02-21-2010
I wonder in that case what my problem is...
I issue the following query in phpMyAdmin:
PHP Code:
SELECT venue, SUM( amount )
FROM IWD
WHERE venue = 'Foxy Hollow'
And get this result:
venue SUM(amount)
NULL NULL
If I take away the single quotes, I get an error:
PHP Code:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Hollow' at line 1
SELECT venue, SUM( amount )
FROM IWD
WHERE venue = Foxy Hollow
If I have entries without spaces (for example FoxyHollow) the query works like a charm..
I guess I could convert the names to replace blanks with underscores.. but I'd rather not. :/
Edit: Damn it, id Didn't work.. but, now I really don't understant this.. it worked yesterday.. I HAVE made some changes somewhere, dut damn me if I could remember what.. there IS something wrong with the query.. my apologies... now.. I just need to find out WHAT is wrong with my query.. :/
Last edited by noratx; 02-21-2010 at 06:14 AM..
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LEARN ABOUT PHP
mysql_db_query
MYSQL_DB_QUERY(3) 1 MYSQL_DB_QUERY(3)
mysql_db_query - Selects a database and executes a query on it
SYNOPSIS
Warning
This function was deprecated in PHP 5.3.0, and will be removed in the future, along with the entirety of the original MySQL exten-
sion. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more
information. Alternatives to this function include:
omysqli_select_db(3) then the query
o PDO::__construct
resource mysql_db_query (string $database, string $query, [resource $link_identifier = NULL])
DESCRIPTION
mysql_db_query(3) selects a database, and executes a query on it.
o $database
- The name of the database that will be selected.
o $query
- The MySQL query. Data inside the query should be properly escaped.
o $
link_identifier -The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect(3) is
assumed. If no such link is found, it will try to create one as if mysql_connect(3) was called with no arguments. If no connection
is found or established, an E_WARNING level error is generated.
Returns a positive MySQL result resource to the query result, or FALSE on error. The function also returns TRUE/ FALSE for INSERT/ UPDATE/
DELETE queries to indicate success/failure.
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.0 | |
| | |
| | This function now throws an E_DEPRECATED notice. |
| | |
+--------+---------------------------------------------------+
Example #1
mysql_db_query(3) alternative example
<?php
if (!$link = mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
echo 'Could not connect to mysql';
exit;
}
if (!mysql_select_db('mysql_dbname', $link)) {
echo 'Could not select database';
exit;
}
$sql = 'SELECT foo FROM bar WHERE id = 42';
$result = mysql_query($sql, $link);
if (!$result) {
echo "DB Error, could not query the database
";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_assoc($result)) {
echo $row['foo'];
}
mysql_free_result($result);
?>
Note
Be aware that this function does NOT switch back to the database you were connected before. In other words, you can't use this
function to temporarily run a sql query on another database, you would have to manually switch back. Users are strongly encouraged
to use the database.table syntax in their sql queries or mysql_select_db(3) instead of this function.
mysql_query(3), mysql_select_db(3).
PHP Documentation Group MYSQL_DB_QUERY(3)