警告：mysqli :: query（）：無法在C：\ xampp \ htdocs \ palo \ graph.php中獲取mysqli

Question

我有不同的選項卡應顯示不同的圖表。第一個選項卡顯示沒有問題的第一個圖表，但我在第二個選項卡中顯示第二個圖表時遇到問題。當我點擊第二個選項卡上，我只是不斷收到這些錯誤：

Warning: mysqli::query(): Couldn't fetch mysqli in C:\xampp\htdocs\palo\graph.php on line 248
Warning: main(): Couldn't fetch mysqli in C:\xampp\htdocs\palo\graph.php on line 248
Warning: main(): Couldn't fetch mysqli in C:\xampp\htdocs\palo\graph.php on line 248
Error code():

這裏是我的代碼：

 <div id="c2017" class="tabcontent"> 
      <center> 
      <?php 
       $strQuery = "SELECT COUNT(*) as student_count, exam_year, result FROM studentsinfo WHERE exam_year = 2017 GROUP BY result ORDER BY student_count DESC"; 
       $result = $dbhandle->query($strQuery) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}"); 
       if ($result) { 
        $arrData = array(
         "chart" => array(
          "caption" => "TEST RESULTS GRAPH FOR SCHOOL YEAR 2017", 
          "showValues" => "0", 
          "theme" => "fint" 
         ) 
        ); 

        $arrData["data"] = array(); 
         while($row = mysqli_fetch_array($result)) { 
         array_push($arrData["data"], array(
          "label" => $row["result"], 
          "value" => $row["student_count"] 
          ) 
         ); 
         } 
        $jsonEncodedData = json_encode($arrData); 
        $columnChart = new FusionCharts("column2D", "myFirstChart" , 700, 400, "chart-1", "json", $jsonEncodedData); 
        $columnChart->render(); 
        $dbhandle->close(); 
       } 
      ?> 
      <br> 
      <div id="chart-1"></div> 
      </center> 
     </div> 

     <div id="c2018" class="tabcontent"> 
      <center> 
      <?php 
       $strQuery2 = "SELECT COUNT(*) as student_count, exam_year, result FROM studentsinfo WHERE exam_year = 2018 GROUP BY result ORDER BY student_count DESC"; 
       $result2 = $dbhandle->query($strQuery2) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}"); 
       if ($result2) { 
        $arrData2 = array(
         "chart" => array(
          "caption" => "TEST RESULTS GRAPH FOR SCHOOL YEAR 2018", 
          "showValues" => "0", 
          "theme" => "fint" 
         ) 
        ); 

        $arrData2["data"] = array(); 

         while($row2 = mysqli_fetch_array($result2)) { 
         array_push($arrData2["data"], array(
          "label" => $row2["result"], 
          "value" => $row2["student_count"] 
          ) 
         ); 
         } 

        $jsonEncodedData2 = json_encode($arrData2); 
        $columnChart2 = new FusionCharts("column2D", "mySecondChart" , 700, 400, "chart-2", "json", $jsonEncodedData2); 
        $columnChart2->render(); 
        $dbhandle->close(); 
       } 
      ?> 
      <br> 
      <div id="chart-2"></div> 
      </center> 
     </div> 

的錯誤是什麼地方在：

$strQuery2 = "SELECT COUNT(*) as student_count, exam_year, result FROM studentsinfo WHERE exam_year = 2018 GROUP BY result ORDER BY student_count DESC"; 
$result2 = $dbhandle->query($strQuery2) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}"); 

我不知道錯誤是什麼意思，因爲我還是這個新手。我希望有人可以看看這個問題，並幫助我修復我的代碼。非常感謝！

Answer 1

那麼，你在第一個圖表之後關閉$dbhandle，所以它和mysqli連接不可用於第二個查詢。

將其保持打開狀態，直至完成所有查詢。因此，僅在最後一次查詢後使用$dbhandle->close();。