我有以下代码:
jQuery的:
function suggestPtype() {
$.ajax({
type:'POST',
url: 'include/suggest_ptype_categ.php',
data:$('#ptype_form').serialize(),
success: function(response) {
$('#ptype_form').find('.ptype_thanks').html(response);
}
});
return false;
}
HTML:
<div id="ptype_box" class="ptype_popup">
<a href="#" class="close"><img src="<?php echo $_SERVER["SITE_HTMLROOT"]?>/images/close.png" class="btn_close" title="Close Window" alt="Close" /></a>
<form method="post" id="ptype_form" onsubmit="return suggestPtype();">
<span>Product Type</span><input class="required" name="suggest_ptype" value="" type="text" autocomplete="off" placeholder="Product Type"><br />
<?php
$namecat=$langid."_name";
$result1 = mysql_query ( "SELECT categories.id, categories.$namecat FROM categories LEFT JOIN ptype_new ON ( ptype_new.catid LIKE concat( '%::', categories.id, '::%' ) ) WHERE ptype_new.id =12" ) or die(mysql_error());
while($row2 = mysql_fetch_array($result1)){
echo '<input type="checkbox" name="suggest_cat[]" value="'.$row2[0].'"><span id="label" style="padding-left:10px;color:#ffffff; font:13px Arial,Helvetica,sans-serif;">'.$row2[1].'</span><br/>';} ?>
<button class="submit button" type="submit" id="doSuggestPtype" name="doSuggestPtype">Save</button>
<div class="ptype_thanks"></div>
</form>
</div>
PHP:
<?php
if(isset($_POST['doSuggestPtype'])) {
$suggest_ptype = $_POST['suggest_ptype'];
echo " Suggested Product Type is:".$suggest_ptype;
?>
class="ptype_popup"
是一个模态窗口。
我想通过ajax调用以相同的形式(弹出窗口)调用表单中的值,但我不管理。
我哪里错了
由于 科斯明