在本文中，我们将给您介绍关于LaravelHTTP测试使用getJson+assertJsonStructure对象数组的详细内容，并且为您解答laravel获取post参数的相关问题，此外，我们还将

在本文中，我们将给您介绍关于Laravel HTTP 测试使用 getJson + assertJsonStructure 对象数组的详细内容，并且为您解答laravel获取post参数的相关问题，此外，我们还将为您提供关于.NET 3.5: 使用DataContractJsonSerializer进行JSON 序列化、c# – DataContractJsonSerializer和JsonConvert给出不同的结果、c# – 使用DataContractJsonSerializer设置JSON对象root、com.alibaba.fastjson JSONObject toJSONString 时出现数组转换错误的知识。

• Laravel HTTP 测试使用 getJson + assertJsonStructure 对象数组（laravel获取post参数）
• .NET 3.5: 使用DataContractJsonSerializer进行JSON 序列化
• c# – DataContractJsonSerializer和JsonConvert给出不同的结果
• c# – 使用DataContractJsonSerializer设置JSON对象root
• com.alibaba.fastjson JSONObject toJSONString 时出现数组转换错误

Laravel HTTP 测试使用 getJson + assertJsonStructure 对象数组（laravel获取post参数）

data 是一个 array，因此，要断言嵌套对象的结构，我们需要使用 * 通配符：

$response->assertJsonStructure([
   'data' => [
         '*' => [
            'createdAt','endAt','name','startAt','startedAt','status','updatedAt','range' => [
                'max','min','type'
            ],]
      ]
]);

.NET 3.5: 使用DataContractJsonSerializer进行JSON 序列化

InASP.NET AJAX Extensions v1.0 for ASP.NET 2.0 there is the JavaScriptSerializer class that provides JSONserialization and deserialization functionality. However,in .NET 3.5 the JavaScriptSerializer has been marked obsolete. The new object to use for JSON serialization in .NET 3.5 is the DataContractJsonSerliaizer object. I'm still new to the DataContractJsonSerializer,but here's a summary of what I've learned so far...

Object to Serialize

Here's a simple Person object with First Name and Last Name properties.

Now in orderto serialize our objectto JSONusing the DataContractJsonSerializer wemust either mark it with the Serializable attribute or the DataContract attribute. If we mark the class with the DataContract attribute,then we must mark each property we want serialized with the DataMember attribute.

Serialization Code

Jere's the most basic code to serialize our object to JSON:

Our resulting JSON looks like this:

As you can see the first serialization with the class markedwith theSerializable attribute isn't quite what we were expecting,but is still JSON.This serialization actually isn't compatible with the client-side JSON Serializer in ASP.NET AJAX.

As you can see the second serialization with the class marked with the DataContract attribute is exactly what we were expecting,and is the same JSON that the old JavaScriptSerializer object would have generated. This is the method of JSON serialization using the DataContractJsonSerializer that you'll need to do if you are going to pass the resulting JSON down to the client to be consumed with ASP.NET AJAX.

Deserialization Code

Here's the most basic code to deserialize our object from JSON:

Controlling the property names in the resulting JSON

When using the DataContract and DataMember attributes,you can tell the DataMember attribute the specific name you want that property to have within the JSON serialization by setting its "Name" named parameter.

Here's an example that will give the name of "First" to the "FirstName" property within the JSON serialization:

The resulting JSON looks like this:

Here's the code for some Helper methods using Generics to do all the dirty work for you

What Assembly References does my application need for this?

From the namespace that containsDataContractJsonSerializer you can probably tell that you need to add a reference to the System.Runtime.Serialization assembly. However,you also need to add a reference to the System.ServiceModel.Web assembly.

c# – DataContractJsonSerializer和JsonConvert给出不同的结果

using (MemoryStream memoryStream = new MemoryStream())
{
    DataContractJsonSerializer dataContractSerializer = new DataContractJsonSerializer(typeof(Message),this.kNowTypes);
    dataContractSerializer.WriteObject(memoryStream,message);

    byte[] byteArray = memoryStream.ToArray();
    memoryStream.Close();
    return byteArray;
}

byte[] byteArray = System.Text.Encoding.UTF8.GetBytes(JsonConvert.SerializeObject(message));
return byteArray;

settings = new JsonSerializerSettings();
 settings.TypeNameAssemblyFormat = System.Runtime.Serialization.Formatters.FormatterassemblyStyle.Full;
 settings.TypeNameHandling = TypeNameHandling.Objects;

DataContractJsonSerializerSettings settings = new DataContractJsonSerializerSettings()
{
    EmitTypeinformation = EmitTypeinformation.Never
};

c# – 使用DataContractJsonSerializer设置JSON对象root

[DataContract]
public class Person
{
    public Person() { }
    public Person(string firstname,string lastname)
    {
        this.FirstName = firstname;
        this.LastName = lastname;
    }

    [DataMember]
    public string FirstName { get; set; }

    [DataMember]
    public string LastName { get; set; }
}

public static string Serialize<T>(T obj)
{
    Json.DataContractJsonSerializer serializer = 
        new DataContractJsonSerializer(obj.GetType());
    MemoryStream ms = new MemoryStream();
    serializer.WriteObject(ms,obj);
    string retVal = Encoding.Default.GetString(ms.ToArray());
    ms.dispose();
    return retVal;
}

{"FirstName":"Jane","LastName":"McDoe"}

{Person: {"FirstName":"Jane","LastName":"McDoe"}}

string retVal = "{Person:" + Encoding.Default.GetString(ms.ToArray()) + "}";

public class StackOverflow_7930629
{
    [DataContract]
    public class Person
    {
        public Person() { }
        public Person(string firstname,string lastname)
        {
            this.FirstName = firstname;
            this.LastName = lastname;
        }

        [DataMember]
        public string FirstName { get; set; }

        [DataMember]
        public string LastName { get; set; }
    }

    public static string Serialize<T>(T obj)
    {
        DataContractJsonSerializer serializer =
            new DataContractJsonSerializer(typeof(T),typeof(T).Name);
        MemoryStream ms = new MemoryStream();
        XmlDictionaryWriter w = JsonReaderWriterFactory.CreateJsonWriter(ms);
        w.WriteStartElement("root");
        w.WriteAttributeString("type","object");
        serializer.WriteObject(w,obj);
        w.WriteEndElement();
        w.Flush();
        string retVal = Encoding.Default.GetString(ms.ToArray());
        ms.dispose();
        return retVal;
    }
    public static void test()
    {
        Console.WriteLine(Serialize(new Person("Jane","McDoe")));
    }
}

com.alibaba.fastjson JSONObject toJSONString 时出现数组转换错误

JSONObject.toJSONString(vo); 改为 JSONObject.toJSON(templateCreateVo).toString();

今天关于Laravel HTTP 测试使用 getJson + assertJsonStructure 对象数组和laravel获取post参数的介绍到此结束，谢谢您的阅读，有关.NET 3.5: 使用DataContractJsonSerializer进行JSON 序列化、c# – DataContractJsonSerializer和JsonConvert给出不同的结果、c# – 使用DataContractJsonSerializer设置JSON对象root、com.alibaba.fastjson JSONObject toJSONString 时出现数组转换错误等更多相关知识的信息可以在本站进行查询。