小编Siq*_*ira的帖子

我的Oracle 12c数据库中有一个表

XML Schema创建:

BEGIN
-- Register the schema
DBMS_XMLSCHEMA.registerSchema('http://www.example.com/fvInteger_12.xsd',
'<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="FeatureVector">
<xs:complexType>
<xs:sequence>
<xs:element name="feature" type="xs:integer" minOccurs="12" maxOccurs="12"/>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>',
   TRUE, TRUE, FALSE);
END;
/

创建表:

CREATE TABLE fv_xml_12_1000 (
   id    NUMBER,
   fv  XMLTYPE)
   XMLTYPE fv STORE AS OBJECT RELATIONAL
      XMLSCHEMA "http://www.example.com/fvInteger_12.xsd"
      ELEMENT "FeatureVector";

表DDL:

SELECT 
DBMS_METADATA.GET_DDL( 'TABLE','FV_XML_12_1000') 
FROM DUAL;

上面的查询结果:

  CREATE TABLE "HIGIIA"."FV_XML_12_1000"
   (    "ID" NUMBER,
    "FV" "SYS"."XMLTYPE"
   ) SEGMENT CREATION IMMEDIATE
  PCTFREE 10 PCTUSED 40 INITRANS 1 MAXTRANS 255
 NOCOMPRESS LOGGING
  STORAGE(INITIAL 65536 …

我正在尝试通过使用 Laravel 构建的 Web 服务中的表单插入数据（日期类型数据）。

slicitacoes.blade.php 代码：

    @extends('layout.app', ["current" => "solicitacaos"])

@section('body')

<div class="container">
<div class="card-border">
    <div class="card-body">
        <h3 class="card-title">Solicitações</h3>
        <table class="table table-striped table-bordered table-hover">
                    <tr>
                        <th> Professor
                        </th>
                        <th>
                            Criação
                        </th>
                        <th>
                            Data Solicitação
                        </th>
                        <th>
                            Nome do Produto
                        </th> 
                        <th>
                            Observações
                        </th>
                        <th>
                            Status
                        </th>
                        <th>
                            Ação
                        </th>           

                    </tr>

                    @foreach($sols as $sol)

                    @php

                        dd($sol)

                    @endphp


                    <tr>
                        <td> {{$sol->nome_professor}}
                        </td>
                        <td> {{$sol->criacao}}
                        </td>
                        <td> {{dd($sol)}}
                        </td>
                        <td> {{$sol->nome_produto}}
                        </td>
                        <td> {{$sol->observacao}}
                        </td>
                        <td> {{$sol->status}}
                        </td>
                        <td>
                            <a href="/solicitacao/apagar/{{$sol->id}}" class="btn …

此链接显示了如何在 Oracle 中获取过程/函数变量的类型：查看变量的类型。

它通过函数“get_plsql_type_name”来实现：

create or replace function get_plsql_type_name
(
    p_object_name varchar2,
    p_name varchar2
) return varchar2 is
    v_type_name varchar2(4000);
begin
    select reference.name into v_type_name
    from user_identifiers declaration
    join user_identifiers reference
        on declaration.usage_id = reference.usage_context_id
        and declaration.object_name = reference.object_name
    where
        declaration.object_name = p_object_name
        and declaration.usage = 'DECLARATION'
        and reference.usage = 'REFERENCE'
        and declaration.name = p_name;

    return v_type_name;
end;
/

alter session set plscope_settings = 'IDENTIFIERS:ALL';

create or replace type my_weird_type is object
(
    a number
);

create or …

我在我的应用程序中添加了使用firebase身份验证的facebook登录,但登录身份验证仅适用于Android API 24.

当我尝试使用较低级别的API时,屏幕会冻结,应用程序将停止工作.

它太高,大多数用户的API级别较低.

实际上,我打算发布较低API级别的应用程序.

我正在使用的代码如下.

public class ConfigurationActivity extends Activity {

    Button b_v_mm,b_som;
    LoginButton loginButton;
    CallbackManager callbackManager;

    // [START declare_auth]
    private FirebaseAuth mAuth;
    // [END declare_auth]


    private void handleFacebookAccessToken(AccessToken token) {
        Log.d("FireBase", "handleFacebookAccessToken:" + token);

        AuthCredential credential = FacebookAuthProvider.getCredential(token.getToken());
        mAuth.signInWithCredential(credential)
                .addOnCompleteListener(this, new OnCompleteListener<AuthResult>() {
                    @Override
                    public void onComplete(@NonNull Task<AuthResult> task) {
                        if (task.isSuccessful()) {
                            // Sign in success, update UI with the signed-in user's information
                            Log.d("FireBase", "signInWithCredential:success");
                            FirebaseUser user = mAuth.getCurrentUser();
                            //updateUI(user);
                        } else {
                            // …

我在 Oracle 模式中有一个类型层次结构：

CREATE OR REPLACE TYPE FV AS OBJECT (
   idno           NUMBER)
NOT FINAL;
/

CREATE TYPE FV_Integer UNDER FV (
   features INTEGER_ARRAY)
   NOT FINAL;
/


CREATE TYPE FV_Number UNDER FV (
   features NUMBER_ARRAY)
   NOT FINAL;
/

我想构建一个 PLSQL 函数来验证哪种类型的层次结构是对象：对于函数虚拟（obj1 FV，obj2 FV）...我如何检查用户正在使用的层次结构的对象类型是什么？

例如，我想打印对象类型名称（该函数用于说明，它不是真正的 pl/sql 代码）：

 dummy(obj1 FV, obj2 FV){
      if obj1%type = FV_INTEGER
          THEN print 'FV_INTEGER'
      endif
      if obj2%type = FV_NUMBER
          THEN print 'FV_NUMBER'
      endif
}

我正在尝试使用rowid数据类型创建一个类型，但是由于要使用的类型而出现此错误：

SQL> CREATE TYPE join_t IS OBJECT (inn  rowid,    out rowid ); 
/

Warning: Type created with compilation errors.

即使我可以使用rowid数据类型创建表：

SQL> create table test_rowid (inn rowid,out rowid);

Table created.

是否可以在join_t上面使用rowid-type属性创建此类型？