ajax请求中的Javascript回调函数

debugcn 发表于 Dev

k1nda

我想在这里用js写一个回调函数。问题是返回值不正确。ajax 调用中的变量“d”包含正确的数据。但是 done(..) 函数中的变量 a 没有。有谁知道如何分配 d 的值？

    function render_confirmation_email(data, cart, delivery_date){
        console.log("Render confirmation email")
        var purchaseTable = "<table>"
        for (var i = 0; i < cart.length; i++) {
            console.log(i);
            var concept = cart[i].name;
            var price = cart[i].price;
            purchaseTable += "<tr>"
            purchaseTable += "<td>" + concept + " - </td>"
            purchaseTable += "</tr>"
            purchaseTable += "<tr>"
            purchaseTable += "<td>" + price + " kr\n</td>"
            purchaseTable += "</tr>"
        }

        purchaseTable += "</table>"
        purchaseTable += "<br> <p>It will be delivered on " + delivery_date + "</p>"

        var tempDom;
        tempDom = $('<div></div>').append(data);
        tempDom.find('#purchaseTable').append(purchaseTable);
        return tempDom.text()
    }


    function get_confirmation_email(cart, delivery_date, render_confirmation_email) {

        return $.ajax({
            type: "GET",
            url:"/confirmation_email",
            async: false,
            success:function(data) {
                console.log("success");
                // render_confirmation_email called when data is ready
                var d = (render_confirmation_email(data, cart, delivery_date));
                console.log("Rendering done")
                console.log(d)
                return d
            }
        });
    }

    var a = get_confirmation_email(JSONcart, form.querySelector('input[name=deliverydate]').value, render_confirmation_email);
    a.done(function(data) {
        console.log("Function done");
        console.log(data);
    });

谢谢！！！

查理特

使用then()每个实例。Areturn什么都不做，success因为它不是承诺链的一部分

基本示例

function doAjax() {
  // sample data that will be returned
  var json = '{"foo":[1,2,3]}'

  return $.ajax({...}).then(function(resp){
     // return to next "then()" in promise chain 
     return resp.foo
  })
}


doAjax().then(function(data){
  console.log(data) // [1,2,3]
})

DEMO

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