I would like to change the submit button on my form. I succeed, and it's pretty easy, but the code breaks. When I fixed the code. The submit button returns to its default image, and it said a send request that I did not initialize. I did not understand how to fix this since. Which is strange.
<?Php
echo "<div style='text-align:right'>
<form action='upd2.php' method='POST'>
<Table border='0' align='right'>
<tr>
<td ><font size='-3'>ID Search</font></td>
<td rowspan =2><input name='image' type='image' id='SUBMIT' src='images/_Images_buttons_search_button.jpg'></td>
</tr>
<tr>
<td><input name='ID_NO' type='text' id='ID_NO'></td>
</tr>
</table><br>
</FORM> </div>";
?>
I pass it
<?php
print "<H2>Update User Info</H2>";
$ID_NO = clean($_POST['ID_NO']);
$_SESSION["ID_NO"] = $ID_NO;
if (isset($_POST['image'])) {
$result = mysql_query("SELECT * FROM user_info WHERE ID_NO = '$ID_NO'");
if(!mysql_num_rows($result)) {
print "<BR><BR>There is no such User with a User Number of $ID_NO <BR><A HREF ='UpdateUser.php'>Go Back</A>";
exit();
}
else {
$row_array = mysql_fetch_array($result, MYSQL_ASSOC);
?>
however, it does not work (the variable wasnt able to get it). I need to change the "image" to "SUBMIT". But if I do, the image will be the default.
so shorter
type = 'SUBMIT' -> I can make the code work, but the image is by default.
type = 'image' → I get a nice image, but the code is not working.
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