passing value into variable from an image?

adamhhh

ok i know of course how to pass values from fields and buttons, but im struggling to pass a value from an image. i mean im trying something like below (Which sucks i know but its getting late) but my brain is a bit slow and i cant figure anything out right now.

test.php--->

<a href="test2.php"><img src="images/abram1.jpg" name="abram" width="102" height="102" border="0" id="abram" value="<?php echo @$theValue ?>"></a>
</p>
$theValue = "abram";

test2.php--->

if ($theValue == 'abram')
{
$wishlist= "yes";
echo $wishlist;
}

all i want is for an image to be clicked and then ''yes'' to be printed out to the screen (so i can further manipulate this), and when a different image is clicked something else happens. any ideas? sorry if its a bit confusing. cheers.

bpat1434

Use an image button...

<?php
$theValue = 'abram';
?>
<button><img src="images/abram1.jpg" name="abram" width="102" height="102" border="0" id="abram" value="<?php echo $theValue; ?>" /></button>

if($_POST['abram'] == 'abram')
{
  echo 'Yes!!';
}

adamhhh

nope been messing about with that all day and it doesnt work.

Right im trying something like below-

test.php

<?php
include ("function.inc");
?>
<html>
<head>

<title>Account details</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body bgcolor="333333" text="e5e5e5" link="e5e5e5" vlink="e5e5e5" alink="e5e5e5">
<?php
$theValue = 'abram';

if (@$POST['abram'] == $theValue) # 7
{
header("Location: test2.php");
}
$theValue = $POST['abram'];
// post $productID??
?>
<?php

include("donuts.inc");

$connection = mysql_connect($host,$user,$password)
or die ("couldn't connect to server");
$db = mysql_select_db($database,$connection)
or die ("Couldn't select database");

$result = mysql_query("select * from Products where productType='Silly Shirts'");

while($r=mysql_fetch_array($result))
{
$productID=$r["productID"];

$productName=$r["productName"];
$productType=$r["productType"];
$description=$r["description"];
$productRRP=$r["productRRP"];
$price=$r["price"];
$extraInformation=$r["extraInformation"];
$officialDescription=$r["officialDescription"];
$image=$r["image"];

$f_price = number_format($row['price'],2);
$f_price = $price;

echo "<table cellspacing='0' border='0' cellpadding='0' width='95%' align='center'>";

echo "<tr>\n";

echo "<td width=\"80%\"><font face=\"Verdana\" color=\"#E5E5E5\" font

size=\"-1\">$productName - $description <b>£$f_price</b></font></td>\n";

echo "<td><a href='../store/listitem.php?id=$productID'
border='0'>
<img src='../store/images/$image' border='0'
width='100' height='80'></a></td>\n";
echo "</tr>\n";
addBasket();

	echo "<tr>";  
  echo "<td colspan=\"2\" td background=\"../store/images/linebreak.gif\">&nbsp;</td>\n";
   echo "</tr>\n";

echo "</table>\n";

}
?>

test2.php-

<?php

if($POST['abram'] == $theValue)
{
echo 'Item has been added to wishlist.';
// go back (javascript).
}
$theValue = $POST['abram'];
// $wishlist = $_POST['wishlist'];
$theValue = $wishlistValue;
$wishlistValue = "yes";

include("donuts.inc");

$connection = mysql_connect($host, $user, $password)
or die ("Can't find database, grr");
$db = mysql_select_db($database, $connection)
or die ("Can't find database, grr");

$wishlistValue=trim($wishlistValue);
$wishlistValue=ucfirst(strtolower(strip_tags($wishlistValue)));

$wishlistValue = strip_tags(trim($wishlistValue));

$productID = mysql_insert_id();

$query = "INSERT INTO Wishlist (productID,wishlist)
VALUES ('$productID','$wishlistValue')";

$result = mysql_query($query)
or die ("Can't find database, hmpph");

// insert into wishlist db, get id first.
?>

so im trying to set the image as a variable and sending that to a new page, where it is put into a table in the database along with its productID, however its simply not working. any ideas?

bpat1434

Where in test2.php do you say what $_POST['abram'] should be equal to? I see $theValue, but that's never set to a specific value....