mysql syntax error only when executed by php

Question

I got this request:

"START TRANSACTION; DELETE FROM `awaiting_auth` WHERE `code` = '06b8465eed00727a1eac49fae89b88f876ded2eb' LIMIT 1; INSERT IGNORE INTO `prsn` SET `login` = 'new_user', `passwd` = '40bd001563085fc35165329ea1ff5c5ecbdbbeef', `color` = '#cbc5f2'; COMMIT;"

And I receive this error:

"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 'DELETE FROM awaiting_auth WHERE code = '06b8465eed00727a1eac49fae89b88f876de' at line 1".

However when executing sql via terminal everything goes well and no errors are thrown.
What is wrong with my request? Thanks in advance.

Use mysqli_error($conn) and see what you get

Mr. Alien
– Mr. Alien

2013-08-12 11:07:20 +00:00
Commented Aug 12, 2013 at 11:07 — Mr. Alien
– Mr. Alien, Commented Aug 12, 2013 at 11:07
how you execute this using php? post relevant code

user1646111
– user1646111

2013-08-12 11:07:38 +00:00
Commented Aug 12, 2013 at 11:07 — user1646111
– user1646111, Commented Aug 12, 2013 at 11:07

vidario · Accepted Answer · 2013-08-12 11:10:01Z

2

You must split the queries in separate ->query() calls. The SQL console does that automatically. E.g.

->query("START TRANSACTION");
->query("DELETE FROM...");

answered Aug 12, 2013 at 11:10

vidario

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Marcin Orlowski · Accepted Answer · 2013-08-12 11:09:40Z

0

One operation per query. So you need to split your query into single queries.

answered Aug 12, 2013 at 11:09

Marcin Orlowski

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Your Common Sense · Accepted Answer · 2013-08-12 11:09:52Z

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Just run these queries one by one, not in single packet.

answered Aug 12, 2013 at 11:09

Your Common Sense

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vicentazo · Accepted Answer · 2013-08-12 11:11:10Z

0

Try to execute each statement individually:

START TRANSACTION;
DELETE FROM `awaiting_auth` WHERE `code` = '06b8465eed00727a1eac49fae89b88f876ded2eb' LIMIT 1;
INSERT IGNORE INTO `prsn`(`login`, `passwd`, `color`) VALUES('new_user','40bd001563085fc35165329ea1ff5c5ecbdbbeef', '#cbc5f2');
COMMIT;

answered Aug 12, 2013 at 11:11

vicentazo

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