Undefined variable when declaring with concatenation [duplicate]

问题

This question already has answers here:

PHP: Right way to declare variable before use in loop (4 answers)

Closed 4 years ago.

I get an undefined variable $ar, $pr and $af.

$sql = mysqli_query($connection, "Select empno, username, password, access_level from personaltab where access_level='ADMIN'");

$cnt = mysqli_num_rows($sql);
$i=0;

while ($r=mysqli_fetch_array($sql)){
$md = md5($r['username']."!@#$%^&*()_+|");
$ar .= $md.", ";
$mdp = md5($r['password']."|+_)(*&^%$#@!<>?:{}[]=-");
$pr .= $mdp.", ";
$af .= $r['empno'].", ";
}

回答1:

You need to initialize the variables before the loop. Otherwise, there's no initial value to concatenate to.

$ar = '';
$pr = '';
$af = '';
while ($r = mysqli_fetch_array($sql)) {
    ...
}

来源：https://stackoverflow.com/questions/29359427/undefined-variable-when-declaring-with-concatenation