php 使用 NOT 運算子評估條件

Question

我有 4 個銷售辦事處，一旦所有 4 個銷售辦事處均收到 20 個銷售線索，則只有 4 號銷售辦事處應該收到當天剩餘的銷售線索。我需要一些幫助來創建條件，以基本上檢查銷售辦公室 1-4 是否均已收到 20 個潛在客戶，然後顯然按照說明繼續進行

首先，我查詢資料庫來檢查我的計數器（是的，我知道這可以更緊湊，並且可以在循環內完成）

$sales_office= 1;
$sql = $conn->prepare("SELECT LeadsReceivedToday FROM Lead_Counter WHERE SalesOffice=?");
$sql->execute([$sales_office]);
$result = $sql->fetch();
$sales_office1_count = $result['LeadsReceivedToday'];

$sales_office= 2;
$sql = $conn->prepare("SELECT LeadsReceivedToday FROM Lead_Counter WHERE SalesOffice=?");
$sql->execute([$sales_office]);
$result = $sql->fetch();
$sales_office2_count = $result['LeadsReceivedToday'];

$sales_office= 3;
$sql = $conn->prepare("SELECT LeadsReceivedToday FROM Lead_Counter WHERE SalesOffice=?");
$sql->execute([$sales_office]);
$result = $sql->fetch();
$sales_office3_count = $result['LeadsReceivedToday'];

$sales_office= 4;
$sql = $conn->prepare("SELECT LeadsReceivedToday FROM Lead_Counter WHERE SalesOffice=?");
$sql->execute([$sales_office]);
$result = $sql->fetch();
$sales_office4_count = $result['LeadsReceivedToday'];

當我執行這個邏輯時，它沒有按預期工作？一旦 sales_office = 1 達到 20 個銷售線索，它就會直接轉移到銷售辦公室 4？

if ($sales_office1_count != 20 && $sales_office2_count != 20 && $sales_office3_count != 20)
{
    //distribute leads to sales office 1-4 in order of 1-4
    
}
else
{
   //only send leads to sales office no 4
}

Answer 1

您目前對所有辦公室計數的邏輯檢查，您必須分別檢查每個辦公室。

if ($sales_office1_count 
此外，您可以透過使用陣列來提高程式碼的可讀性：
$officesLeadsCount = [1 => 0, 2 => 0, 3 => 0, 4 => 0];

foreach ($officesLeadsCount as $number => $value) {
  $sql = $conn->prepare("SELECT LeadsReceivedToday FROM Lead_Counter WHERE SalesOffice=?");
  $sql->execute([$number]);
  $result = $sql->fetch();
  $officesLeadsCount[$number] = $result['LeadsReceivedToday'];    
}


if ($officesLeadsCount[1]