我对存储过程相当新。我天真地认为我可以建立一个如下的选择语句。我不能,而且你们中的一些人会对我提出的事情嗤之以鼻。
我怎么做我想做的事情呢?
提前致谢。
CREATE PROCEDURE GET_CUSTOMER_FOR_BORROWER_LETTER (
IN APPLICATION_ID INTEGER,
IN GET_GUARANTOR INTEGER,
IN GET_PREFERRED_CONTACT INTEGER
)
DYNAMIC RESULT SETS 1
READS SQL DATA
P1:BEGIN
DECLARE selectStmt VARCHAR(800);
DECLARE selectStmtPreferred VARCHAR(400);
DECLARE selectStmtApplicants VARCHAR(400);
DECLARE selectStmtGuarantor VARCHAR(400);
DECLARE cursor1 CURSOR WITH RETURN FOR
selectStmt -- will define this later, conditionally (babe in the woods :) )
OPEN cursor1;
set selectStmtPreferred = 'select "preferred applicant" as recipient_type, app.APPLICATION_ID, cust.KEY from application app, customer cust, application_detail appd where app.application_id = 407634 and app.APPLICATION_ID = appd.APPLICATION_ID and appd.PREFERRED_CONTACT_ID = cust.KEY';
set selectStmtApplicants = 'select "applicant" as recipient_type, app.APPLICATION_ID, cust.KEY from application app, applicant applc, customer cust where app.application_id = 407634 and applc.APPLICATION_ID = app.APPLICATION_ID and applc.CUST_ID = cust.CUST_ID';
set selectStmtGuarantor = ' union select "guarantor" as recipient_type ,app.APPLICATION_ID, cust.KEY from application app, application_guarantor appg, customer cust where app.application_id = 407634 and appg.APPLICATION_ID = app.APPLICATION_ID and appg.CUST_ID = cust.CUST_ID';
IF GET_PREFERRED_CONTACT = 1 THEN
IF GET_GUARANTOR = 1 THEN
SET selectStmt = concat (selectStmtPreferred,selectStmtGuarantor);
ELSE
SET selectStmt = selectStmtPreferred;
END IF;
ELSE
IF GET_GUARANTOR = 1 THEN
SET selectStmt = concat (selectStmtApplicants,selectStmtGuarantor);
ELSE
SET selectStmt = selectStmtApplicants;
END IF;
END IF;
selectStmt = concat (selectStmtPreferred,";");
END P1@
答案 0 :(得分:3)
下次需要构建一些动态SQL语句时尝试一下。
DECLARE SELECT_STATEMENT VARCHAR(8000);
DECLARE cursor1 CURSOR WITH RETURN FOR SQL_STATEMENT;
...build dynamic sql here...
PREPARE SQL_STATEMENT FROM SELECT_STATEMENT;
OPEN cursor1;
要明确,SQL_STATEMENT是您想要的任何名称。只需确保它在CURSOR声明和PREPARE语句中是相同的。
答案 1 :(得分:2)
我解决了这个问题。它很难看,但它已经解决了。
P1:BEGIN
DECLARE preferredWithGuarantor CURSOR WITH RETURN FOR
select 'preferred applicant' as recipient_type, app.APPLICATION_ID, cust.KEY from application app, customer cust, application_detail appd where app.application_id = 407634 and app.APPLICATION_ID = appd.APPLICATION_ID and appd.PREFERRED_CONTACT_ID = cust.KEY union select 'guarantor' as recipient_type ,app.APPLICATION_ID, cust.KEY from application app, application_guarantor appg, customer cust where app.application_id = 407634 and appg.APPLICATION_ID = app.APPLICATION_ID and appg.CUST_ID = cust.CUST_ID;
DECLARE preferred CURSOR WITH RETURN FOR
select 'preferred applicant' as recipient_type, app.APPLICATION_ID, cust.KEY from application app, customer cust, application_detail appd where app.application_id = 407634 and app.APPLICATION_ID = appd.APPLICATION_ID and appd.PREFERRED_CONTACT_ID = cust.KEY;
DECLARE applicantWithGuarantor CURSOR WITH RETURN FOR
select 'applicant' as recipient_type, app.APPLICATION_ID, cust.KEY from application app, applicant applc, customer cust where app.application_id = 407634 and applc.APPLICATION_ID = app.APPLICATION_ID and applc.CUST_ID = cust.CUST_ID union select 'guarantor' as recipient_type ,app.APPLICATION_ID, cust.KEY from application app, application_guarantor appg, customer cust where app.application_id = 407634 and appg.APPLICATION_ID = app.APPLICATION_ID and appg.CUST_ID = cust.CUST_ID;
DECLARE applicant CURSOR WITH RETURN FOR
select 'applicant' as recipient_type, app.APPLICATION_ID, cust.KEY from application app, applicant applc, customer cust where app.application_id = 407634 and applc.APPLICATION_ID = app.APPLICATION_ID and applc.CUST_ID = cust.CUST_ID;
IF GET_PREFERRED_CONTACT = 1 THEN
IF GET_GUARANTOR = 1 THEN
open preferredWithGuarantor;
ELSE
open preferred;
END IF;
ELSE
IF GET_GUARANTOR = 1 THEN
open applicantWithGuarantor;
ELSE
open applicant;
END IF;
END IF;
END P1@