개발/WEB PROGRAMMING

[ JAVASCRIPT ] ajax를 통해 서버 처리 후 특정 레이어에 replace

itaekwon class 2020. 11. 12.
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function ajaxReplace(callUrl, jsonData, layerId, successFunc, errFunc,
		completeFunc) {
	$("#ly-loading").show();
	
	if (getType(jsonData) == "object") {
		jsonData.jdcid = (new Date()).getTime();
	} else if (getType(jsonData) == "string") {
		jsonData += "&jdcid=" + (new Date()).getTime() + "&ajax_yn=Y";
	}

	
	$.ajax({
		url : callUrl,
		data : jsonData,
		contentType : "application/x-www-form-urlencoded; charset=UTF-8",
		type : 'POST',

		success : function(data) { 
			$(layerId).html(data);
			appendHtmlAfterWork(layerId);
			if (successFunc) {
				successFunc();
			}
			if (typeof sessionTimeRefresh != "undefined") {
				sessionTimeRefresh();
			}
		},
		error : function(data) {
			if (errFunc) {
				errFunc();
			}
		},
		complete : function(data) {
			$("#ly-loading").hide();
			if (completeFunc) {
				completeFunc();
			}
			if (typeof sessionTimeRefresh != "undefined") {
				sessionTimeRefresh();
			}
		}
	});
}
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