Backbone保存模型成功与错误

Question

我有这个骨干代码,创建一个视图和模型,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();
});

用于接收数据的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);

?>

我遇到的一个问题是,当调用错误函数时,从服务器返回的模型仍然具有修改的属性.我想知道为什么会发生这种情况,我怎样才能确保如果发生错误,模型保持不变.

另一个问题是在php代码中,当sql查询成功时,如果我回显200或'200',主干会调用成功,但如果我回显一个字符串,主干会调用错误,我想知道背后的逻辑是什么它.

Answer 1

从骨干文档:

如果您想在模型上设置新属性之前等待服务器,请传递{wait:true}.

如果您不希望模型更新,直到保存成功完全传递wait: true为选项.

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
    });

Answer 2

骨干

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

返回一个承诺.您可以使用

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

就像你处理承诺一样.done()保证在服务器返回内容后执行该方法.

有关更多信息,请参阅AJAX.jqXHR的jQuery API文档.