mysql_fetch_array ?

Anon

I am trying to get Julie Meloni's mysql example to work on p. 225 of her book "PHP Fast & Easy Web Development."

Here's the source code:

$db_name = "my_db";
$table_name = "my_music";

$conn = mysql_connect("mywebsite, db_name, pword") or die ("Couldn't connect.");

$db = mysql_select_db($db_name, $conn) or die ("Couldn't select database.");

$sql = "
SELECT id, format, title, artist_fn, artist_ln, rec_label, my_notes, date_acq
FROM $table_name
ORDER BY id
";

$result = mysql_query($sql, $conn) or die("Couldn't execute query.");

while ($row = mysql_fetch_array($result))
{
$id = $row['id'];
$format = $row['format'];
.
.
.
}

The query executes, but the problem is, the $row array ends up containing the field names, not the actual stored values.

Is this fetch array call correct? No errata is listed for this page.

Anon

It's a key value issue. I generally reference the array returned by mysql_fetch_array by an index number. Such as:

while ($row = mysql_fetch_array($result))
{
$id = $row[0];
$format = $row[1];
//etc.

then display the info if you'd like
echo "The CD id is $id, the format is $format ya da ya da.... <br> \n";

.
.
}

Anon

Thanks, but even using constants for the keys it still just prints out the names of the fields, not their values.

I'm assuming that the table was populated correctly (at least MySql says it was). This is one thing I miss about a database like MS Access here, you can't "view" the table as a form to see what's actually in there.

Anon

It was G.I.G.O. that was causing the problem. I didn't have enough checks on the data entry to tell me that garbage data had been entered (forgot to prefix the variables with the '$'), thus the array did the best it could while reading it back.

One final question, if I may. If you knew the number of fields and their order in the table, your scheme of using numbers seems like it would be better than using the typical 'id' 'date' 'format' etc., indexing, no?