Show table data

Hello
I am new to php and I wanted to know how to show all the value of my table. I am using mysql.

Here is my code:

PHP Code:

---


<? 
// db variables 
// inserts variables accordingly 
$host = "misterp2d.dyndns.org"; 
$user = "bye"; 
$pass = "allo"; 
$db = "Jeux"; 
$table = "Usager"; 

// connecting to the db 
$link = mysql_connect ($host, $user, $pass) 
      or die("Impossible de se connecter : " . mysql_error()); 
       
// the db query 
$q = "SELECT * from $table"; 

$results = mysql_select_db ($db); 

// fetching the results 
    $result = mysql_query("SELECT id_Usager, name_Usager FROM Usager"); 

    while ($row = mysql_fetch_array($result)) 
   { 
        printf ("ID: %s  Name: %s", $row[0], $row[1]);  
    } 

    mysql_free_result($result); 

// closing the db link 
mysql_close ($link); 
?>

---

Here is the errors I got:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp\www\test.php on line 29

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp\www\test.php on line 34

it work now!!!:)

Since I don't agree with some of the stuff in that code, I thought I'd give my thoughts:

PHP Code:

---

<?php //#1
    // db variables 
    // inserts variables accordingly 
    $host = "misterp2d.dyndns.org"; 
    $user = "bye"; 
    $pass = "allo"; 
    $db = "Jeux"; 
    $table = "Usager"; 

    // connecting to the db 
    $link = mysql_connect ($host, $user, $pass) 
        or die("Impossible de se connecter : " . mysql_error()); 
    mysql_select_db($db, $link); //#2

    // the db query 
    //$q = "SELECT * from $table"; //#3

    // fetching the results 
    $result = mysql_query("SELECT id_Usager, name_Usager FROM Usager"); 

    while ($row = mysql_fetch_array($result)) 
    { 
        printf ("ID: %s  Name: %s", $row[0], $row[1]);  
    } 

    mysql_free_result($result); //#4

    // closing the db link 
    mysql_close ($link); //#5
?>

---

[list=1][*]It is recommended that you not use short tags (<?). See here.[*]Since mysql_select_db returns a BOOL, all you're doing is storing true and false, and since you're not checking whether it returned true or false, there's no need to assign it to a value.

If you are connection to more than one database, you might want to get in the habit of specifying a resource identifier in this function as well.[*]This is not used in the code. Why have it?[*]This is generally not required in projects unless you are worried about memory. All results are freed when the script finishes execution.[*]This is not required. The resource is closed when the script finishes execution.[/list=1]

Thx Hobo,
PHP is so new to me, I do not understand all, thx man :) I will check all my code to clear it :)

actually, to add on to hobo's insight,

mysql_select_db($db, $link); //#2

that is generally used if you connect to more than one database at a time. other than that, all you need is

mysql_select_db($db);

Quote:

---

Originally posted by phpman
actually, to add on to hobo's insight,

mysql_select_db($db, $link); //#2

that is generally used if you connect to more than one database at a time. other than that, all you need is

mysql_select_db($db);

---

I already said that:

Quote:

---

If you are connection to more than one database, you might want to get in the habit of specifying a resource identifier in this function as well.

---

:p

sorry, I didn't even see that line.