HI ALL,

question 1:

below is my code

 

TABLE FILE SQLOUT
PRINT *
END

the result is 

ID    Name    address
1      a       aaa
2      b       bbb
3      c       ccc
4      d       ddd

 

how can I get the all Name? like below ,

-SET &ALLNAME = 'a,b,c,d';


question 2:

the code like above.

I can get the max ID,but I can't get the address with MAX.ID all together

below is my code;

 

TABLE FILE SQLOUT
SUM
MAX.ID
ON TABLE SAVE AS TEST1
END
-RUN

-READ TEST1 &ID.A1.
-TYPE ID===&ID;

 

How can I get the MAX.ID 's address ?

I try to
SUM
MAX.ID
address

the address is the frist address , not max ...


Thanks,
James

This message has been edited. Last edited by: Yangyang,

There are many ways to do this, some have been discussed in this forum before.

Here is an example.

TABLE FILE CAR
SUM MAX.CAR
BY COUNTRY
ON TABLE HOLD AS TMP1
END
TABLE FILE TMP1
SUM COMPUTE
    MAXCNTRY/A10 = MAX.COUNTRY ; NOPRINT
PRINT CAR
BY COUNTRY
WHERE TOTAL COUNTRY EQ MAXCNTRY
END
DEFINE FILE TMP1
 Cntr/I9 = LAST Cntr + 1 ;
 All/A100V = IF Cntr EQ 1 THEN COUNTRY ELSE All || ',' || COUNTRY ;
END
TABLE FILE TMP1
SUM LST.All
END

Thanks Waz,

this work for me , I select from forum before... but I can't get the key words


Thanks,
James

If each of your IDs is unique, you should also be able to obtain the max ID and its address with this code:

TABLE FILE SQLOUT
 PRINT Address
 BY HIGHEST 1 ID
END

Thanks DAN,

I will try this.

Thanks,
James