Question

运行php文件时出现此错误

警告：mysqli_query（）要求参数1为mysqli，第7行的C：\ wamp \ www \ ims \ graphdata.php中给出的字符串



警告：mysqli_fetch_array（）要求参数1为mysqli_result，在第15行的C：\ wamp \ www \ ims \ graphdata.php中给出null

这是我的php代码：

<?php

$sql = "SELECT YEAR(borrowDate) as year_val, MONTH(borrowDate) as month_val ,COUNT(*) as total FROM itemrecord GROUP BY YEAR(borrowDate), MONTH(borrowDate);";

$result = mysqli_query($sql, $conn);

//start the json data in the format Google Chart js/API expects to receive it
$data = array('cols' => array(array('label' => 'Year', 'type' => 'number'),
                              array('label' => 'Month', 'type' => 'number'),
                              array('label' => 'Total', 'type' => 'number')),
              'rows' => array());

while($row = mysqli_fetch_array($result)){
  array_push($data['rows'], array('c' => array(
  array('v' => $row['Year']),
  array('v' => $row['Month']),
  array('v' => $row['Total']))));
  }  

$jsonData = json_encode($data);  
echo $jsonData; 

?>

Answer 1

您使用mysqli_query功能错误。见http://php.net/manual/en/mysqli.query.php

在你的情况下，你使用程序风格，但不要提供正确位置的链接变量。将其添加为第一个参数或使用OO样式。

编辑：程序样式的示例：

$result=mysqli_query($conn,$sql);

mysqli_query（）期望参数1和mysqli_fetch_array（）期望参数1

1 个答案: