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I have 2 Tables

Table 1 has Id, Name
Table 2 has Id, Col1_IdVal, Col2_IdVal,Col3_IdVal,Col4_IdVal

I would like simple solution for

Count(T1.Id) WHERE T1.Id In (T2.Col1_IdVal,T2.Col2_IdVal,T2.Col3_IdVal,T2.Col4_IdVal)

Something I have tried does not return correct result set , too few in the result :

SELECT COUNT(T1.Id) FROM Table1 As T1     
LEFT JOIN Table2 AS T2_1 ON T2_1.T2Column_1 = T1.Id     
LEFT JOIN Table2 AS T2_2 ON T2_2.T2Column_2 = T1.Id    
LEFT JOIN Table2 AS T2_3 ON T2_3.T2Column_3 = T1.Id    
LEFT JOIN Table2 AS T2_4 ON T2_4.T2Column_4 = T1.Id    
LEFT JOIN Table2 AS T2_5 ON T2_5.T2Column_5 = T1.Id;

Table 1 has Unique Values - it is Primary Key, Table 2 can have any number of values MANY. Table2 columns are not foreign keys of Table1 - they are values that must exist in Table1 but other than that - they are not FKeys.

So basically Count(Id) Where Id Value Is in T2, Cols 1..n

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  • What column is IN (...) ? Perhaps you mean T1.id IN (...)?
    – Rick James
    Commented Jul 15, 2019 at 17:09
  • @RickJames Yes - I missed that ..
    – StixO
    Commented Jul 15, 2019 at 17:10
  • Make the col names consistent. Shorten the question to only 3 tables total. Shorten the column names. You say "not correct" -- too high? Too low? Are the tables 1:many? (or what)? Switch from LEFT JOIN to EXISTS?
    – Rick James
    Commented Jul 15, 2019 at 17:13
  • You probably wanted inner joins, not outer.
    – mustaccio
    Commented Jul 15, 2019 at 17:15
  • @mustaccio No I do not want inner joins . With that it combines the joins and the resulting count is 0, I want to get a count for Id where it exists in the other T2.Columns 1 through x ..
    – StixO
    Commented Jul 15, 2019 at 17:22

1 Answer 1

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Look for

SELECT COUNT(t1.Id) FROM Table1 As t1     
{ INNER | LEFT } JOIN Table2 AS t2 ON t1.Id IN (t2.T2Column_1,
                                                t2.T2Column_2,
                                                t2.T2Column_3,
                                                t2.T2Column_4,
                                                t2.T2Column_5);
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  • I will test this tomorrow when I am back at work.
    – StixO
    Commented Jul 16, 2019 at 5:32

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