Error: No matching signature for function IF for argument types: BOOL. Supported signature: IF(BOOL, ANY, ANY)

I’m trying to count the number of times a “rating” is above 9. Working in Google BigQuery with StandardSQL

I previously was using a CASE WHEN function but wanted to change it to an IF as there is only one case. Previous code which worked fine:

SELECT 
COUNT(CASE WHEN (survey_responses.survey_rating  >= 9) THEN 1 ELSE NULL END) 
   AS survey_responses_num_positives

Current code

SELECT 
COUNT(IF(survey_responses.survey_rating  >= 9),1,NULL)
   AS survey_responses_num_positives

The current code spits out the error:

No matching signature for function IF for argument types: BOOL. Supported signature: IF(BOOL, ANY, ANY) at [11:39]

Don’t really understand it as the IF statement is a BOOL value no? Would appreciate any help on this. Thanks!

Answer

You missplaced a parenthesis:

SELECT 
COUNT(IF(survey_responses.survey_rating  >= 9,1,NULL))
   AS survey_responses_num_positives

IF takes 3 arguments, the condition, if true, and else.