PHP variable in query doesn't work

Question

I have tried "everything" to get it working, but I just doesn't :-(

$maxcount = 3;
$conn->query("INSERT INTO getal (getal, aantalmaal) SELECT getal, aantalmaal FROM v_topgetallen limit '.$maxcount.'");

Do you guys have any idea? Thanks a lot!

Yes, use a prepared statement; all quote, concatenation and sql injection problems solved at once. — jeroen, Sep 20 '17 at 19:06
You're syntax of the `INSERT INTO` seems wrong, see: https://dev.mysql.com/doc/refman/5.7/en/insert.html — KIKO Software, Sep 20 '17 at 19:07
Hi KIKO, when use 3 instead of $maxcount, everything works fine. — franky555, Sep 20 '17 at 19:08
@KIKOSoftware https://dev.mysql.com/doc/refman/5.7/en/insert-select.html — jeroen, Sep 20 '17 at 19:09

score 1 · Accepted Answer · answered Sep 20 '17 at 19:11

1

Try this:

$maxcount = 3;

$conn->query("INSERT 
              INTO getal (getal, aantalmaal) 
              SELECT getal, aantalmaal 
              FROM v_topgetallen 
              LIMIT $maxcount");

but please use prepared statements.

answered Sep 20 '17 at 19:11

KIKO Software

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Yes! That works!! I'll try to find out if I can use prepared statements. Thanks a lot! – franky555 Sep 20 '17 at 19:14
@franky555 You can find information on [Prepared Statements](https://en.wikipedia.org/wiki/Prepared_statement) for [MySQLi](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) or [PDO](http://php.net/manual/en/pdo.prepared-statements.php). – GrumpyCrouton Sep 20 '17 at 19:25
Thanx, I'll read about it. – franky555 Sep 20 '17 at 19:29

PHP variable in query doesn't work

1 Answers1