警告：mysqli :: query（）：无法在C：\ xampp \ htdocs \ palo \ graph.php中获取mysqli [重复]

Question

这个问题在这里已有答案：

• mysqli::query(): Couldn't fetch mysqli 4个答案

我有不同的选项卡，应显示不同的图表。第一个选项卡显示第一个图表没有问题，但我在第二个选项卡中显示第二个图表时遇到问题。当我点击第二个标签时，我只是继续收到这些错误：

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 ():

这是我的代码：

        <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>

错误在某处：

 $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}");

我不知道错误意味着什么，因为我还是新手。我希望有人可以看看这个问题，并帮我修复我的代码。非常感谢！

Answer 1

好吧，你在第一个图表后关闭$dbhandle，所以它和mysqli连接不能用于第二个查询。

在所有查询完成之前保持打开状态。因此，仅在最后一次查询后使用$dbhandle->close();。