我有 4 个销售办事处，一旦所有 4 个销售办事处每个都收到 20 条线索，那么只有 4 号销售办事处应该收到当天剩余的线索。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 是否每个都收到了 20 个潜在客户，然后显然按照说明继续
首先，我查询数据库以检查我的计数器（是的，我知道这可以更紧凑，而是在循环中完成）
$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 = $result['LeadsReceivedToday'];当我执行此逻辑时，它没有按预期工作？只要 sales_office = 1 达到 20 个潜在客户，它就直接移动到销售办公室 4？
if ($sales_office1_count != 20 && $sales_office2_count != 20 && $sales_office3_count != 20)
{
//distribute leads to sales office 1-4 in order of 1-4
}
else
{
//only send leads to sales office no 4
}回答1
您当前对所有办公室计数的逻辑检查，您必须分别检查它们中的每一个。
if ($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
}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] < 20 || $officesLeadsCount[2] < 20 || $officesLeadsCount[3] < 20)
{
//distribute leads to sales office 1-4 in order of 1-4
}
else
{
//only send leads to sales office no 4
}