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Jquery-。单击“提交时不触发”按钮

ajax mysql jquery php

Renari · 技术社区 · 14 年前

嗯,我好像不明白为什么这不管用(底部有错误说明)

<script type="text/javascript">
$(document).ready(function(){
 $('#submit').click(function(){
  $('input[type=checkbox]:checked').each(function(){
   ajax=$.ajax({
    type: "POST",
    url: "rollcall_submitvote.php",
    data: 'division='+ $(this).val,(),
    cache: false
   }
  });
 });
});
</script>
<?php
/**
 * Include vB core.
 */ 
require_once('./global.php');
require_once(DIR . '/includes/functions_bigthree.php');
require_once(DIR . '/includes/functions_forumlist.php');

/**
 * Grab the divisions from the database.
 */
$query = $db->query_read("SELECT division FROM rollcall_divisions");
while($array = $db->fetch_array($query)){
 echo "<input type='checkbox' value=".$array[division].">".$array[division]."<br>";
}
?>
<input type="submit" id="submit">

<?php
/**
 * Include vB core.
 */ 
require_once('./global.php');
require_once(DIR . '/includes/functions_bigthree.php');
require_once(DIR . '/includes/functions_forumlist.php');

$db->query_write("INSERT INTO rollcall_users VALUES ('".$_POST["division"]."', '".$vbulletin->userinfo['username']."')")or die(mysql_error());
?>

1 回复 | 直到 14 年前

Louis 14 年前

$(document).ready(function(){
 $('#submit').click(function(){
  $('input[type=checkbox]:checked').each(function(){
   ajax=$.ajax({
    type: "POST",
    url: "rollcall_submitvote.php",
    data: 'division='+ $(this).val,(),
    cache: false
   });
  });
 });
});

没有看到错误消息,但您丢失了 );