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theponcer
Quartz | Level 8

Is there a reason I can't use variables when I am defining a hash table? For example, the code below works:

 

 

data out_table;
set base_table;
if 0 then set lookup_table;

if _n_=1 then do;

dcl hash h(dataset: 'lookup_table');
h.definekey('Var1');
h.definedata('Var1','Var2');
h.definedone();
end;

rc_h=h.find();
run;

 

 

But THIS doesn't:

 

 

%let table_variable = lookup_table;


data out_table;
set base_table;
if 0 then set &table_variable.;

if _n_=1 then do;

dcl hash h(dataset: '&table_variable.');
h.definekey('Var1');
h.definedata('Var1','Var2');
h.definedone();
end;

rc_h=h.find();
run;

 

 

Why not?

1 ACCEPTED SOLUTION

Accepted Solutions
r_behata
Barite | Level 11

Macro variables will not resolve in singe quotes . Try

 

%let table_variable = lookup_table;


data out_table;
set base_table;
if 0 then set &table_variable.;

if _n_=1 then do;

dcl hash h(dataset: "&table_variable.");
h.definekey('Var1');
h.definedata('Var1','Var2');
h.definedone();
end;

rc_h=h.find();
run;

View solution in original post

1 REPLY 1
r_behata
Barite | Level 11

Macro variables will not resolve in singe quotes . Try

 

%let table_variable = lookup_table;


data out_table;
set base_table;
if 0 then set &table_variable.;

if _n_=1 then do;

dcl hash h(dataset: "&table_variable.");
h.definekey('Var1');
h.definedata('Var1','Var2');
h.definedone();
end;

rc_h=h.find();
run;

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