Concatenation php variables:

Question

I have a problem in creating variable in a loop and accessing third variable value, i did try many way but now i don't know how to do that.... the code is..

$rand_1 =       random_username($_POST['txtuser_name']);
$rand_2 =       random_username($_POST['txtuser_name']);
$rand_3 =       random_username($_POST['txtuser_name']);

$username   =   "";

for($i=1; $i<=3; ++$i){
   $name   =    "rand_".$i;
   $username .= $name."<br />";
}

echo $username;

Any suggestion.....

alex · Accepted Answer · 2011-05-11 05:58:26Z

3

Try $$name, which is a variable variable.

Still, when you see var_1 etc, generally it means you should be using an array.

Then you could make your code...

$rand = array();

foreach(range(0, 2) as $index) {
    $rand[] = random_username($_POST['txtuser_name']);
}

$username = join('<br />', $rand) . '<br />';

edited May 11, 2011 at 5:58

answered May 11, 2011 at 5:38

alex

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1 Comment

alex Over a year ago

@Ibu Just an easier way to loop through numbers than with a normal for loop.

Gaurav · Accepted Answer · 2011-05-11 05:53:02Z

1

use $username .= $$name."<br />"; instead of $username .= $name."<br />";

But better approach may be

$user=array();

for($i=1; $i<=3; ++$i){
   $user[] =  random_username($_POST['txtuser_name']);
}

echo implode("<br/>", $user)."<br />";

edited May 11, 2011 at 5:53

answered May 11, 2011 at 5:39

Gaurav

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3 Comments

Imran Khan Over a year ago

its all work with $$name,, and the second method is also good.

alex Over a year ago

Your better approach won't produce the same output as the OP's code.

Imran Khan Over a year ago

@lex, hey man you both are right but in loop i also want to check in db. so thats why it is not perfect to implode or join all username, but simply check in loop if not exist in d.b then join with username otherwise go next....so $query1 = "select count(*) from member_details where username ='".$$name."'"; $nameCount = dbQuery($query1, 'count'); if($nameCount==0){ $username .= $$name."<br />"; }

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