John Davidson

Message:

I have two variables, number and user, which are not fixed. When I iterate through the array below, how do I determine if the number and user criteria are met to get the appropriate ID

Example:

If the value of amount is 100 and the value of user is 6, then the obtained ID is 1；

If the value of amount is 50000 and the value of user is 10, the obtained ID is also 1;

If the value of amount is 50000 and the value of user is 12, the obtained ID is also 2;

Why do I always get 1?

$amount = 10000;

$user = 12;

$id = null;

foreach ($array as $item) {

  if ($amount >= $item['value']['amount']['min'] && $user >= $item['value']['user']['min']) {

    $id = $item['id'];

    break;

  }

}

echo $id;

The array as follows：

$array = [

  [

    'id'    => 1,

    'value' => [

      'amount' => [

         'min' => 100,

         'max' => 9999 

       ],

      'user'   => [

         'min' => 5,

         'max' => 10 

       ],

      'ratio'  => 10

    ]

  ],

  [

    'id'    => 2,

    'value' => [

      'amount' => [

         'min' => 10000,

         'max' => 49999 

       ],

      'user'   => [

         'min' => 11,

         'max' => 30 

       ],

      'ratio'  => 30

    ]

  ],

  [

    'id'    => 3,

    'value' => [

      'amount' => [

         'min' => 50000,

       ],

      'user'   => [

         'min' => 50,

       ],

      'ratio'  => 35

    ]

  ] 

];

---

My Answer:
= $item['amount'] && $user == $item['user']) {
$id = $item['id'];
break;
}
}

This code snippet iterates through the array and checks if the current item's 'amount' and 'user' values match the given $amount and $user variables. If a match is found, the corresponding 'id' value is assigned to the $id variable and the loop is exited using the 'break' statement.

If you are always getting 1 as the result, it could be due to the fact that the conditions in the if statement are not being met for the given $amount and $user values. Make sure that the values in the array match the values you are checking for in the loop.