click on picture

Hello friends Can U Help Me

I used This Method

$cardname=$_POST[card];

echo "CardValue- $cardname";

echo"<input border='0' src='image/1.jpg' name='card' width='102' height='97' type='image' value='1'>";

My Problem Is that When i click on picture It does not show Or Take The value of Picture in IE But Properly Work In Firefox .
Can any one help me out
thanks in advance

Try using the attribute ID. So that

Code:

---

echo"<input border='0' src='image/1.jpg' name='card' width='102' height='97' type='image' value='1'>";

---

will look like this

Code:

---

echo"<input border='0' src='image/1.jpg'  id='card' name='card' width='102' height='97' type='image' value='1'>";

---

Is that all your code? This might sound stupid, but is your <input> in a <form> tag?

You also shouldn't be using echo to produce HTML.

echo vs print.

http://www.faqts.com/knowledge_base/...l/aid/1/fid/40

You shouldn't use print either.

expanding on visualad....

you should never use php to show HTML code, use html to show html... example:

PHP Code:

---

<?php
if ($imcool == True) {
?>
     <b><i><whatevertag>Dylan is cool</whatevertag></i></b>
<?php
} else {
?>
     <b><i><whatevertag>Dylan is <font color='FF0000'>NOT</font> cool</whatevertag></i></b>
<?php
{
?>

---

NOTE: it looks disorganized, but with a good editor, you can make it look neat.

PHP Code:

---

<?php if (something): ?>
  <something />
<?php else: ?>
  <something-else />
<?php endif; ?>

---