而不是
data从
success:返回来传递
data给函数。
var view_data;$.ajax({ url:"/getDataFromServer.json", //async: false, type: "POST", dataType: "json", success:function(response_data_json) { view_data = response_data_json.view_data; console.log(view_data); //Shows the correct piece of information doWork(view_data); // Pass data to a function } });function doWork(data){ //perform work here}


