Reputation: 1983
I need help with the displayed image below. The two buttons "edit and back" button is not being displayed on line. The code for both the button is as below. They are both inside the same <td></td>
<tr><td>
<input type="hidden" name="token" value="<?php echo $token; ?>"/>
<input type="submit" name="edit" value="Edit"/>
<?php
//create a back button that takes user back to referer url on click
if (isset($HTTP_REFERER)) {
echo '<a href="$HTTP_REFERER"><img src="images/buttons/back.gif" alt="back"></a>';
} else {
echo '<a href="javascript:history.back()"><img src="images/buttons/back.gif" alt="back"></a>';
}
?>
</td></tr>
I can't upload the image directly so here is the link where I uploaded the image image source: http://i39.tinypic.com/2miab.png
Upvotes: 2
Views: 2183
Reputation: 531
well I noticed that some solutions might work differently in different browsers putting your buttons in another small table without any attributes will keep them in a single line
<table><tr> <td><input type="hidden" name="token" value="<?php echo
$token; ?>"/></td> <td> <?php
//create a back button that takes user back to referer url on click
if (isset($HTTP_REFERER)) {
echo '<a href="$HTTP_REFERER"><img1 src="images/buttons/back.gif" alt="back"></a>';
} else {
echo '<a href="javascript:history.back()"><img1 src="images/buttons/back.gif" alt="back"></a>';
}
?> </td> </tr></table>
hope this is what you wanted
Upvotes: 2
Reputation: 2530
try this:
<td>
<input type="hidden" name="token" value="<?php echo $token; ?>"/>
<input type="submit" name="edit" value="Edit"/>
</td>
<td>
<?php
//create a back button that takes user back to referer url on click
if (isset($HTTP_REFERER)) {
echo '<a href="$HTTP_REFERER"><img src="images/buttons/back.gif" alt="back"></a>';
} else {
echo '<a href="javascript:history.back()"><img src="images/buttons/back.gif" alt="back"></a>';
}
?>
</td>
Upvotes: 1