Re: [PHP] String variable

2009-01-11 Thread Eric Butera 
On Sun, Jan 11, 2009 at 8:59 AM, MikeP mpel...@princeton.edu wrote:
 Hello,
 I am trying yo get THIS:
 where ref_id = '1234'
 from this.
 $where=where ref_id=.'$Reference[$x][ref_id]';

 but i certainly have a quote problem.

 Any help?
 Thanks
 Mike




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Looks like you're missing the single tics around ref_id.  In your
example you'd want one of these:

$where=where ref_id=.'{$Reference[$x]['ref_id']};
$where=where ref_id=.$Reference[$x]['ref_id'];
$where=sprintf(where ref_id=%d, (int)$Reference[$x]['ref_id']); --
use this one or else! (sql injection)

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Re: [PHP] String variable

2009-01-11 Thread Ashley Sheridan 
On Sun, 2009-01-11 at 08:59 -0500, MikeP wrote:
 Hello,
 I am trying yo get THIS:
 where ref_id = '1234'
 from this.
 $where=where ref_id=.'$Reference[$x][ref_id]';
 
 but i certainly have a quote problem.
 
 Any help?
 Thanks
 Mike
 
 
 
 

It should look like this:

$where=where ref_id='{$Reference[$x][ref_id]}';


Ash
www.ashleysheridan.co.uk


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Re: [PHP] String variable

2009-01-11 Thread Ashley Sheridan 
On Sun, 2009-01-11 at 14:36 +, Ashley Sheridan wrote:
 On Sun, 2009-01-11 at 08:59 -0500, MikeP wrote:
  Hello,
  I am trying yo get THIS:
  where ref_id = '1234'
  from this.
  $where=where ref_id=.'$Reference[$x][ref_id]';
  
  but i certainly have a quote problem.
  
  Any help?
  Thanks
  Mike
  
  
  
  
 
 It should look like this:
 
 $where=where ref_id='{$Reference[$x][ref_id]}';
 
 
 Ash
 www.ashleysheridan.co.uk
 
 
Sorry, it should look like this:

$where=where ref_id='{$Reference[$x][ref_id]}';

I missed taking an extra quote mark out


Ash
www.ashleysheridan.co.uk


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Re: [PHP] String variable

2009-01-11 Thread Nathan Rixham 

Ashley Sheridan wrote:

On Sun, 2009-01-11 at 14:36 +, Ashley Sheridan wrote:

On Sun, 2009-01-11 at 08:59 -0500, MikeP wrote:

Hello,
I am trying yo get THIS:
where ref_id = '1234'
from this.
$where=where ref_id=.'$Reference[$x][ref_id]';

but i certainly have a quote problem.

Any help?
Thanks
Mike





It should look like this:

$where=where ref_id='{$Reference[$x][ref_id]}';


Ash
www.ashleysheridan.co.uk



Sorry, it should look like this:

$where=where ref_id='{$Reference[$x][ref_id]}';

I missed taking an extra quote mark out


Ash
www.ashleysheridan.co.uk



actually unless ref_id is a constant (which i doublt) you may be best 
going with:


$where = WHERE ref_id=' . $Reference[$x]['ref_id'] . ';

keep the php and sql seperate and you'll find it much easier to see in 
you're editor


(imho) :)

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Re: [PHP] String variable

2009-01-11 Thread Lars Torben Wilson 
2009/1/11 Ashley Sheridan a...@ashleysheridan.co.uk:
 On Sun, 2009-01-11 at 14:36 +, Ashley Sheridan wrote:
 On Sun, 2009-01-11 at 08:59 -0500, MikeP wrote:
  Hello,
  I am trying yo get THIS:
  where ref_id = '1234'
  from this.
  $where=where ref_id=.'$Reference[$x][ref_id]';
 
  but i certainly have a quote problem.
 
  Any help?
  Thanks
  Mike
 
 
 
 

 It should look like this:

 $where=where ref_id='{$Reference[$x][ref_id]}';


 Ash
 www.ashleysheridan.co.uk


 Sorry, it should look like this:

 $where=where ref_id='{$Reference[$x][ref_id]}';

 I missed taking an extra quote mark out

Closer, but still not quite there. For encapsulation in the string, it
should look like:

$where = where ref_is='{$Reference[$x]['ref_id']}';

Someone else mentioned casting to int first as well to sanitize, which
is also a good idea.


Torben

 Ash
 www.ashleysheridan.co.uk




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