accessing information from a db

darkhalf · March 6th, 2004, 05:29 PM

Hi,

I'm new w/ php/mysql... I trying to display information from a db.... Here is the code in question.. Below that is how I created the db, it went through ok, then the error I get when I try it out... Any help would be great.

<html>
<body>
<?
 $db = mysql_connect("localhost", "mysql");
 mysql_select_db("mydb",$db);
 $result = mysql_query("SELECT * FROM employees",$db);

 printf("First Name: %s \n", mysql_result($result,0,"first"));
 printf("Last Name: %s \n", mysql_result($result,0,"last"));
 printf("Address: %s \n", mysql_result($result,0,"address"));
 printf("Position: %s \n", mysql_result($result,0,"position"));
?>
</body>
</html>

CREATE TABLE employees ( id tinyint(4) DEFAULT '0' NOT NULL AUTO_INCREMENT, first varchar(20), last varchar(20), address varchar(255), position varchar(50), PRIMARY KEY (id), UNIQUE id (id));INSERT INTO employees VALUES (1,'Bob','Smith','128 Here St, Cityname','Marketing Manager');
INSERT INTO employees VALUES (2,'John','Roberts','45 There St , Townville','Telephonist');
INSERT INTO employees VALUES (3,'Brad','Johnson','1/34 Nowhere Blvd, Snowston','Doorman');

mysql -u mysql mydb < mydb.dump

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/local/apache/htdocs/php/db_test.php on line 25
First Name:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/local/apache/htdocs/php/db_test.php on line 29
Last Name:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/local/apache/htdocs/php/db_test.php on line 33
Address:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/local/apache/htdocs/php/db_test.php on line 37
Position:

nikolai · March 12th, 2004, 04:22 PM

for starters, you should be checking whether the previous mysql functions are successful. Use mysql_error() to determine this.

For example:

$db = mysql_connect("localhost", "mysql");
if ($db == FALSE)
{
 die("Couldn't connect to MySQL. MySQL said: " . mysql_error());
}

$link = mysql_select_db("mydb",$db);
if ($link == FALSE)
{
 die("Couldn't select database. MySQL said: " . mysql_error());
}

$result = mysql_query("SELECT * FROM employees",$db);
if ($result == FALSE)
{
 die("Query failed. MySQL said: " . mysql_error());
}

?>

Also, you should use mysql_fetch_array() instead of several calls to mysql_result() if you're going to be displaying more than one column's data from each row.

 $row = mysql_fetch_array($result);

 echo "First Name: {$row['first']} \n";
 echo "Last Name: {$row['last']} \n";
 echo "Address: {$row['address']} \n";
 echo "Position: {$row['position']} \n";

It's difficult to give you more insight because the line numbers in your error messages do not correspond to the code you posted.

I strongly suggest reading through the MySQL section of the PHP manual:
 http://www.php.net/mysql

Take care,

Nik
http://www.bigaction.org/