* Standard COBOL (file "CDR004.SCO") calling SQL * procedures in file "CDR004.MCO"
**************************************************************** * * COMMENT SECTION * * DATE 1989/04/07 STANDARD COBOL LANGUAGE * NIST SQL VALIDATION TEST SUITE V6.0 * DISCLAIMER: * This program was written by employees of NIST to test SQL * implementations for conformance to the SQL standards. * NIST assumes no responsibility for any party's use of * this program. * * CDR004.SCO * WRITTEN BY: SUN DAJUN * TRANSLATED AUTOMATICALLY FROM EMBEDDED COBOL BY CHRIS SCHANZLE * * THIS ROUTINE TESTS THE <CHECK> CONSTRAINT. * * REFERENCES * AMERICAN NATIONAL STANDARD database language - SQL * with Integrity Enhancement * * SECTION 6.8 * <check constraint definition>::= * CHECK (<search condition>) ****************************************************************
******************** BEGIN TEST0310 ****************** *This procedure tests if x NOT IN and NOT x IN of the *CHECK specification has the same meaning . * CHECK condition * NOT GRADE IN (5,22) on STAFF12
******************** BEGIN TEST0311 ******************* *This test check the constraint NOT NULL in column *definition constraint. * CHECK condition * NOT NULL in EMPNAME def. of STAFF15 * INSERT NULL INTO EMPNAME explicitly
* EXEC SQL INSERT INTO STAFF15 * VALUES('E2',NULL,52,'Newyork') END-EXEC CALL"SUB20"USINGSQLCODE MOVESQLCODETO SQL-COD MOVESQLCODETO code1 * EXEC SQL SELECT COUNT(*) INTO :count2 * FROM STAFF15 END-EXEC CALL"SUB21"USINGSQLCODE count2 MOVESQLCODETO SQL-COD DISPLAY" The correct answer is :" DISPLAY" SQLCODE of second INSERT not equal to 0." DISPLAY" count1 is 1" DISPLAY" count2 is 1" DISPLAY" Your answer is :" DISPLAY" SQLCOD of second INSERT is ", code1 DISPLAY" count1 is ", count1 DISPLAY" count2 is ", count2 if (code1 NOT = 0 AND count1 = 1 AND count2 = 1) then * EXEC SQL INSERT INTO TESTREPORT * VALUES('0311','pass','MCO') END-EXEC CALL"SUB22"USINGSQLCODE MOVESQLCODETO SQL-COD DISPLAY" *** pass *** " else * EXEC SQL INSERT INTO TESTREPORT * VALUES('0311','fail','MCO') END-EXEC ADD 1 TO errcnt CALL"SUB23"USINGSQLCODE MOVESQLCODETO SQL-COD DISPLAY" cdr004.sco *** fail *** " END-IF DISPLAY"================================================"
DISPLAY" " * EXEC SQL COMMIT WORK; CALL"SUB24"USINGSQLCODE MOVESQLCODETO SQL-COD **************** END TEST 0311 **********
******************** BEGIN TEST0312 ******************* *This test checks NOT NULL in table definition con- *straint. * CHECK condition * EMPNAME IS NOT NULL. on STAFF15 * INSERT VALUE INTO all columns except EMPNAME
* EXEC SQL INSERT INTO STAFF15(EMPNUM,GRADE,CITY) * VALUES('E2',52,'Newyork') END-EXEC CALL"SUB28"USINGSQLCODE MOVESQLCODETO SQL-COD MOVESQLCODETO code1 * EXEC SQL SELECT COUNT(*) INTO :count2 * FROM STAFF15 END-EXEC CALL"SUB29"USINGSQLCODE count2 MOVESQLCODETO SQL-COD DISPLAY" The correct answer is :" DISPLAY" SQLCODE of second INSERT not equal to 0." DISPLAY" count1 is 1" DISPLAY" count2 is 1" DISPLAY" Your answer is :" DISPLAY" SQLCOD of second INSERT is ", code1 DISPLAY" count1 is ", count1 DISPLAY" count2 is ", count2 if (code1 NOT = 0 AND count1 = 1 AND count2 = 1) then * EXEC SQL INSERT INTO TESTREPORT * VALUES('0312','pass','MCO') END-EXEC CALL"SUB30"USINGSQLCODE MOVESQLCODETO SQL-COD DISPLAY" *** pass *** " else * EXEC SQL INSERT INTO TESTREPORT * VALUES('0312','fail','MCO') END-EXEC ADD 1 TO errcnt CALL"SUB31"USINGSQLCODE MOVESQLCODETO SQL-COD DISPLAY" cdr004.sco *** fail *** " END-IF DISPLAY"================================================"
DISPLAY" " * EXEC SQL COMMIT WORK; CALL"SUB32"USINGSQLCODE MOVESQLCODETO SQL-COD **************** END TEST 0312 **********
**** TESTER MAY CHOOSE TO INSERT CODE FOR errcnt > 0 STOPRUN.
* **** Procedures for PERFORM statements
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