Question

请帮我，我怎么可以把异步：假这段代码

$.getJSON("http://192.168.1.100:8080/m-help/apps/json_doctor.php", function(data) { 
    $.each(data.result, function(){ 
     $("#list").append("<li><a href='doctor-details.html?id=" +this['id']+ "'><span class='img'>  <img src='http://192.168.1.100:8080/m-help/images/upload/"+this['images']+"' alt=''/></span>" +this['doclname']+ ", " +this['docFname']+ "</a></li>"); 
    }); 
}); 

Answer 1

你不能。 jQuery的简写ajax函数对他们提供的定制非常有限。改为使用.ajax。

function success (data) { 
    $.each(data.result, function(){ 
     $("#list").append("<li><a href='doctor-details.html?id=" +this['id']+ "'><span class='img'>  <img src='http://192.168.1.100:8080/m-help/images/upload/"+this['images']+"' alt=''/></span>" +this['doclname']+ ", " +this['docFname']+ "</a></li>"); 
    }); 
}); 

var url = "http://192.168.1.100:8080/m-help/apps/json_doctor.php"; 

$.ajax({ 
    dataType: "json", 
    url: url, 
    success: success, 
    async: false // This is horrible and will lock up the UI while it runs 
});