是否有一种方法可以在jQuery中传递更多的数据到回调函数?
我有两个函数,我想回调到$。例如,使用post来传递AJAX调用的结果数据和一些自定义参数
function clicked() {
var myDiv = $("#my-div");
// ERROR: Says data not defined
$.post("someurl.php",someData,doSomething(data, myDiv),"json");
// ERROR: Would pass in myDiv as curData (wrong)
$.post("someurl.php",someData,doSomething(data, myDiv),"json");
}
function doSomething(curData, curDiv) {
}
我希望能够将自己的参数传递给回调,以及从AJAX调用返回的结果。
$(document).on('click','[action=register]',function(){
registerSocket(registerJSON(),registerDone,second($(this)));
});
function registerSocket(dataFn,doneFn,second){
$.ajax({
type:'POST',
url: "http://localhost:8080/store/public/register",
contentType: "application/json; charset=utf-8",
dataType: "json",
data:dataFn
}).done ([doneFn,second])
.fail(function(err){
console.log("AJAX failed: " + JSON.stringify(err, null, 2));
});
}
function registerDone(data){
console.log(JSON.stringify(data));
}
function second(element){
console.log(element);
}
次要途径:
function socketWithParam(url,dataFn,doneFn,param){
$.ajax({
type:'POST',
url:url,
contentType: "application/json; charset=utf-8",
headers: { 'Authorization': 'Bearer '+localStorage.getItem('jwt')},
data:dataFn
}).done(function(data){
doneFn(data,param);
})
.fail(function(err,status,xhr){
console.log("AJAX failed: " + JSON.stringify(err, null, 2));
});
}
$(document).on('click','[order-btn]',function(){
socketWithParam(url,fakeDataFn(),orderDetailDone,secondParam);
});
function orderDetailDone(data,param){
-- to do something --
}
你可以使用JavaScript的闭包:
function wrapper( var1, var2,....) // put here your variables
{
return function( data, status)
{
//Handle here results of call
}
};
当你可以这样做时:
$.post("someurl.php",data,wrapper(var1, var2, etc...),"html");