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click on picture
Hello friends Can U Help Me
I used This Method
$cardname=$_POST[card];
echo "CardValue- $cardname";
echo"<input border='0' src='image/1.jpg' name='card' width='102' height='97' type='image' value='1'>";
My Problem Is that When i click on picture It does not show Or Take The value of Picture in IE But Properly Work In Firefox .
Can any one help me out
thanks in advance
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Re: click on picture
Try using the attribute ID. So that
Code:
echo"<input border='0' src='image/1.jpg' name='card' width='102' height='97' type='image' value='1'>";
will look like this
Code:
echo"<input border='0' src='image/1.jpg' id='card' name='card' width='102' height='97' type='image' value='1'>";
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Re: click on picture
Is that all your code? This might sound stupid, but is your <input> in a <form> tag?
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Re: click on picture
You also shouldn't be using echo to produce HTML.
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Re: click on picture
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Re: click on picture
You shouldn't use print either.
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Re: click on picture
expanding on visualad....
you should never use php to show HTML code, use html to show html... example:
PHP Code:
<?php
if ($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.
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Re: click on picture
PHP Code:
<?php if (something): ?>
<something />
<?php else: ?>
<something-else />
<?php endif; ?>