Question

Do you see any issues with this MySQL query? It does not execute.
Any other better ideas?

INSERT INTO registration 
   ( m_ID,  e_ID,   STATUS) 
SELECT 
   :m_ID, :e_ID, 1
WHERE 
    EXISTS ( SELECT 1 FROM tableb WHERE ID = :e_ID)
AND EXISTS ( SELECT 1 FROM tablec WHERE m_ID = :m_ID)

Passing array with values for e_id and m_ID (PDO)

Basically the goal is to check if both IDs exists in different tables before inserting onto another table.

ERROR:

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL    
syntax; check the manual that corresponds to your MySQL server version for the right  
syntax to use near 'WHERE EXISTS ( SELECT 1 FROM tableb WHERE ID = '276') AND EX' at 
line 4

Answer 1

if both IDs exists in different tables

Yet your query searches tableb twice. Unless you really meant “in a single different table” (and I don’t think so, you’d have used a single EXISTS), then this might be the error you wanted.

Also, PDO should have some sort of error logging. The query might not execute, but the reason should appear – or it should be possible to make it appear – somewhere.

UPDATE

I’m afraid I was wrong, the syntax is not correct. MySQL doesn’t allow SELECT ... WHERE unless a table is specified.

Try changing

SELECT 
   :m_ID, :e_ID, 1
WHERE EXISTS

in

SELECT * FROM
   ( SELECT :m_ID, :e_ID, 1 ) AS dummy
WHERE EXISTS

TEST:

mysql> SELECT 42;
+----+
| 42 |
+----+
| 42 |
+----+
1 row in set (0.00 sec)

mysql> SELECT 42 WHERE true;
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE true' at line 1


mysql> SELECT * FROM ( SELECT 42 ) AS answer WHERE true;
+----+
| 42 |
+----+
| 42 |
+----+
1 row in set (0.00 sec)