1) {
		echo "No Images To Display";
		}
}
mysql_close();
?>

     

    But this is not working, the page has a table with a dynamic link in it, this links address is http://hayley.awardspace.com?image_id(then the image id number)

    but when you click on the link it has the same content, it shouldn't.

    the URL changes in the address bar, but the page remains unchanged .

    any ideas?

    [SOLVED] how can i get the name an uploaded file (and the extension e.g. jpg)?

    in PHP Coding Help

    Posted

    how can i get the name of an uploaded file,

    this is my code

    <?php
if(isset($_POST['upload']))
{
$target = "images/";
$image =  $_POST['uploaded'];
$imagename = $_POST['imagename'];
$imagecaption = $_POST['imagecaption'];
$user_check = mysql_num_rows(mysql_query("SELECT * FROM `hayleyimages` WHERE image_name = 1 "));
$target = $target . basename( $_FILES['uploaded']['name']) ; 
$ok=1;

//This is our size condition
if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}

//This is our limit file type condition
if ($uploaded_type =="text/php")
{
echo "No PHP files<br>";
$ok=0;
}

//Here we check that $ok was not set to 0 by an error
if ($ok==0)
{
Echo "Sorry your file was not uploaded";
}

if($user_check >= 1)
{
echo "$imagename already exists, please rename it.";
}

//If everything is ok we try to upload it
else
{
//if($image && $imagename && $imagecaption)
//{
mysql_query("INSERT INTO `hayleyimages` (image_link,  image_name, image_caption, image_id) VALUES 	  ( '$image','$imagename','$imagecaption', '')")
or die('Error ' . mysql_error());
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded";
}
else
{
echo "Sorry, there was a problem uploading your file.";
} 
//}
//else
//{
//echo "<br>Please make sure all fields are filled in.";
//}
}
}
?>
<form enctype="multipart/form-data" action="<?php $_SERVER['PHP_SELF'] ?>" method="POST">Image Name: <input type="text" name="imagename" /><br />Image Caption: <input type="text" name="imagecaption" /><br />
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Submit" name="upload" />
</form>

     

    this

    "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded";

    only echos

     

    The file has been uploaded

     

     

    [SOLVED] upload form not working properly

    in PHP Coding Help

    Posted

    ok, i solved that part

     

    now for this problem

    <?php
if(isset($_POST['upload']))
{
$target = "images/";
$image =  basename( $_FILES['uploadedfile']['name']);
$imagename = $_POST['imagename'];
$imagecaption = $_POST['imagecaption'];
$image_thumb = '<img src='.$image.' width="200px"/>';
$image_link = '<a href='.$image.' rel="lightbox" title='.$imagecaption.'>'.$imagename.'</a>'; 
$user_check = mysql_num_rows(mysql_query("SELECT * FROM `hayleyimages` WHERE image_name=1 "));
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;

//This is our size condition
if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}

//This is our limit file type condition
if ($uploaded_type =="text/php")
{
echo "No PHP files<br>";
$ok=0;
}

//Here we check that $ok was not set to 0 by an error
if ($ok==0)
{
Echo "Sorry your file was not uploaded";
}

if($user_check >= 1)
{
echo "$imagename already exists, please rename it.";
}

//If everything is ok we try to upload it
else
{
//if($image && $imagename && $imagecaption)
//{
mysql_query("INSERT INTO `hayleyimages` (image_link,  image_name, image_caption, image_id, image_thumb) VALUES 	  ('$image_link','$imagename','$imagecaption', '', '$image_thumb')")
or die('Error ' . mysql_error());
	if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
	{
	echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded";

	}
		else
		{
		echo "Sorry, there was a problem uploading your file.";
		}
//}
//else
//{
//echo "<br>Please make sure all fields are filled in.";
//}
}
}
?>
<form enctype="multipart/form-data" action="<?php $_SERVER['PHP_SELF'] ?>" method="POST">Image Name: <input type="text" name="imagename" /><br />Image Caption: <input type="text" name="imagecaption" /><br />
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Submit" name="upload" />
</form>
</body>
</html>

     

    these two bits

    $image_thumb = '<img src='.$image.' width="200px"/>';
$image_link = '<a href='.$image.' rel="lightbox" title='.$imagecaption.'>'.$imagename.'</a>';

    are being inserted to the database as

    <a href= rel="lightbox" title=test>test</a> <img src= width="200px"/>

    without links , and title=test should be title="test"

    does anyone know how to fix this

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