Add MySQL result into PHP array

Question

I would like to add results into array and print on screen. Here's my code and nothing is printing...could someone help me take a look.

include('config.php');

$con = mysql_connect($host, $username, $password);
if (!$con){
    die('Could not connect: ' . mysql_error());
}

mysql_select_db('members', $con) or die(mysql_error()) ;

$sql = "SELECT * FROM people where status like '%married%' "; 
$result = mysql_query($sql);

while($row = mysql_fetch_array($result)){
    $result_array[] = $row['id']; // <-------**here**             
}

return $result_array;       

echo($result_array);
mysql_close($con);

Marc B · Accepted Answer · 2011-10-20 14:54:22Z

5

You're doing a return before you do the echo/mysql_close, so the echo/close calls never get executed. Unless you've got more code around this snippet, there's no point in having that return call, as you're not actually in a function, so the return is effectively acting as an "exit()" call, since you're already at the top of the execution stack.

answered Oct 20, 2011 at 14:54

Marc B

362k44 gold badges433 silver badges508 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

999999 Over a year ago

Not to mention using echo on arrays won't give you much.

daiscog · Accepted Answer · 2011-10-20 15:02:17Z

Firstly, change:

mysql_select_db(mtar, $con) or die(mysql_error());

To:

mysql_select_db('mtar', $con) or die(mysql_error());

Then remove return $result_array; - you don't need it and when used outside of a function it just halts execution of the script.

Finally, change

echo($result_array);

to:

print_r($result_array);

EDIT: Some additional thoughts:

You don't need parentheses around the argument to echo - it's actually more efficient to leave them out: echo $var; is quicker than echo($var);

If you're only ever going to use the id column, then don't select the whole row: use SELECT id FROM instead of SELECT * FROM.

Are you sure you need the wildcards either side of "married"? You may well do (depends on what the possible values of status are), but you probably don't. So

$sql = "SELECT id FROM people where status = 'married' ";

May be better,

@daiscog - Thanks for that - it stopped me from cursing myself :)

Collectives™ on Stack Overflow

Add MySQL result into PHP array

2 Answers 2

1 Comment

2 Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

2 Comments

Related