Question

这是我在class.php中的代码：

 public function select(){
        $stmt = $this->conn->prepare("SELECT country FROM `country`") or die($this->conn->error);
        if($stmt->execute()){
            $result = $stmt->get_result();
            return $result;
        }
    }


public function read(){
        $stmt = $this->conn->prepare("SELECT segment FROM `segments`") or die($this->conn->error);
        if($stmt->execute()){
            $result = $stmt->get_result();
            return $result;
        }
    }

现在在index.php中我这样做了：

                            <th class="text-center">country</th>

                            <th class="text-center">segments</th>
            </thead>
            <tbody>
            <?php

                require 'class.php';

                $conn = new db_class();
                $read = $conn->select();
                                    $test = $conn->read();

                while($fetch = $read->fetch_array(MYSQLI_ASSOC)&& $fetch1 = $test->fetch_array(MYSQLI_ASSOC)){ 
                                        foreach ($fetch1 as $field => $values) {


                                      foreach($fetch as $field=>$value){
   echo '<tr><td>' . $value . '</td>'
           . '<td>'. $values .'</td>';
}

                                        }
}       


            ?>




            </tbody>
        </table>

我只是试图从2个表中获取数据并将它们放在一个html表中

我得到这个错误6次：

Warning: Invalid argument supplied for foreach() in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\segments\countrySegment.php on line 59

我只是试图从2个表中获取数据并将它们放在一个html表中如果有人知道如何，我希望每个国家都有一个带有分段值的下拉菜单任何的想法？谢谢你提前

Answer 1

这是因为数组指针问题

foreach ($fetch1 as $field => $values) {
   foreach($fetch as $field=>$value){
      echo '<tr><td>' . $value . '</td>'
           . '<td>'. $values .'</td>';
   }
}

第一次第二次foreach打印下一次没有任何值的东西它超出了值..

因此您需要使用重置

重置阵列

foreach ($fetch1 as $field => $values) {
   reset($fetch)
   foreach($fetch as $field=>$value){
      echo '<tr><td>' . $value . '</td>'
           . '<td>'. $values .'</td>';
   }
}

或者您可以将其保存为临时变量

foreach ($fetch1 as $field => $values) {
   $fetchtmp = $fetch;
   foreach($fetchtmp as $field=>$value){
      echo '<tr><td>' . $value . '</td>'
           . '<td>'. $values .'</td>';
   }
}

foreach里面的foreach错误

1 个答案: