Question

我有一个名称和地址表，想返回一个完整的名称和地址列表，并带有一个额外的列，用于标识该地址是重复地址还是唯一地址。运行更简单的语句时，我可以看到几个重复的地址条目-

SELECT PersonAddressLine1, Count (*)
FROM CompanyTable1
GROUP BY PersonAddressLine1

或

SELECT PersonAddressLine1, Count (*)
FROM CompanyTable1
GROUP BY PersonAddressLine1
HAVING COUNT (*) > 1

但是我需要返回带有列的完整表-

PersonID, PersonName, PersonAddressLine1, AddressVerification (Duplicate / Unique), CompanyID, CompanyName.

PersonID - 1
PersonID - 2
PersonID - 3
PersonName - Ryan
PersonName - Andrew
PersonName - Ben
PersonAddressLine1 - 100 Avenue
PersonAddressLine1 - 100 Avenue
PersonAddressLine1 - 200 Avenue

如果我使用以下case语句，则组条件会影响PersonAddressLine1的计数结果。它显示所有地址都是唯一的。

SELECT PersonID, PersonName, PersonAddressLine1, CompanyID, CompanyName, 

CASE WHEN count(*) > 1 
THEN CONCAT(PersonName,' ','Address Line 1 is duplicate')
    ELSE CONCAT(PersonName,' ','Address Line 1 is unique')
END AS 'Person Address Verification'

FROM CompanyTable1

GROUP by PersonAddressLine1, PersonID, PersonName, CompanyID, CompanyName

Answer 1

您可以使用窗口函数public class TestRules { static KieContainer kieContainer; StatelessKieSession sessionStateless = null; KieSession sessionStatefull = null; @BeforeClass public static void beforeClass() { kieContainer = KnowledgeSessionHelper.createRuleBase(); } @Test public void testCuentasCoactivoLimite(){ sessionStatefull= KnowledgeSessionHelper .getStatefulKnowledgeSession(kieContainer,"ksession-rules"); Data datos=new Data(); sessionStatefull.setGlobal("datos", datos); Embargo embargo=new Embargo("e1",LocalDate.of(2018, 11, 24),TipoEmbargo.COACTIVO,new BigDecimal(18000000)); sessionStatefull.insert(embargo); ArrayList<Persona> personas=new ArrayList<>(); personas.add(new Persona("p1", "e1",TipoIdentificacion.NATURAL, new BigDecimal(6000000))); personas.add(new Persona("p2", "e1",TipoIdentificacion.NATURAL, new BigDecimal(6000000))); personas.add(new Persona("p3", "e1",TipoIdentificacion.JURIDICA, new BigDecimal(6000000))); personas.stream().forEach(x->sessionStatefull.insert(x)); ArrayList<Cuenta> cuentas=new ArrayList<>(); //p1 cuentas.add(new Cuenta("c1","p1","e1",TipoCuenta.AHORROS,SubtipoCuenta.BASICA,LocalDate.of(2014, 05, 26), new BigDecimal(17000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c2","p1","e1",TipoCuenta.AHORROS,SubtipoCuenta.BASICA,LocalDate.of(2015, 05, 26), new BigDecimal(3000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c3","p1","e1",TipoCuenta.CORRIENTE,SubtipoCuenta.BASICA,LocalDate.of(2016, 05, 26), new BigDecimal(2000000),EstadoCuenta.ACTIVA)); //p2 cuentas.add(new Cuenta("c4","p2","e1",TipoCuenta.AHORROS,SubtipoCuenta.BASICA,LocalDate.of(2016, 05, 26), new BigDecimal(3000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c5","p2","e1",TipoCuenta.CDT,SubtipoCuenta.BASICA,LocalDate.of(2016, 05, 26), new BigDecimal(1000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c6","p2","e1",TipoCuenta.CDAT,SubtipoCuenta.BASICA,LocalDate.of(2016, 05, 26), new BigDecimal(1000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c7","p2","e1",TipoCuenta.ELECTRONICOS,SubtipoCuenta.PUBLICO,LocalDate.of(2016, 05, 26), new BigDecimal(1000000),EstadoCuenta.ACTIVA)); //p3 cuentas.add(new Cuenta("c8","p3","e1",TipoCuenta.AHORROS,SubtipoCuenta.BASICA,LocalDate.of(2014, 05, 26), new BigDecimal(37000000),EstadoCuenta.ACTIVA)); cuentas.add(new Cuenta("c9","p3","e1",TipoCuenta.AHORROS,SubtipoCuenta.BASICA,LocalDate.of(2015, 05, 26), new BigDecimal(3000000),EstadoCuenta.ACTIVA)); cuentas.stream().forEach(x->sessionStatefull.insert(x)); sessionStatefull.fireAllRules(); imprimir(embargo, personas, cuentas); }来计算重复次数

COUNT(*) OVER (PARTITION BY PersonAddressLine1)

您当前的查询现在正在做什么，不仅在检查SELECT PersonID, PersonName, PersonAddressLine1, CompanyID, CompanyName, CASE WHEN COUNT(*) OVER (PARTITION BY PersonAddressLine1) > 1 THEN CONCAT(PersonName,' ','Address Line 1 is duplicate') ELSE CONCAT(PersonName,' ','Address Line 1 is unique') END AS 'Person Address Verification' FROM CompanyTable1，而且还检查了以下所有内容 PersonAddressLine1

如何编写案例声明以返回重复记录？

1 个答案: