echoing users

boburmum1

i have a simple code to echo out my users in a table.

<?php

include "./global.php";

$sql = "SELECT * FROM `users`";
$res = mysql_query($sql) or die(mysql_error());

while($row = mysql_fetch_assoc($res)){
	echo "Users: " . $row . "<br> \n";
}

if(mysql_num_rows($res) == 0){
	echo "There are no users to look at!\n";
}

?>

but when it comes up, it comes up with the right amount of users but not there names

it comes up with

User:Array
User:Array

and so on. i wrote this myself but i only just started about two weeks ago.

laserlight

That is because mysql_fetch_assoc() returns an array. You want a particular element from this array.

boburmum1

laserlight;10895911 wrote:
That is because mysql_fetch_assoc() returns an array. You want a particular element from this array.

i'm sorry, what?

laserlight

Read the PHP manual on [man]mysql_fetch_assoc/man. There are examples that demonstrate how to use it correctly.

djjjozsi

<?php

include "./global.php";

$sql = "SELECT * FROM `users`";
$res = mysql_query($sql) or die(mysql_error());


if(mysql_num_rows($res) > 0){

	while($row = mysql_fetch_assoc($res)){
		/*  you can get the $row array's elements with the table field's name */
		echo "Users: " . $row["username"] ." - ".$row["user_id"]. "<br /> \n";
	}
	}
	else
		echo "There are no users to look at!\n";

	/* there was a little mistake in my code, sorry about that :)*/
?>

laserlight

Of course, djjjozsi is making the assumption that your columns are named "username" and "user_id". The example code also has a typo error where "ekse" should be "else".

EDIT:
Okay, there is actually yet another typo error. See? Go read the PHP manual as I suggested.

boburmum1

i fixed this bare time ago.