Closed GoogleCodeExporter closed 9 years ago
Hi. Maybe this is because you are enclosing the select in parenthesis.
please try:
$parser = new PHPSQLParser("INSERT INTO newTablename SELECT field1, field2,
field3 FROM oldTablename where field1 > 100;", true);
Original comment by greenlion@gmail.com
on 13 Dec 2012 at 8:31
Although I think, parenthesis should be accepted/parsed, you are right and it
works fine without parenthesis.
For me that is now o.k. and solved, I can change our sql to adapt for
php-sql-parser.
Thanks!
Original comment by i...@enterprise-technologies.de
on 14 Dec 2012 at 5:49
The revision r1004 contains code to fix this issue.
Original comment by pho...@gmx.de
on 13 Jan 2014 at 10:10
Original issue reported on code.google.com by
i...@enterprise-technologies.de
on 13 Dec 2012 at 8:08