注意:数组到字符串的转换在第74行未知列'在'其中条款'51 - Notice: Array to string conversion in line 74 Unknown column 'Array' in 'where clause'51

Notice: Array to string conversion in line 74 Unknown column 'Array' in 'where clause'51

当我试图在statusUpdate函数中传递我的id时，我得到了这个错误。UPDATE查询显示了这个错误。有人能帮我解决这个问题吗?提前谢谢。

 <select class="form-control" name="setstatus">                    
  <option value="0">Not Solved</option>
  <option value="1">Solved</option>
</select>
<?php
$id = $_GET['bugid'];
if($_SERVER['REQUEST_METHOD'] == 'POST'){ 
   $setstatus = $_POST['setstatus'];
   $insertreport = $pd->statusUpdate($_POST, $id ); 
 }
?>
public function statusUpdate($id){
    $setstatus = $_POST['setstatus'];
    $query     = "UPDATE `bugstable` SET `status`= $setstatus WHERE `bugid`= $id ";
    $updated_rows = $this->db->update($query);
     if ($updated_rows) {
      echo "<span class='success'>Successfully.</span>";
    }else {
     echo "<span class='error'>Not Updated !</span>";
     }
}

首先，你必须在你调用的函数中传递状态参数:

$setstatus = intval($_POST['setstatus']);
$insertreport = $pd->statusUpdate($setstatus,$id);

第二，你只需要在你定义的函数中添加第一个参数作为状态:

public function statusUpdate($setstatus,$id){
    $query     = "UPDATE `bugstable` SET `status`= $setstatus WHERE `bugid`= $id ";
    /** Your rest code **/