嘿傢伙我試圖通過這種方式工作,但我有一個問題,甚至進入我的功能表單提交在JQuery中,我已經設置了幾個console.logs,但它從來沒有真正 進入我的第一個功能,有誰知道我做錯了什麼?JQuery提交永不提交?
代碼
<script src='http://www.google.com/jsapi'></script>
<script> google.load('jquery', '1.7.1'); </script>
<script>
console.log('outside function');
$("form").submit(function() {
console.log('submit happened');
queryJsonServer($(this), "class/");
return false;
});
function queryJsonServer(form, path) {
var inputs = $('input:radio').serializeArray();
var params = createAction("saveFormData", inputs);
var url = path + "json.php";
$.post(url, params, function(data) {
console.log(data);
$.("form").submit();
});
}
</script>
HTML
<form>
<fieldset>
<legend>Select Orders</legend>
<table id='master'></table>
</div> <!-- end input1 -->
<div>
<button name="select" type="submit" id="btnLoad" value="load">Refresh</button>
<button name="select" type="submit" id="btnJson" value="down">Download JSON</button>
<button name="select" type="submit" id="btnView" value="view">View/Enter</button>
</div>
</fieldset>
</form>
就是這樣。非常感謝。等待批准;) – ehime 2012-02-20 17:38:54