Mysql比较值

Question

我希望从我的下拉列表中获取值，并将其与我的departmentId行进行比较，因此在生成报告时，我只需从departments表中调用departmendId行即可。

<div class="form-group category">
        <label for="category">Departamento: </label>
        <select  class="form-control form-element" name='department'>
        <option selected disabled> --Piliin ang Departamento-- </option>
        <h3><optgroup label="ADMINISTRATIVE SERVICES">  
        <option class='drop-down' value='1'> Office of the City Mayor </option> 
        <option class='drop-down' value='2'> Office of the City Administrator </option>                                         
        <option class='drop-down' value='3'> Human Resources Management Office </option>                                                
        <option class='drop-down' value='4'> Central Records Management Office </option>                                                    
        <option class='drop-down' value='5'> General Services Office </option>
        <option class='drop-down' value='6'> Procurement Office  
    </option>   
        <option class='drop-down' value='7'> Public Information Office </option>
        <option class='drop-down' value='8'> Anti-Graft Board  </option>
        </optgroup>
        </select>
        </div>

//这是我的下拉列表

 if (isset($_GET['department']))
        {
            $department=$_GET['department'];
                              }

//这是我从dropdown获取值的代码

 $depId = "SELECT FROM departments WHERE 'departmentId' = $department";
  $result = mysql_query($depId);
  if($department == $result) {
   $query = "INSERT INTO complain (department,name,address,barangay,complainee,subject,message) VALUES('$result','$fullname','$address','$barangay','$complainee','$subject','$message')";
   $res = mysql_query($query);
   }

//这是我将dropdown的值与departmentId行进行比较时的代码，departmentId的值必须在抱怨表的部门行中，当我查看我的抱怨表部门行时，它什么也没有显示。

Answer 1

第一个正确的SQL SELECT查询必须是

SELECT * FROM departments WHERE departmentId = $department;

然后，您需要使用mysqli_fetch()语句之一获取departmentId的值，最后将其与下面的$department.As进行比较。

 $depId = "SELECT * FROM departments WHERE departmentId = $department"
 $result = mysqli_query($connection,$depId);//$connection is database connection object
 while($row = mysqli_fetch_assoc($result)){
 if($department == $row['departmentId']) {
     $fullname = $row['fullname'];
     $address = $row['address'];
     $barangay = $row['barangay'];
     $complainee = $row['complainee'];
     $subject = $row['subject'];
     $message = $row['message'];
     $query = "INSERT INTO complain (department,name,address,barangay,complainee,subject,message) VALUES('$department','$fullname','$address','$barangay','$complainee','$subject','$message')";
     mysqli_query($connection,$query);
 }
}

Answer 2

您在查询中遇到更改select查询问题，如下所示：

SELECT * 
FROM departments 
WHERE departmentId = $department;

Answer 3

更改此行：

$depId = "SELECT FROM departments WHERE 'departmentId' = $department";

使用

$depId = "SELECT * FROM departments WHERE departmentId = '$department'";

使用

var_dump(...);die();

为了方便起见，您的代码的每一层。例如：

 $depId = "SELECT * FROM departments WHERE 'departmentId' = $department";
 var_dump($depId);die();