topic Error using LIKE operator in SAS Programming

Error using LIKE operator

Babloo — Fri, 06 Sep 2019 12:58:14 GMT

I'm trying to resolve the error as mentioned below but I'm unable to do it. SCHDLD_OBJ_NM is the variable which I'm creating in the same datastep where I placed the below if condition and I was asked to achieve this one single step.

4219       if SCHDLD_OBJ_NM like %"_AP_"% then FNCTL_CMPNT_NM="AP" else "LIS";
                            ____                                     ____
                            388                                      388
                            202                                      202
ERROR 388-185: Expecting an arithmetic operator.

ERROR 202-322: The option or parameter is not recognized and will be ignored.

4220       run;

Re: Error using LIKE operator

PeterClemmensen — Fri, 06 Sep 2019 13:01:50 GMT

You use the Like Operator in a Where Statement. If you want to apply the logic in an If Statement, then read

Sample 43303: Using the equivalent of CONTAINS and LIKE in an IF statement

Re: Error using LIKE operator

Jagadishkatam — Fri, 06 Sep 2019 13:08:38 GMT

You can also try IFC function

FNCTL_CMPNT_NM=ifc(indexw(SCHDLD_OBJ_NM,"_AP_"),'AP','LIS');

Re: Error using LIKE operator

hashman — Fri, 06 Sep 2019 21:59:52 GMT

@Babloo:

LIKE doesn't like IF, plus your double quotes are extraneous.

Use the FIND function instead. Also, if FNCTL_CMPNT_NM is a new variable not yet seen by the compiler, either declare it in the LENGTH statement as $3 beforehand or reverse the order in which the compiler sees "AP" and "LIS"; otherwise your "LIS" will be truncated to "LI" when "_AP_" is not found in SCHDLD_OBJ_NM:

if find (SCHDLD_OBJ_NM, "_AP_") = 0 then FNCTL_CMPNT_NM = "LIS" ;                                                                                                                                                                                               
else FNCTL_CMPNT_NM = "AP" ;

Kind regards

Paul D.