ooops, forgot to reply to list too.
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Dan Barton
Terrestrial Program Biologist
Asst. Data Manager
Point Reyes Bird Observatory
http://www.prbo.org
[EMAIL PROTECTED]
[EMAIL PROTECTED]
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Hola,
Check out the php manual for an explanation of resources and arrays etc.
"Resource ID #6" IS the result of your query - PHP stores it as a
"Resource." I've only used PostgreSQL and MySQL with PHP, and what you need
to do with a query result "resource" using these two databases is "fetch"
the data from it using a command such as mysql_fetch_array or mysql
_fetch_assoc.
From http://www.php.net/manual/en/ref.oracle.php it looks like you would
need to use ora_fetch to convert the "resource" into a PHP array. On this
page, there's a code example of how to use PHP's Oracle functions to extract
data from a database and output HTML.
Cheers,
db
John Kolvereid wrote:
> Hi,
> I am relatively new to Oracle PHP and trying to figure out the
> Oracle functions. I am able to connect and open a db. But that's the
> extent of what I can do w/o further explanation. Within the PHP docs I
> can view all the Oracle functions, but their explanations are sparse to
> say the least. I tried the following:
> $i = ora_do($conn, "Select count(*) from patients");
> echo "returned $i\n";
> The statement is valid, I use it all the time in SQL+ and DBI. However
> in PHP it returns 'Resource ID #6' I am not sure that is a result of
> the echo statement or the ora_do function. Please advise, and let me
> know where a better Oracle/PHP document resides. Thanks.
>
> --
> John Kolvereid
> http://www.odinfo.com
> http://www.kolvereid.com
> [EMAIL PROTECTED]
> 1.610.296.4485
>
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--
----------
Dan Barton
Terrestrial Program Biologist
Asst. Data Manager
Point Reyes Bird Observatory
http://www.prbo.org
[EMAIL PROTECTED]
[EMAIL PROTECTED]
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