JAVA/API 如何获得正确的 JSON 格式

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英文:

JAVA/API how to get the correct JSON format

问题

  1. {
  2.   "usuario": {
  3.     "ID_USUARIO": 0,
  4.     "CORREO": "maaridano12345@CORREO.com"
  5.   },
  6.   "persona": {
  7.     "ID_USUARIO": 0,
  8.     "NOMBRE_COMPLETO": "MaARIANO2 GARFEL",
  9.     "domicilios": [
  10.       {
  11.         "COLONIA": "CaOLONIA23",
  12.         "CALLE": "SaTREET"
  13.       }
  14.     ]
  15.   },
  16.   "asociado": {
  17.     "FOTO": "",
  18.     "ID_USUARIO": 0,
  19.     "NOMBRE_EMPRESA": "EMPRESAMARIANO2",
  20.     "tipoPagos": [
  21.       {
  22.         "ID_TIPO_PAGO": 1
  23.       }
  24.     ],
  25.     "SERVICIOS": [
  26.       {
  27.         "ID_SERVICIO": 2
  28.       }
  29.     ]
  30.   }
  31. }

Code to add Objects:

  1. usuario = new User(0,emailInput,passwordInput,nameInput,false,false,true,false);
  2.     
  3. ArrayList<Adress> Domicilios = new ArrayList<Adress>();
  4. domicilio = new Adress("MONARCAS","CALLETEST",34,12,83550,1,0,2);
  5. Domicilios.add(domicilio);
  7. persona = new Person(0, "MARIANO GARFEL","MARIANO","GARFEL","GARCIA", "9876347586","8575757575",
  8.                 "GARFSONAL@GMAIL.COM","CORREO2@FDS.COM", Domicilios);
  10. ArrayList<Services> Servicios = new ArrayList<Services>();
  11. tipo_servicio = new Services(2);
  12. Servicios.add(tipo_servicio);
  13. ArrayList<PaymentType> TipoPago = new ArrayList<PaymentType>();
  14. tipo_pago = new PaymentType(2);
  15. TipoPago.add(tipo_pago);
  16. asociado = new Associate("",0,"LaEmpresaChida",0,0,"La mejor empresa",2, Servicios, TipoPago );
  18. ArrayList<mRegistro> RegistroAsociado = new ArrayList<mRegistro>();
  19. ArrayList<Associate> asociadoList = new ArrayList<Associate>();
  20. asociadoList.add(asociado);
  21. ArrayList<Person> personList = new ArrayList<Person>();
  22. personList.add(persona);
  23. ArrayList<User> userList = new ArrayList<User>();
  24. userList.add(usuario);
  25. Registro = new mRegistro(userList, personList, asociadoList);
英文:

Good afternoon, my question may be very simple for some, but I am learning and I cannot get a good result.

I have an API which requires obtaining a json object with the following structure:

  1. {
  2. &quot;usuario&quot;: {
  3.     &quot;ID_USUARIO&quot;: 0,
  4.     &quot;CORREO&quot;: &quot;maaridano12345@CORREO.com&quot;,
  6.     
  7. },
  8. &quot;persona&quot;: {
  9.     &quot;ID_USUARIO&quot;: 0,
  10.     &quot;NOMBRE_COMPLETO&quot;: &quot;MaARIANO2 GARFEL&quot;,
  11.     &quot;domicilios&quot; : [{
  12.             &quot;COLONIA&quot; : &quot;CaOLONIA23&quot;,
  13.             &quot;CALLE&quot; : &quot;SaTREET&quot;,
  14.     }]
  16. },
  17. &quot;asociado&quot;: {
  18.     &quot;FOTO&quot;: &quot;&quot;,
  19.     &quot;ID_USUARIO&quot;: 0,
  20.     &quot;NOMBRE_EMPRESA&quot;: &quot;EMPRESAMARIANO2&quot;,
  21.     &quot;tipoPagos&quot; : [{
  22.     &quot;ID_TIPO_PAGO&quot; : 1
  23.     }],
  24.     &quot;SERVICIOS&quot; : [{
  25.     &quot;ID_SERVICIO&quot; : 2
  26.     }]
  27. }
  28. }

but, when I create my JSON object:

  1. Registro = new mRegistro(userList, personList, asociadoList);
  3.     Gson gson = new Gson();
  4.     json = gson.toJson(Registro);

the structure that creates me is the following

  1. {
  2. &quot;usuario&quot;: [{
  3.     &quot;ID_USUARIO&quot;: 0,
  4.     &quot;CORREO&quot;: &quot;maaridano12345@CORREO.com&quot;,
  6.     
  7. }],
  8. &quot;persona&quot;: [{
  9.     &quot;ID_USUARIO&quot;: 0,
  10.     &quot;NOMBRE_COMPLETO&quot;: &quot;MaARIANO2 GARFEL&quot;,
  11.     &quot;domicilios&quot; : [{
  12.             &quot;COLONIA&quot; : &quot;CaOLONIA23&quot;,
  13.             &quot;CALLE&quot; : &quot;SaTREET&quot;,
  14.     }]
  16. }],
  17. &quot;asociado&quot;: [{
  18.     &quot;FOTO&quot;: &quot;&quot;,
  19.     &quot;ID_USUARIO&quot;: 0,
  20.     &quot;NOMBRE_EMPRESA&quot;: &quot;EMPRESAMARIANO2&quot;,
  21.     &quot;tipoPagos&quot; : [{
  22.     &quot;ID_TIPO_PAGO&quot; : 1
  23.     }],
  24.     &quot;SERVICIOS&quot; : [{
  25.     &quot;ID_SERVICIO&quot; : 2
  26.     }]
  27. }]
  28. }

I want the same structure but in the User, Person and Associate parameters I do not want it to have the square bracket, I know that it is there because it is a list, it is like this because my class "mRegistro" contains the classes "usuario, persona and asociado"

Code to add Objects:

  1. usuario = new User(0,emailInput,passwordInput,nameInput,false,false,true,false);
  3.     ArrayList&lt;Adress&gt; Domicilios = new ArrayList&lt;Adress&gt;();
  4.     domicilio = new Adress(&quot;MONARCAS&quot;,&quot;CALLETEST&quot;,34,12,83550,1,0,2);
  5.     Domicilios.add(domicilio);
  7.     persona = new Person(0, &quot;MARIANO GARFEL&quot;,&quot;MARIANO&quot;,&quot;GARFEL&quot;,&quot;GARCIA&quot;, &quot;9876347586&quot;,&quot;8575757575&quot;,
  8.             &quot;GARFSONAL@GMAIL.COM&quot;,&quot;CORREO2@FDS.COM&quot;, Domicilios);
  10.     ArrayList&lt;Services&gt; Servicios = new ArrayList&lt;Services&gt;();
  11.     tipo_servicio = new Services(2);
  12.     Servicios.add(tipo_servicio);
  13.     ArrayList&lt;PaymentType&gt; TipoPago = new ArrayList&lt;PaymentType&gt;();
  14.     tipo_pago = new PaymentType(2);
  15.     TipoPago.add(tipo_pago);
  16.     asociado = new Associate(&quot;&quot;,0,&quot;LaEmpresaChida&quot;,0,0,&quot;La mejor empresa&quot;,2, Servicios, TipoPago );
  18.     ArrayList&lt;mRegistro&gt; RegistroAsociado = new ArrayList&lt;mRegistro&gt;();
  19.     ArrayList&lt;Associate&gt; asociadoList = new ArrayList&lt;Associate&gt;();
  20.     asociadoList.add(asociado);
  21.     ArrayList&lt;Person&gt; personList = new ArrayList&lt;Person&gt;();
  22.     personList.add(persona);
  23.     ArrayList&lt;User&gt; userList = new ArrayList&lt;User&gt;();
  24.     userList.add(usuario);
  25.     Registro = new mRegistro(userList, personList, asociadoList);

答案1

得分: 0

看一下这段代码：

  1. ArrayList<Associate> asociadoList = new ArrayList<Associate>();
  2. asociadoList.add(asociado);
  4. ArrayList<Person> personList = new ArrayList<Person>();
  5. personList.add(persona);
  7. ArrayList<User> userList = new ArrayList<User>();
  8. userList.add(usuario);
  10. Registro = new mRegistro(userList, personList, asociadoList);

你在这里创建了不必要的 ArrayLists，这就是为什么在你的 Json 输出中会得到列表对象。

你需要做的是将上面的代码更改为：

  1. Registro = new mRegistro(usuario, persona, asociado);

但是，这样做也会导致错误，因为 mRegistro 类需要 ArrayLists 作为参数，所以将其构造函数的参数从

  1. mRegistro(ArrayList<User> userList, ArrayList<Person> personList, ArrayList<Associate> asociadoList)

更改为

  1. mRegistro(User user, Person person, Associate asociado)
英文:

See this code:

  1. ArrayList&lt;Associate&gt; asociadoList = new ArrayList&lt;Associate&gt;();
  2. asociadoList.add(asociado);
  4. ArrayList&lt;Person&gt; personList = new ArrayList&lt;Person&gt;();
  5. personList.add(persona);
  7. ArrayList&lt;User&gt; userList = new ArrayList&lt;User&gt;();
  8. userList.add(usuario);
  10. Registro = new mRegistro(userList, personList, asociadoList);

You're creating unnecessary ArrayLists here because of which you're getting list object in your Json output.

What you've to do here is change above code as:

  1. Registro = new mRegistro(usuario, persona, asociado);

But, this will also give an error as the Class mRegistro needs ArrayLists as parameters so change its constructor's arguments from

  1. mRegistro(ArrayList&lt;User&gt; userList, ArrayList&lt;Person&gt; personList, ArrayList&lt;Associate&gt; asociadoList)

to mRegistro(User user, Person person, Associate asociado).

huangapple
  • 本文由 发表于 2020年8月25日 01:04:38
  • 转载请务必保留本文链接：https://go.coder-hub.com/63565559.html
  • android
  • api
  • java
  • json
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