Question

我遇到了一个不寻常的问题，一个我真的无法相信的问题实际上正在发生，但它正在发生并且无法找到解决方法。如果有人知道我做错了什么，请帮助我。

我需要一个数组$EmpExists=array();，它在检查数据库中是否有与之关联的员工的所有业务后显示结果。我使用PHP从数据库中获取一些值，并检查值是否为空。我有以下表格和数据行：

Table `EmployeeList`
Columns   EmpID       BusinessID
Row 1 `emp-000001`,`business-000001`
Row 2 `emp-000002`,`business-000002`



Table `BusinessList`
Columns    BusinessID
Row 1   `business-000001`
Row 2   `business-000002`
Row 3   `business-000003`

我正在使用以下php代码来调用商家列表：

<?php
$BusinessIDforthis = array();
$select_BusinessIDs = "SELECT BusinessID FROM BusinessList ORDER BY BusinessID ASC;";
$select_BusinessIDs_query = mysqli_query($connection, $select_BusinessIDs);
if (!$select_BusinessIDs_query) {
    die ("Database query for searching BusinessID failed.");
}
while ($BusinessIDs_array = mysqli_fetch_assoc($select_BusinessIDs_query)) {
    $BusinessIDforthis[] = $BusinessIDs_array["BusinessID"];
}

这给了我一个BusinessID的数组，然后我使用以下php代码来获取EmpID的{{1}}员工

BusinessID

现在$EmpID = ''; $EmpExists = array(); $EmpIDRecord = array(); foreach ($BusinessIDforthis as $x) { $select_EmpID = "SELECT EmpID FROM EmployeeList WHERE BusinessID='{$x}';"; $select_EmpID_query = mysqli_query($connection, $select_EmpID); if (!$select_EmpID_query) { die ("Database query for searching EmpID failed."); } while ($EmpID_array = mysqli_fetch_assoc($select_EmpID_query)) { $EmpID = $EmpID_array["EmpID"]; if (empty($EmpID)) { array_push($EmpExists, 'EmpNotExists'); array_push($EmpIDRecord, 'Employee does not exist.'); } else { array_push($EmpExists, 'EmpExists'); array_push($EmpIDRecord, $EmpID); } $EmpID = ''; } }显示一个包含以下答案的数组：

$EmpExists=array();

和Array ( [0]=> EmpExists [1]=> EmpExists [2]=> EmpExists )显示一个包含以下答案的数组：

$EmpIDRecord=array();

和Array ( [0]=> emp-000001 [1]=> emp-000002 )显示一个包含以下答案的数组：

$BusinessIDforthis=array();

问题是，我需要它将Array ( [0]=> business-000001 [1]=> business-000002 [2]=> business-000003 )中的最后一项显示为$EmpExists=array();，因为来自EmpNotExists的最后一个BusinessID没有任何内容在BusinessList 中记录。我怎么能这样做，请尽可能指导我？

Answer 1

$select_EmpID_query = mysqli_query($connection, $select_EmpID);将始终返回true，除非资源错误，这意味着它将循环次数与其真实次数（BusinessList = 3次循环中的3行）。您应该使用mysqli_num_rows()代替。

真的，你应该使用JOIN来获取你的数据：

SELECT * FROM BusinessList AS BL
LEFT JOIN EmployeeList as EL ON BL.BusinessID = EL.BusinessID
ORDER BY BL.BusinessID ASC

另外，$EmpID = $EmpID_array["EmployeeID"];这不应该是$EmpID = $EmpID_array["EmpID"];？

Answer 2

要检查上述问题的空状态mysqli_fetch_assoc总是产生true，可以使用mysqli_num_rows来代替上述问题，方法如下：

$EmpID = '';
$EmpExists = array();
$EmpIDRecord = array();
foreach ($BusinessIDforthis as $x) {
  $select_EmpID = "SELECT EmpID FROM EmployeeList WHERE BusinessID='{$x}';";
  $select_EmpID_query = mysqli_query($connection, $select_EmpID);
  if (!$select_EmpID_query) {die ("Database query for searching EmpID failed.");}
 else
 {$EmpIDrows = mysqli_num_rows($select_EmpID_query);
  if ($EmpIDrows===0) {
    array_push($EmpExists, 'EmpNotExists');
    array_push($EmpIDRecord, 'Employee does not exist.');
   } else {
    array_push($EmpExists, 'EmpExists');
    array_push($EmpIDRecord, $EmpID);
   }
    $EmpIDrows= '';
   }
   }

php empty（）无法正常工作

2 个答案: