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$.ajax为何自定义的options无法传进去给回调?

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2016-06-13 12:16:02873browse

$.ajax为什么自定义的options无法传进去给回调????

本帖最后由 jianye112 于 2015-03-09 23:53:25 编辑 $.ajax为什么自定义的options无法传进去给回调????

//单击执行AJAX请求操作<br />function clickSubmit(sendUrl, data, _sucmsg, _errmsg){<br />	$.ajax({<br />		type: "POST",<br />		url: sendUrl,<br />		dataType: "json",<br />		timeout: 20000,<br />		data: data,<br />		sucmsg: _sucmsg,		//????????????????????????<br />		errmsg: _errmsg,          //????????????????????????<br />		success: function(data, textStatus) {<br />			if (data.status == 1){<br />				$.dialog.tips(this.sucmsg ? this.sucmsg : data.msg, 2, "32X32/succ.png", function(){<br />					//location.reload();<br />			    });<br />			} else {<br />				$.dialog.alert(this.errmsg ? this.errmsg : data.msg);<br />			}<br />		},<br />		error: function (XMLHttpRequest, textStatus, errorThrown) {<br />			$.dialog.alert("状态:" + textStatus + ";出错提示:" + errorThrown);<br />		}<br />	});<br />}<br />

------解决思路----------------------
可以的!
你看看回调函数的 data 参数是什么
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