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delayedinsanity · 05-01-2008, 05:59 PM

According to the documentation, $result should be an object returned by the query, and the actual location of the variable would be $result->affected_rows. $this->mysqli->affected_rows just returns 0, and $this->mysqli->num_rows doesn't exist (as you're supposed to use $result->num_rows). It works just fine in my other database queries, which is why it's wierding me out that it's not returning $result for this one.

I wound up changing it to;

PHP Code:


			
$q = sprintf("UPDATE " . TBL_USERS . " SET password='%s' WHERE username='%s'", $szPassword, $szUsername);

if ($this->mysqli->query($q)) {
    return 1;
} else {
    return 0;
}

but I'm still curious as to why that one query won't return an object to $result.
-m

05-01-2008, 05:59 PM	#3 (permalink)
delayedinsanity is cute and cuddly Join Date: Mar 2008 Location: Vegas, Baby Posts: 963 Thanks: 31	According to the documentation, $result should be an object returned by the query, and the actual location of the variable would be $result->affected_rows. $this->mysqli->affected_rows just returns 0, and $this->mysqli->num_rows doesn't exist (as you're supposed to use $result->num_rows). It works just fine in my other database queries, which is why it's wierding me out that it's not returning $result for this one. I wound up changing it to; PHP Code: `$q = sprintf("UPDATE " . TBL_USERS . " SET password='%s' WHERE username='%s'", $szPassword, $szUsername); if ($this->mysqli->query($q)) { return 1; } else { return 0; }` but I'm still curious as to why that one query won't return an object to $result. -m