Backbone.js 主干存储模型成功与错误

我有以下主干代码，创建视图和模型，并调用save方法将数据保存到数据库：

var form = document.forms[0];

var RetailerModel = Backbone.Model.extend({
    urlRoot: ' retailer.php',
    defaults: {
        name: 'company-name',
        address: 'company-address',
        phone: 'company-phone',
        icon: 'http://localhost/icon.png'
    }
});

var RetailerCollection = Backbone.Collection.extend({

});

var RetailerView = Backbone.View.extend({

    className: 'retailer',

    template: _.template($('#retailer-template').html()),

    initialize: function() {

        var obj = {
            name: form.name.value,
            address: form.address.value,
            phone: form.phone.value
        };

        var o = this;

        this.model.save(obj, {
            success: function(model, response) {
                console.log(model);
                console.log(response);
                o.render();
                console.log('success');
            },
            error: function(model, response) {
                console.log(model);
            }
        });
    },

    render: function() {
        $('#retailer-list').append(this.$el.html(this.template(this.model.toJSON())));

        return this;
    }
});

var RetailerViews = Backbone.View.extend({

});

$('#submit').click(function(e){
    var retailer_model = new RetailerModel();
    var retailer_view = new RetailerView({model: retailer_model});
    form.reset();
});

接收数据的php代码如下：

<?php
$connect = mysql_connect('127.0.0.1','root','xxxxxx');
if (!$connect) {
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("retail", $connect);
if($_SERVER['REQUEST_METHOD'] == 'POST') //POST GET PUT DELETE
{
    $data = json_decode(file_get_contents('php://input'), true);
}

$name       = $data['name'];
$address    = $data['address'];
$phone      = $data['phone'];
$icon       = $data['icon'];

if(!(mysql_query("INSERT INTO retailer (name, address, phone, icon)VALUES ('".$name."', '".$address."','$phone', '".$icon."')")))
{

    echo 200;
}
else{
    echo 'record has not been insert to db';
}

mysql_close($connect);

?>

来自

如果要在设置之前等待服务器，请传递{wait:true} 模型上的新属性

如果在保存成功之前不想更新模型，请选择全通

等待：true

this.model.save(obj, {
        success: function(model, response) {
            console.log(model);
            console.log(response);
            o.render();
            console.log('success');
        },
        error: function(model, response) {
            console.log(model);
        },
        wait: true // Add this
    });

骨干

save( so are others like fetch, update...) 

回报一个承诺。你可以用

save().done( 
    function( data ) {}, // success
    function( err ) {}   // fail
)

就像你如何处理承诺一样。

done（）

方法保证在服务器返回内容后执行

---

有关更多信息，请参阅。主干网返回承诺

这就是我要让它工作的地方

save({wait: true}).success(function(data){
   console.log(data); //returned data from server
}).error(function(error){
   console.log(error); //error returned from server
});

---

.done

不是

.success

的同义词吗？好像你在找

。总是