Question

我正在尝试通过Ajax将一个单独文件中的JSON数组添加到主页中的另一个数组中。由于某种原因，我的Ajax无法正常工作，有4个数组，我需要将它们中的每一个存储在主页中的单独数组中。这就是我得到的。加载文件：

<?php 

  session_start();

  if(!isset($_SESSION['usersId']))
  {
    header("Location: ../index.php");
    exit();
  }
  else
  {
    include_once 'includes/dbh.inc.php';
  }

  $id = $_SESSION['userId']; 
  $dBname = "infosensor";
  $conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBname);

  $sql = "SELECT sensor1, sensor2, sensor3 FROM `$id`;";

  $result = mysqli_query($conn, $sql);
  $jsonsensor1 = array();
  $jsonsensor2 = array();
  $jsonsensor3 = array();
  $jsonsensorsum = array();
  if (mysqli_num_rows($result) > 0)
  {
    while ($row = mysqli_fetch_assoc($result))
    {
      $jsonsensor1[] = intval($row['sensor1'] * ($p = pow(10, 2))) / $p;
      $jsonsensor2[] = intval($row['sensor2'] * ($p = pow(10, 2))) / $p;
      $jsonsensor3[] = intval($row['sensor3'] * ($p = pow(10, 2))) / $p;
      $jsonsensorsum[] = intval(($row['sensor1'] + $row['sensor2'] + $row['sensor3']) * ($p = pow(10, 2))) / $p;
    } 
  }

  echo json_encode($jsonsensor1);
  echo json_encode($jsonsensor2);
  echo json_encode($jsonsensor3);
  echo json_encode($jsonsensorsum);
?>

加载文件的输出：

[5,10,10.99,10.99,13,5,14.31,1,1,5,5,5,1,5,3,3,5,5,1,5,10.32,10.32,5,8,5,10,5,5,19,5,7.36,7.36,5,12.2,12.2,2.2,2.2,23.3,5,10.87,6.87,6.87,5,5,10,10,10,10,5,5,5,5,5,0,5,5]

[8,12.5,12.5,12.53,12.53,8,10.11,1,1,8,8,8,1,8,3,3,8,8,1,8,12.83,32.32,8,8,8,10,8.31,8,10,8,18.2,18.2,8,10.3,10.3,2.29,2.29,12.3,8,8.23,2.23,2.23,8,8,10,10,10,20,5,5,5,5,8,0,8,2]

[6,8.86,8.86,8.87,8.87,6,8.33,1,2,6,2,3,1,6,3,8,6,6,1,6,8.32,7.32,6,8,6,10,3.31,6,12,6,12.3,12.3,6,11.1,11.1,4.09,4.09,33.1,6,5.16,12.16,2.16,6,6,10,20,30,30,30,30,5,0,6,0,6,5]

[19,31.36,32.36,32.4,34.4,19,32.76,3,4,19,15,16,3,19,9,14,19,19,3,19,31.47,49.96,19,24,19,30,16.62,19,41,19,37.86,37.86,19,33.6,33.6,8.6,8.6,68.7,19,24.26,21.26,11.26,19,19,30,40,50,60,40,40,15,10,19,0,19,12]

我在主页上尝试过的操作：

 $(document).ready(function(){
      $.getJSON("loadchart.php", function(jsonsensor1){
        var sensor1 = [$jsonsensor1]; 
        var sensor2 = [$jsonsensor2];
        var sensor3 = [$jsonsensor3];
        var sensorsum = [$jsonsensorsum];
      });
    });

Answer 1

不要使用多个编码/回显调用。这只会创建无效的json。而是将数据放入单个容器（对象或数组）中并对其进行编码。

例如使用普通数组：

$data = [$jsonsensor1,$jsonsensor2,$jsonsensor3,$jsonsensorsum];
echo json_encode($data);

在前端相应地访问它们

$.getJSON("loadchart.php", function(data){
    //get the arrays based on what position you put them 
    //in the array in the backend
    //change this if you end up using a different container, 
    //eg php associative array or object
    var sensor1 = data[0];
    var sensor2 = data[1];
    //and so on
 });

Answer 2

替换此代码：

 echo json_encode($jsonsensor1);
  echo json_encode($jsonsensor2);
  echo json_encode($jsonsensor3);
  echo json_encode($jsonsensorsum);

使用此代码：

    $data=array("jsonsensor1"=>$jsonsensor1,"jsonsensor2"=>$jsonsensor2,"jsonsensor3"=>$jsonsensor3,"jsonsensorsum"=>$jsonsensorsum);
echo json_encode($data);

通过Ajax将JSON数组获取到另一个文件

2 个答案: