Question

我有一个小脚本查询我的数据库＆amp;建立一个谷歌图表。

效果很好，我看不出任何问题。但是，错误日志显示不同的故事。我得到以下内容：

错误记录 mysqli_fetch_array（）期望参数1为mysqli_result，在第186行的/var/www/html/header.php中给出null

包含错误的代码行：

这是脚本。这是第186行：＆＃34; while（$ row = mysqli_fetch_array（$ exec））{＆＃34; - 大约是下面的一半。

完整脚本

<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
    google.load("visualization", "1", {packages:["corechart"]});
    google.setOnLoadCallback(drawChart);
    function drawChart() {
     var data = google.visualization.arrayToDataTable([

       ['Date', 'Total Orders'],
    <?php 
              $memberid = getLoggedMemberID();
    $query = "SELECT date as complain_Date1, IFNULL(sum(count),0) as count
    FROM (
    SELECT Date_Closed as complain_Date1, count(id) as count
    FROM Complaints
    INNER JOIN membership_userrecords ON Complaints.id =   
    membership_userrecords.pkValue
    where membership_userrecords.memberID = '$memberid' AND 
    membership_userrecords.tableName='Complaints' AND Date_Closed IS NOT 
    NULL
    GROUP BY Date_Closed
    ) AS t
    RIGHT JOIN dates ON t.complain_Date1 = dates.date
    Where date between DATE_SUB(NOW(), INTERVAL 30 DAY) and CURRENT_DATE()
    GROUP BY date  
    ORDER BY date asc";

     $exec = mysqli_query($conn,$query);
     while($row = mysqli_fetch_array($exec)){

     echo "['".$row['complain_Date1']."',".$row['count']."],";
    }
 ?>

 ]);

 var options = {
                'chartArea': {'width': '85%', 'height': '70%'},
legend: {position: 'none'},
 colors: ['#2ab77e'],

  is3D: true,
backgroundColor: { fill:'transparent' }

 };
 var chart = new google.visualization.AreaChart(document.getElementById("complaints"));
 chart.draw(data, options);
 }
 </script>

有人能解决我的问题吗？

我在这里试过这个解决方案： error mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given

但它只会使我的整个页面变白，所以必须有错误

Answer 1

替换

 while($row = mysqli_fetch_array($exec)){

与

while($row = mysqli_fetch_array($exec, MYSQLI_ASSOC)){

参考 https://www.tutorialspoint.com/php/mysql_select_php.htm了解详情

如果您需要进一步的帮助，请告诉我

期望参数1是mysqli_result错误消息

1 个答案: