Re: [PHP-DB] Firebird return wrong value

2013-01-18 Thread Lester Caine
Berko Bubu wrote: INSERT INTO PRICE (ID, "NAME", COST) You should not use double quotes around column names. Try to use backtick operator instead: INSERT INTO `PRICE` (`ID`, `NAME`, `COST`) Don't know if that solves the problem, but it's atleast good practice. Second, do you really need firebi

Re: [PHP-DB] Firebird return wrong value

2013-01-18 Thread Lester Caine
Matijn Woudt wrote: Dear List! > >i have a strange problem. > >I have a Firebird database (dialect 3). Firebird server: 2.0.6 >I create a table, and insert a row like that: >CREATE TABLE PRICE ( > ID INTEGER NOT NULL, > "NAME" VARCHAR(10), > COST NUMERIC(15, 2)); > >INSERT INTO PRICE (ID, "

Re: [PHP-DB] Firebird return wrong value

2013-01-18 Thread Berko Bubu
> > INSERT INTO PRICE (ID, "NAME", COST) > > > > You should not use double quotes around column names. Try to use backtick > operator instead: > INSERT INTO `PRICE` (`ID`, `NAME`, `COST`) > Don't know if that solves the problem, but it's atleast good practice. > Second, do you really need firebird

Re: [PHP-DB] Firebird return wrong value

2013-01-18 Thread Matijn Woudt
On Fri, Jan 18, 2013 at 12:49 PM, wrote: > Dear List! > > i have a strange problem. > > I have a Firebird database (dialect 3). Firebird server: 2.0.6 > I create a table, and insert a row like that: > CREATE TABLE PRICE ( > ID INTEGER NOT NULL, > "NAME" VARCHAR(10), > COST NUMERIC(15, 2));

Re: [PHP-DB] Firebird return wrong value

2013-01-18 Thread Lester Caine
vbe...@mail.com wrote: result: Array ( [0] => Array ( [COST] => -0.00 ) ) As a starting point ... can you look at the database using Flamerobin and check what it shows has been stored. I don't see anything wrong with what you have done except that 'cost' should perhaps be -1.0, so once we kno

[PHP-DB] Firebird return wrong value

2013-01-18 Thread vberko
Dear List! i have a strange problem. I have a Firebird database (dialect 3). Firebird server: 2.0.6 I create a table, and insert a row like that: CREATE TABLE PRICE (   ID INTEGER NOT NULL,   "NAME" VARCHAR(10),   COST NUMERIC(15, 2)); INSERT INTO PRICE (ID, "NAME", COST) VALUES (2, 'my price2',