我想从表中选择,其中表名是一个局部变量,列名是另一个局部变量,但我得到以下错误:
指定的表ll_tb_name
不在数据库中
CREATE PROCEDURE rmg_request_file ( al_task_code integer)
RETURNING integer as byte;
DEFINE ll_task_code integer;
DEFINE ll_pdf_column varchar(50);
DEFINE ll_tb_name varchar(60);
DEFINE ll_request_file REFERENCES BYTE;
SELECT pdf_column,table_name INTO ll_pdf_column ,ll_tb_name FROM rnmtask WHERE task_code = al_task_code;
FOREACH
SELECT ll_pdf_column into ll_request_file FROM ll_tb_name
return ll_request_file with resume ;
end foreach;
end procedure
最佳答案
假设您有一个足够新的informix(11.70)版本,那么您应该能够在spl中使用动态sql,如下所示:
BEGIN;
CREATE TABLE rnmtask
(
pdf_column VARCHAR(32) NOT NULL,
table_name VARCHAR(32) NOT NULL,
task_code INTEGER NOT NULL PRIMARY KEY
);
INSERT INTO rnmtask VALUES("symbol", "elements", 1);
INSERT INTO rnmtask VALUES("name", "elements", 2);
INSERT INTO rnmtask VALUES("atomic_number", "elements", 3);
CREATE PROCEDURE rmg_request_file(al_task_code INTEGER)
RETURNING VARCHAR(255) AS colval;
DEFINE ll_pdf_column VARCHAR(50);
DEFINE ll_tb_name VARCHAR(60);
DEFINE stmt VARCHAR(255);
DEFINE result VARCHAR(255);
SELECT pdf_column, table_name
INTO ll_pdf_column, ll_tb_name
FROM rnmtask
WHERE task_code = al_task_code;
LET stmt = "SELECT " || ll_pdf_column || " FROM " || ll_tb_name;
PREPARE p FROM stmt;
DECLARE C CURSOR FOR p;
OPEN C;
WHILE sqlcode = 0
FETCH C INTO result;
IF sqlcode != 0 THEN
EXIT WHILE;
END IF;
RETURN result WITH RESUME;
END WHILE;
CLOSE C;
FREE C;
FREE p;
END PROCEDURE;
EXECUTE PROCEDURE rmg_request_file(1);
EXECUTE PROCEDURE rmg_request_file(2);
EXECUTE PROCEDURE rmg_request_file(3);
ROLLBACK;
这假设您在数据库中有一个方便的元素表:
CREATE TABLE elements
(
atomic_number INTEGER NOT NULL PRIMARY KEY CONSTRAINT c1_elements
CHECK (atomic_number > 0 AND atomic_number < 120),
symbol CHAR(3) NOT NULL UNIQUE CONSTRAINT c2_elements,
name CHAR(20) NOT NULL UNIQUE CONSTRAINT c3_elements,
atomic_weight DECIMAL(8, 4) NOT NULL,
period SMALLINT NOT NULL
CHECK (period BETWEEN 1 AND 7),
group CHAR(2) NOT NULL
-- 'L' for Lanthanoids, 'A' for Actinoids
CHECK (group IN ('1', '2', 'L', 'A', '3', '4', '5', '6',
'7', '8', '9', '10', '11', '12', '13',
'14', '15', '16', '17', '18')),
stable CHAR(1) DEFAULT 'Y' NOT NULL
CHECK (stable IN ('Y', 'N'))
);
INSERT INTO elements VALUES( 1, 'H', 'Hydrogen', 1.0079, 1, '1', 'Y');
INSERT INTO elements VALUES( 2, 'He', 'Helium', 4.0026, 1, '18', 'Y');
INSERT INTO elements VALUES( 3, 'Li', 'Lithium', 6.9410, 2, '1', 'Y');
INSERT INTO elements VALUES( 4, 'Be', 'Beryllium', 9.0122, 2, '2', 'Y');
INSERT INTO elements VALUES( 5, 'B', 'Boron', 10.8110, 2, '13', 'Y');
INSERT INTO elements VALUES( 6, 'C', 'Carbon', 12.0110, 2, '14', 'Y');
INSERT INTO elements VALUES( 7, 'N', 'Nitrogen', 14.0070, 2, '15', 'Y');
INSERT INTO elements VALUES( 8, 'O', 'Oxygen', 15.9990, 2, '16', 'Y');
INSERT INTO elements VALUES( 9, 'F', 'Fluorine', 18.9980, 2, '17', 'Y');
INSERT INTO elements VALUES( 10, 'Ne', 'Neon', 20.1800, 2, '18', 'Y');
INSERT INTO elements VALUES( 11, 'Na', 'Sodium', 22.9900, 3, '1', 'Y');
INSERT INTO elements VALUES( 12, 'Mg', 'Magnesium', 24.3050, 3, '2', 'Y');
INSERT INTO elements VALUES( 13, 'Al', 'Aluminium', 26.9820, 3, '13', 'Y');
INSERT INTO elements VALUES( 14, 'Si', 'Silicon', 28.0860, 3, '14', 'Y');
INSERT INTO elements VALUES( 15, 'P', 'Phosphorus', 30.9740, 3, '15', 'Y');
INSERT INTO elements VALUES( 16, 'S', 'Sulphur', 32.0650, 3, '16', 'Y');
INSERT INTO elements VALUES( 17, 'Cl', 'Chlorine', 35.4530, 3, '17', 'Y');
INSERT INTO elements VALUES( 18, 'Ar', 'Argon', 39.9480, 3, '18', 'Y');
关于sql - 如何从表名称为局部变量的表中选择(informix),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14746540/