php 使用 NOT 运算符评估条件

Question

我有 4 个销售办事处，一旦所有 4 个销售办事处均收到 20 个销售线索，则只有 4 号销售办事处应该收到当天剩余的销售线索。我需要一些帮助来创建条件，以基本上检查销售办公室 1-4 是否均已收到 20 个潜在客户，然后显然按照说明继续进行

首先，我查询数据库来检查我的计数器（是的，我知道这可以更紧凑，并且可以在循环内完成）

$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'];

当我执行这个逻辑时，它没有按预期工作？一旦 sales_office = 1 达到 20 个销售线索，它就会直接转移到销售办公室 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

您当前对所有办公室计数的逻辑检查，您必须分别检查每个办公室。

if ($sales_office1_count 
此外，您可以通过使用数组来提高代码的可读性：
$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]