Using variable names stored in database fields.

[bofh@visi.com]

I'm working on a project where I need to store a CGI query in a
database field. The query contains variables which will be substitued
at runtime (e.g., today's date, key to select upon, etc), and may be
pointed to different URLs depending upon the table key.

The variable substitution works fine when hardcoding into the script,
e.g.:


$var1='data1';
$var2='data2';
$a="http://www.webserver.c om/query.cgi?a=$va r1&b=$var2";
$fd=fopen($a,"r ");
...

However, the substitution fails when the query is pulled from a database
record, e.g.

//Only one record returned for testing purposes.
//
//location field (varchar) is:
//"http://www.webserver.c om/query.cgi?a=$va r1&b=$var2"

$var1='data1';
$var2='data2';
$dbconn=yadda yada;
$dbquery=pg_que ry($dbconn,"sel ect key,location from schema.table");
$row=pg_fetch_a ssoc($dbquery);
$fd=fopen($row['location'],"r");
...

In this case, the variable substitution doesn't occur in the query (i.e.,
$var1 remains the string $var1, instead of changing into the value of say,
'03-23-2004'). I've tried different variations, and treating as a
variable variable (with $$row['url'] and ${$row['url']}. No joy.

Any ideas would be appreciated. Thanks.

-------------------------------------------------
Only in America will someone |
order a Big Mac, large fries, | bofh@visi.com
and a *Diet* Coke... |

[Jeffrey Silverman]

Re: Using variable names stored in database fields.

On Tue, 23 Mar 2004 11:33:16 -0600, bofh wrote:
[color=blue]
> I'm working on a project where I need to store a CGI query in a
> database field. The query contains variables which will be substitued
> at runtime (e.g., today's date, key to select upon, etc), and may be
> pointed to different URLs depending upon the table key.
>
> The variable substitution works fine when hardcoding into the script,
> e.g.:
>
>
> $var1='data1';
> $var2='data2';
> $a="http://www.webserver.c om/query.cgi?a=$va r1&b=$var2";
> $fd=fopen($a,"r ");
> ...
>
> However, the substitution fails when the query is pulled from a database
> record, e.g.
>
> //Only one record returned for testing purposes.
> //
> //location field (varchar) is:
> //"http://www.webserver.c om/query.cgi?a=$va r1&b=$var2"
>
> $var1='data1';
> $var2='data2';
> $dbconn=yadda yada;
> $dbquery=pg_que ry($dbconn,"sel ect key,location from schema.table");
> $row=pg_fetch_a ssoc($dbquery);
> $fd=fopen($row['location'],"r");
> ...
>
> In this case, the variable substitution doesn't occur in the query (i.e.,
> $var1 remains the string $var1, instead of changing into the value of say,
> '03-23-2004'). I've tried different variations, and treating as a
> variable variable (with $$row['url'] and ${$row['url']}. No joy.
>
> Any ideas would be appreciated. Thanks.[/color]


It's not working because the SQL result set is just a string that has no
relation whatsoever to the variables $var1 and $var2. I.e. the string is
not treated as PHP.

You could use eval().

Or you could use a regex replace. example:

preg_replace("/$var1/", $var1, $row['location']);
(...repeat for $var2...)

Or, if the query string is always going to include "a=$var1" and "b=$var2"
you could recreate the query string from scratch:

(assuming $row['location'] contains only the URL without the query
string):

$url = $row['location'] . "?a=$var1&b=$va r2";

There are probably other ways to solve this problem, as there are always
at least two more ways to do something beyond even the half dozen you may
have thought of. good luck...

--
Jeffrey D. Silverman | jeffrey AT jhu DOT edu
Website | http://www.wse.jhu.edu/newtnotes/

[bofh@visi.com]

Re: Using variable names stored in database fields.

I kind of figured it was only being seen as a string, and I've thought of
a few ways around it (explode() and implode() being the most likely
candidates at present). I was just hoping that I had overlooked
something, and that some complexity could be engineered out.



On Tue, 23 Mar 2004, Jeffrey Silverman wrote:
[color=blue]
> On Tue, 23 Mar 2004 11:33:16 -0600, bofh wrote:
>[color=green]
> > I'm working on a project where I need to store a CGI query in a
> > database field. The query contains variables which will be substitued
> > at runtime (e.g., today's date, key to select upon, etc), and may be
> > pointed to different URLs depending upon the table key.
> >
> > The variable substitution works fine when hardcoding into the script,
> > e.g.:
> >
> >
> > $var1='data1';
> > $var2='data2';
> > $a="http://www.webserver.c om/query.cgi?a=$va r1&b=$var2";
> > $fd=fopen($a,"r ");
> > ...
> >
> > However, the substitution fails when the query is pulled from a database
> > record, e.g.
> >
> > //Only one record returned for testing purposes.
> > //
> > //location field (varchar) is:
> > //"http://www.webserver.c om/query.cgi?a=$va r1&b=$var2"
> >
> > $var1='data1';
> > $var2='data2';
> > $dbconn=yadda yada;
> > $dbquery=pg_que ry($dbconn,"sel ect key,location from schema.table");
> > $row=pg_fetch_a ssoc($dbquery);
> > $fd=fopen($row['location'],"r");
> > ...
> >
> > In this case, the variable substitution doesn't occur in the query (i.e.,
> > $var1 remains the string $var1, instead of changing into the value of say,
> > '03-23-2004'). I've tried different variations, and treating as a
> > variable variable (with $$row['url'] and ${$row['url']}. No joy.
> >
> > Any ideas would be appreciated. Thanks.[/color]
>
>
> It's not working because the SQL result set is just a string that has no
> relation whatsoever to the variables $var1 and $var2. I.e. the string is
> not treated as PHP.
>
> You could use eval().
>
> Or you could use a regex replace. example:
>
> preg_replace("/$var1/", $var1, $row['location']);
> (...repeat for $var2...)
>
> Or, if the query string is always going to include "a=$var1" and "b=$var2"
> you could recreate the query string from scratch:
>
> (assuming $row['location'] contains only the URL without the query
> string):
>
> $url = $row['location'] . "?a=$var1&b=$va r2";
>
> There are probably other ways to solve this problem, as there are always
> at least two more ways to do something beyond even the half dozen you may
> have thought of. good luck...
>
> --
> Jeffrey D. Silverman | jeffrey AT jhu DOT edu
> Website | http://www.wse.jhu.edu/newtnotes/
>
>[/color]

-------------------------------------------------
Only in America will someone |
order a Big Mac, large fries, | bofh@visi.com
and a *Diet* Coke... |

[bofh@visi.com]

Re: Using variable names stored in database fields.

Tried out eval() on a lark, and that works just fine.

Thanks much.


On Tue, 23 Mar 2004, Jeffrey Silverman wrote:
[color=blue]
> On Tue, 23 Mar 2004 11:33:16 -0600, bofh wrote:
>[color=green]
> > I'm working on a project where I need to store a CGI query in a
> > database field. The query contains variables which will be substitued
> > at runtime (e.g., today's date, key to select upon, etc), and may be
> > pointed to different URLs depending upon the table key.
> >
> > The variable substitution works fine when hardcoding into the script,
> > e.g.:
> >
> >
> > $var1='data1';
> > $var2='data2';
> > $a="http://www.webserver.c om/query.cgi?a=$va r1&b=$var2";
> > $fd=fopen($a,"r ");
> > ...
> >
> > However, the substitution fails when the query is pulled from a database
> > record, e.g.
> >
> > //Only one record returned for testing purposes.
> > //
> > //location field (varchar) is:
> > //"http://www.webserver.c om/query.cgi?a=$va r1&b=$var2"
> >
> > $var1='data1';
> > $var2='data2';
> > $dbconn=yadda yada;
> > $dbquery=pg_que ry($dbconn,"sel ect key,location from schema.table");
> > $row=pg_fetch_a ssoc($dbquery);
> > $fd=fopen($row['location'],"r");
> > ...
> >
> > In this case, the variable substitution doesn't occur in the query (i.e.,
> > $var1 remains the string $var1, instead of changing into the value of say,
> > '03-23-2004'). I've tried different variations, and treating as a
> > variable variable (with $$row['url'] and ${$row['url']}. No joy.
> >
> > Any ideas would be appreciated. Thanks.[/color]
>
>
> It's not working because the SQL result set is just a string that has no
> relation whatsoever to the variables $var1 and $var2. I.e. the string is
> not treated as PHP.
>
> You could use eval().
>
> Or you could use a regex replace. example:
>
> preg_replace("/$var1/", $var1, $row['location']);
> (...repeat for $var2...)
>
> Or, if the query string is always going to include "a=$var1" and "b=$var2"
> you could recreate the query string from scratch:
>
> (assuming $row['location'] contains only the URL without the query
> string):
>
> $url = $row['location'] . "?a=$var1&b=$va r2";
>
> There are probably other ways to solve this problem, as there are always
> at least two more ways to do something beyond even the half dozen you may
> have thought of. good luck...
>
> --
> Jeffrey D. Silverman | jeffrey AT jhu DOT edu
> Website | http://www.wse.jhu.edu/newtnotes/
>
>[/color]

-------------------------------------------------
Only in America will someone |
order a Big Mac, large fries, | bofh@visi.com
and a *Diet* Coke... |

[Pedro Graca]

Re: Using variable names stored in database fields.

bofh@visi.com top-posted (corrected):[color=blue]
> On Tue, 23 Mar 2004, Jeffrey Silverman wrote:[color=green]
>> On Tue, 23 Mar 2004 11:33:16 -0600, bofh wrote:[color=darkred]
>> > However, the substitution fails when the query is pulled from a database
>> > record, e.g.[/color][/color][/color]
[color=blue][color=green][color=darkred]
>> > In this case, the variable substitution doesn't occur in the query (i.e.,
>> > $var1 remains the string $var1, instead of changing into the value of say,
>> > '03-23-2004'). I've tried different variations, and treating as a
>> > variable variable (with $$row['url'] and ${$row['url']}. No joy.[/color][/color][/color]
[color=blue][color=green]
>> It's not working because the SQL result set is just a string that has no
>> relation whatsoever to the variables $var1 and $var2. I.e. the string is
>> not treated as PHP.
>>
>> You could use eval().[/color][/color]
[color=blue]
> I kind of figured it was only being seen as a string, and I've thought of
> a few ways around it (explode() and implode() being the most likely
> candidates at present). I was just hoping that I had overlooked
> something, and that some complexity could be engineered out.[/color]

As Jeffrey told you, eval() is your friend

http://www.php.net/eval

For example:

<?php
$data = 'URL?a=$var1&b= $var2';
$var1 = 'foobar';
$var2 = 'quux';

// you want to transform $data to 'URL?a=foobar&b =quux'
eval('$data = "' . $data . '";');
echo $data;
?>
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