Query Problem - Printable Version

Query Problem - Printable Version

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+--- Thread: Query Problem (/showthread.php?tid=52800)

Pages: 1 2

Query Problem - El Forum - 06-27-2012

[eluser]tolgay[/eluser]
Hi

I am using this code

Code:
$this->db->select("*");

$this->db->from('arkadaslar');

$this->db->join('uyeler',"uyeler.id=(arkadaslar.uid+arkadaslar.fid)-2");

$this->db->get();

This is returning error

Quote: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 '' at line 3

SELECT *, `arkadaslar`.`uid` as i1, `arkadaslar`.`fid` as i2 FROM (`arkadaslar`) JOIN `uyeler` ON `uyeler`.`id`=(`i1+i2)`

But if I use this code

Code:
SELECT *FROM `arkadaslar` join uyeler on uyeler.id=(arkadaslar.uid+arkadaslar.fid)-2

This is working.Problem is the nails.How can I resolve this ?

Query Problem - El Forum - 06-27-2012

[eluser]jmadsen[/eluser]
you the problem with the protecting ` , i think?

Code:
(`i1+i2)`

I find ActiveRecord to be a bit tricky when trying to work with more complicated queries. I would recommend just using:

Code:
$sql = "SELECT *FROM `arkadaslar` join uyeler on uyeler.id=(arkadaslar.uid+arkadaslar.fid)-2";

$query = $this->db->query($sql);

Query Problem - El Forum - 06-27-2012

[eluser]tolgay[/eluser]
[quote author="jmadsen" date="1340789447"]you the problem with the protecting ` , i think?

Code:
(`i1+i2)`

I find ActiveRecord to be a bit tricky when trying to work with more complicated queries. I would recommend just using:

Code:
$sql = "SELECT *FROM `arkadaslar` join uyeler on uyeler.id=(arkadaslar.uid+arkadaslar.fid)-2";

$query = $this->db->query($sql);

[/quote]

I am not want use query function.Is there another way ?

Query Problem - El Forum - 06-27-2012

[eluser]jmadsen[/eluser]
try

Code:
$this->db->select("*", FALSE );

but I find it to be unreliable at times

Query Problem - El Forum - 06-27-2012

[eluser]tolgay[/eluser]
[quote author="jmadsen" date="1340793814"]try

Code:
$this->db->select("*", FALSE );

but I find it to be unreliable at times[/quote]

I used this

Code:
$this->db->select("*",false);

$this->db->from('arkadaslar');

$this->db->join('uyeler','uyeler.id=(arkadaslar.uid+arkadaslar.fid)-1)');

$this->db->get();

But it didn't work.

Quote: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 '' at line 3

SELECT * FROM (`arkadaslar`) JOIN `uyeler` ON `uyeler`.`id`=(`arkadaslar`.`uid+arkadaslar`.`fid)-1)`

Query Problem - El Forum - 06-27-2012

[eluser]InsiteFX[/eluser]
Did you count your opening and closing () ?

Code:
$this->db->join('uyeler','uyeler.id=(arkadaslar.uid+arkadaslar.fid)-1)');

Query Problem - El Forum - 06-27-2012

[eluser]tolgay[/eluser]
[quote author="InsiteFX" date="1340806904"]Did you count your opening and closing () ?

Code:
$this->db->join('uyeler','uyeler.id=(arkadaslar.uid+arkadaslar.fid)-1)');

[/quote]

I tried this but didn't work.

Query Problem - El Forum - 06-27-2012

[eluser]InsiteFX[/eluser]

Code:
$this->db->join('uyeler','uyeler.id=(arkadaslar.uid+arkadaslar.fid)-1');

Query Problem - El Forum - 06-27-2012

[eluser]tolgay[/eluser]
[quote author="InsiteFX" date="1340813407"]

Code:
$this->db->join('uyeler','uyeler.id=(arkadaslar.uid+arkadaslar.fid)-1');

[/quote]

Didn't Work Sad

Query Problem - El Forum - 06-27-2012

[eluser]InsiteFX[/eluser]

Code:
$this->db->join('uyeler','uyeler.id = arkadaslar.uid + arkadaslar.fid-1', 'left');

After the query add this and it will show you what is going on.

Code:
echo $this->db->last_query();

exit;