Question

我在对象内有一个数组，我想向该数组添加多个值，但是我的代码开始将它们分开。响应应如下所示：

{
    "requestTime": "1",
    "clients": [{
        "name": "Peter",
        "id": 905
    }]
}

但与此相反，它看起来像这样：

{
    "requestTime": "1",
    "clients": [{
        "name": "Peter"
    }, {
        "id": 905
    }]
}

我的代码：

$myObj = new stdClass();

$myObj->requestTime = $reqtime;
$myObj->clients[]->id = $id;
$myObj->clients[]->name = $name;

$myJSON = json_encode($myObj);

echo $myJSON;

Answer 1

一次性构建数组，而不是分两步生成两个数组。

<!Doctype html>
<html>
<body>
<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $database = "daylight";
    $conn = new mysqli($servername, $username, $password, $database);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }
        $startdate=time()-(time()%(24*3600))-7200;
        $enddate=$startdate+365*24*3600;
        //echo $startdate."<br>".$enddate;
        $s=strftime("%Y-%m-%d", $startdate);
        $e=strftime("%Y-%m-%d", $enddate);

        $Off = 12;  //Offset
        $A = 4;     //Amplidute
        $light=0;

        for($i=$startdate; $i<=$enddate; $i+=$day)  //genData (works)
        {...}

        $sql_request = "SELECT * FROM `db` WHERE `Datum` >= '$s' And `Datum` <= '$e'";
        $daylight_array=array();
        $date_array=array();
        foreach($conn->query($sql_request) as $row)
        {
            $daylight_array[]=$row["Tageslicht"];
            $date_array[]=$row["Datum"];
            //print $row["Tageslicht"];
        }
        // echo json_encode($date_array);
        // echo json_encode($daylight_array);

?>
<div id="chart" style="width: 900px; height: 500px"></div>

<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
google.charts.load('current', {'packages':['corechart']});
google.charts.setOnLoadCallback(drawChart);

    var date = <?php echo json_encode($date_array); ?>;
    var daylight = <?php echo json_encode($daylight_array); ?>;

    function drawChart() {
        var data = google.visualization.arrayToDataTable();
        data.addColumn('string', 'Jahr-Monat-Tag');
        data.addColumn('number', 'Sonnenstunden');

        for(i = 0; i < date.length; i++)
            data.addRow([date[i], daylight[i]]);

        var options = {

          title: 'Sonnenstunden über das Jahr',
          curveType: 'function',
          legend: { position: 'bottom' }
        };

        var chart = new google.visualization.LineChart(document.getElementById('chart'));

        chart.draw(data, options);
  }
</script>
</body>
</html>

Answer 2

尝试做这样的事情：

enterBut.addActionListener(new ActionListener()
        {
               public void actionPerformed(ActionEvent e)
               {

                switch(click)
                    {

                      case 1:
                          double subtotal = Double.parseDouble(Atext.getText());
                          double total = 0.05*Math.round(1.13*subtotal/0.05);
                          answer.setText("The total will be $" + total + " (tax included). Enter the amount of money the customer gave.");
                          int click = 2;

                      case 2:
                          double money = Double.parseDouble(Atext.getText());
                          click = 3;
                          if (money<total)
                          {
                              JOptionPane.showMessageDialog(window, "That is not enough money.");
                              click = 2;
                          }
                    }

               }

        });

Answer 3

如果我根据您的要求输出了解了您的要求，那么这将对您有效，请使用变量代替静态id，name和{{1} }我使用的变量。

requestTime

输出：

<?php  
 $myObj = new stdClass();
 $myObj->requestTime = 1;
 $myObj->clients[] = ['id' => 905, 'name' => 'Peter'];
 $myJSON = json_encode($myObj);
 echo $myJSON;
?>

演示： https://3v4l.org/T9W88

PHP stdclass数组内的多个元素（JSON）

3 个答案: