在documentation之后,我做了:
var collection = new Backbone.Collection.extend({
model: ItemModel,
url: '/Items'
})
collection.fetch({ data: { page: 1} });
网址原来是:http://localhost:1273/Items?[object%20Object]
我期待http://localhost:1273/Items?page=1
那么如何在fetch方法中传递params?
答案 0 :(得分:212)
改变:
collection.fetch({ data: { page: 1} });
到:
collection.fetch({ data: $.param({ page: 1}) });
因此,如果不执行此操作,则会将{data: {page:1}}
对象调用为options
Backbone.sync = function(method, model, options) {
var type = methodMap[method];
// Default JSON-request options.
var params = _.extend({
type: type,
dataType: 'json',
processData: false
}, options);
// Ensure that we have a URL.
if (!params.url) {
params.url = getUrl(model) || urlError();
}
// Ensure that we have the appropriate request data.
if (!params.data && model && (method == 'create' || method == 'update')) {
params.contentType = 'application/json';
params.data = JSON.stringify(model.toJSON());
}
// For older servers, emulate JSON by encoding the request into an HTML-form.
if (Backbone.emulateJSON) {
params.contentType = 'application/x-www-form-urlencoded';
params.processData = true;
params.data = params.data ? {model : params.data} : {};
}
// For older servers, emulate HTTP by mimicking the HTTP method with `_method`
// And an `X-HTTP-Method-Override` header.
if (Backbone.emulateHTTP) {
if (type === 'PUT' || type === 'DELETE') {
if (Backbone.emulateJSON) params.data._method = type;
params.type = 'POST';
params.beforeSend = function(xhr) {
xhr.setRequestHeader('X-HTTP-Method-Override', type);
};
}
}
// Make the request.
return $.ajax(params);
};
因此,它会将“数据”发送给jQuery.ajax,这会尽力将params.data
附加到网址。
答案 1 :(得分:71)
您还可以将processData设置为true:
collection.fetch({
data: { page: 1 },
processData: true
});
Jquery会自动将数据对象处理为param字符串,
但在Backbone.sync函数中, Backbone将关闭processData,因为Backbone将使用其他方法来处理数据 在POST,UPDATE ......
在Backbone来源:
if (params.type !== 'GET' && !Backbone.emulateJSON) {
params.processData = false;
}
答案 2 :(得分:1)
如果您使用钛合金的另一个例子:
collection.fetch({
data: {
where : JSON.stringify({
page: 1
})
}
});
答案 3 :(得分:-2)
try {
// THIS for POST+JSON
options.contentType = 'application/json';
options.type = 'POST';
options.data = JSON.stringify(options.data);
// OR THIS for GET+URL-encoded
//options.data = $.param(_.clone(options.data));
console.log('.fetch options = ', options);
collection.fetch(options);
} catch (excp) {
alert(excp);
}