返回JSON从MySQL数据库PHP

Question

我在我的数据库中的以下数据：

我想在下面的JSON格式的日期：

{ 
    "labels": ["12.11.2016", "13.11.2016", "14.11.2016", ...], 
    "temperature": ["12", "35", "27", ...], 
    "humidity": ["56", "70", "87", ...] 
} 

我当前的代码是：

$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password); 
     // set the PDO error mode to exception 
     $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     $conn->setAttribute(PDO::ATTR_CASE, PDO::CASE_UPPER); 

     //Select the requested data 
     $sql = "SELECT date, temperature, humidity FROM `TestData` ORDER BY date ASC "; 

     $statement = $conn->prepare($sql); 
     $statement->execute(); 
     $result = $statement->fetchAll(PDO::FETCH_ASSOC); 

     //Values returned for JavaScript 
     $labels = array(); 
     $temperature = array(); 
     $humidity = array(); 
     $data = array(); 

     foreach ($result as $row) 
     { 
      array_push($labels, $row['date']); 
      array_push($temperature, $row['temperature']); 
      array_push($humidity, $row['humidity']); 
     } 
     $result = null; 

     //Load data in one single array 
     $data['labels'] = $labels; 
     $data['temperature'] = $temperature; 
     $data['humidity'] = $humidity; 
     //echo $data; 
     //echo json_encode($labels , JSON_FORCE_OBJECT); 
     echo json_encode($data); 

但这只给我以下sh * t：

有没有人有想法吗？即使我已经在网上搜索了5个小时，我也无法正常工作：/

Answer 1

尝试：

array_push($labels, $row['DATE']); 
array_push($temperature, $row['TEMPERATURE']); 
array_push($humidity, $row['HUMIDITY']); 

看起来像你的列名大写因为$conn->setAttribute(PDO::ATTR_CASE, PDO::CASE_UPPER);。

Answer 2

提取的输出是值对的关键。关键是行号。

此外，您的列名似乎是您发布的变量转储的大写。

试试这个正确读取值：

foreach ($result as $key => $row) 
{ 
    array_push($labels, $row['DATE']); 
    array_push($temperature, $row['TEMPERATURE']); 
    array_push($humidity, $row['HUMIDITY']); 
}