I've been looking at the pg_fetch_* functions but do not understand one
little bit.
Using your suggestion gives a blank screen but var_dump($myarray) gives
"bool(false)"
And a var_dump($results) returns "resource(3) of type (pgsql result)
while print_r($results) returns "Resource id#3"
I'm totally lost. :(
Perhaps my query isn't well formulated?
What I am trying to locate (or print out) is the string found the
"description" column whre the row in the "name" column is Alcohol.
item_id || glossary_id || name || description
2 || 2 || Alcohol || One of thetwo major....
blah...blah.. blah
Spiegelberg, Greg wrote:
> You need to fetch the results. See the pg_fetch_* family of functions
> for the one of your liking.
>
> $myarray = pg_fetch_all($results);
> print_r($myarray);
>
> Greg
>
> ------------------------------------------------------------------------
> *From:* pgsql-php-owner@postgresql.org on behalf of PJ
> *Sent:* Wed 7/30/2008 10:18 AM
> *To:* pgsql-php@postgresql.org
> *Subject:* [PHP] display query results
>
> I am trying to learn postgresql with psql - using FreeBS D7.0,
> postgresql 8.3.3 php 5.2.6, apache 2.2.9
>
> How can I echo or print on screen the results; all I get from this code
> is this:
> Resource id #3
> item_idglossary_idnamedescription
> which is the field names run together
>
> I am expecting to see several long text sentences.
>
> Here is the snippet and the connection does work and the database is
> there and functioning.
>
> $db = pg_connect("host=localhost port=5432 dbname=med user=med
> password=0tscc71");
>
> if (!$db)
> {
> die("Could not open connection to database server");
> }
>
> // generate and execute a query
> $query = "SELECT description FROM glossary_item WHERE
> name='Alcohol'";
> $result = pg_query($db, $query) or die("Error in query: $query.
> " . pg_last_error($db));
>
> // Print result on screen
> echo "$result";
>
> pg_close($db);
>
> What am I doing wrong?
>