Re: display query results

I want to thank everyone for their contribution. As usual, the solution
turns out to be simple. Remeber the KISS principle.
see the note within the text: the LIKE condition did it. But I will have
more questions as I grope further in the dark. :))

Andy Shellam wrote:
> Just shooting in the dark here, but I'm thinking there might be some
> extra spaces around the column values.  For some reason PostgreSQL is
> not returning any rows for that query, which is where the root of your
> problem lies.
>
> Do you have PgAdmin?  If so try the exact same query against the same
> database and server.
> Also try changing your query as follows:
>
> SELECT description FROM glossary_item WHERE name LIKE '%Alcohol%'
Great aim for shooting in the dark. :)) That was it.
Now, could someone explain why I need the LIKE statement?
I will try to check the documentation...
>
> and see what you get.
>
> Regards,
>
> Andy
>
>
> PJ wrote:
>> Annotated within text below:
>>
>> Andy Shellam wrote:
>>> PJ wrote:
>>>> Lynna Landstreet wrote:
>>>> Well, it does explain things a little. Unfortunately, I have tried
>>>> about everything imaginable before posting except the right thing.
>>>> I can not visualize what it is that my query is returning. Here is
>>>> what the code is:
>>>>
>>>> Whatever  I enter as values for pg_fetch_result, the screen output
>>>> is :
>>>>
>>>> resource(3) or type (pgsql result)
>>>> *Warning*: pg_fetch_result() [function.pg-fetch-result
>>>> <http://biggie/k2/function.pg-fetch-result>]: Unable to jump to row
>>>> 1 on PostgreSQL result index 3 in
>>>> */usr/local/www/apache22/data/k2/test1_db.php* on line *29*
>>>
>>> This suggests that there is no row 1 in your result-set.  I believe
>>> it is zero-based, so try fetching row 0 if your query only returns 1
>>> row.
>> Been there, done that. No change.
>>>
>>>
>>>> I don't understand what $resuts is returning - if it is an entire
>>>> row, the one that the field is in that I am looking for, then why
>>>> do I not get a printout of the text that is in that field? The row
>>>> in the table is the second row and the field I am trying to
>>>> retrieve is the 4th field.
>>> $results as explained previously is just a pointer to a recordset.
>>> This analogy isn't brilliant, but think of your database table as a
>>> book.  Each row on a page within that book is a record, and the
>>> words in that row are the data in the table's columns.
>>>
>>> When you run a query, think of yourself looking at the book's index
>>> for a given word.  The index will tell you the pages that word is
>>> on.  That's your $results - simply a pointer to your data.
>>>
>>> You then need to turn to that page in the book (pg_fetch_*
>>> functions) to start examining the lines for the word you want.  Once
>>> you've got your line, you can find the word (column/data, from your
>>> array) you're looking for.
>>>
>>> Now turn that into PHP and SQL.  You run your query (looking in the
>>> book's index) and the PostgreSQL driver will save the results into a
>>> block of memory in your server's RAM, returning a resource
>>> identifier.  This is literally just saying "resource #3 is located
>>> at this location in the computer's memory."  When you look up a
>>> record from that result-set, PHP then knows where to look for the data.
>>>
>>> I never really use the "or die" syntax, I tend to explicitly check
>>> the return values of the functions.  Try this:
>>>
>>> <?php
>>>       $db = pg_connect("host=localhost port=5432 dbname=med user=med
>>> password=0tscc71");
>>>
>>>       // Note: according to
>>> http://uk2.php.net/manual/en/language.types.boolean.php a resource
>>> always evaluates to true,
>>>       // therefore !$db may not evaluate to false when connection
>>> fails.
>>>        if ($db === false)
>>>        {
>>>            die("Could not open connection to database server");
>>>        }
>>>
>>>         // generate and execute a query
>>>        $query = "SELECT description FROM glossary_item WHERE
>>> name='Alcohol'";
>>>        $results = pg_query($db, $query);
>>>        var_dump ($results);
>>>
>>>       if ($results === false)
>>>       {
>>>          die("SQL query failed: " . pg_last_error($db));
>>>       }
>>>       else if (pg_num_rows($results) == 0)
>>>       {
>>>          // Only do this if you were expecting at least 1 row back
>>>          die("SQL query returned no rows");
>>>       }
>>>
>>>       $results_formatted = pg_fetch_all($results);
>>>       echo "<pre>"; // need this to show output better in a HTML page
>>>       var_dump($results_formatted);
>>>       echo "</pre>"; // need this to show output better in a HTML page
>>>
>>>       /*
>>>          $results_formatted will then be set out like follows:
>>>
>>>          $results_formatted[row_index][column_name] = column_value
>>>       */
>>>              pg_close($db);
>>>       ?>
>> Tried your coding - returns: resource(3) of type (pgsql result) SQL
>> query returned no rows
>> The row is there... isn't that what were asking for?
>> To go by the book, I even changed the description to * as noted
>> before below.
>> Something is rotten in Denmark.
>> This is getting ridiculous - I have followed the instructions as
>> specified in the Postgresql documentation and examples - and it just
>> doesn't work.
>>
>> The db is like this..
>> int4        ||int4              || varchar(32)|| text
>> _item_id  ||  glossary_id ||  name         || description _
>> 2           ||        1          || Alcohol      || One of thetwo
>> major.... blah...blah.. blah
>>>> Am I querying correctly? The table is "glossary_item", the row I
>>>> want is the one that is unique in containing the word "Alcohol" in
>>>> the column "name"
>>>>
>>>> I changed:   $query = "SELECT * FROM glossary_item WHERE name=
>>>> 'Alcohol'";
>>>> same result
>>>>
>>>> Picture me tearing out my hair...
>>>>
>>>
>>> Regards,
>>>
>>> Andy
>>>
>>
>>
>