php - 如何在json中编码php数组并解析为jquery

Question

我正在使用 ajax 调用通过下拉获取记录，但问题是它只获取了一条记录，而数据库中有三条记录

这是我的 PHP 代码:

<?php
include 'config/dbconfig.php';

$genid      = $_POST['id'];
$operatorId = $_POST['operatorId'];

$query = mysqli_query($con, "SELECT * FROM generatorrun WHERE generatorId='$genid' AND operatorId='$operatorId'");
while($result = mysqli_fetch_array($query)) {

    $turnOn           = $result['startTime'];
    $turnOff          = $result['endTime'];
    $datetime1        = new DateTime($turnOn);
    $datetime2        = new DateTime($turnOff);
    $interval         = $datetime1->diff($datetime2);
    $datedifference   = $interval->format('%Y-%m-%d %H:%i:%s');
    $startReading     = $result['startReading'];
    $endReading       = $result['endReading'];
    $dailyConsumption = $endReading - $startReading;

    $postData = array(
        "turnOn"           => $turnOn,
        "turnOff"          => $turnOff,
        "runningTime"      => $datedifference,
        "startReading"     => $startReading,
        "endReading"       => $endReading,
        "dailyConsumption" => $dailyConsumption,
    );
}

echo json_encode($postData);
?>

我必须从 MySQL 中获取值并将其存储在关联数组中，然后使用 json_encode() 函数对其进行编码。

这是在 jquery 中获取记录的代码:

<script>
$(document).ready(function () {
    $(".bg-yellow").hide();
    $(".bg-red").hide();
    $("#getGen").change(function () {

        var id = $('#getGen').val();
        var operatorId = $(".opid").val();
        $.ajax({
            type: "POST",
            url: 'getGenerator.php',
            data: {id: id, operatorId: operatorId},
            success: function (response) {
                var data = jQuery.parseJSON(response);
                $(".turnOn").html(data.turnOn);
                $(".turnOff").html(data.turnOff);
                $(".running").html(data.runningTime);
                $(".startReading").html(data.startReading);
                $(".endReading").html(data.endReading);
                $(".dailyConsumption").html(data.dailyConsumption);
                $(".bg-yellow").show();
                $(".bg-red").show();
            }
        });
    });
});
</script>

问题是它只获取了一条记录，我使用 while 循环遍历了 MySQL 表中的所有记录，但它只获取了一条记录

Answer 1

在您的 while 循环中，您正在分配变量。所以每个循环都会覆盖这些值。您必须使用 array_push

$postData = array();
while($result=mysqli_fetch_array($query)){
$turnOn=$result['startTime'];
$turnOff=$result['endTime'];
$datetime1 = new DateTime($turnOn);
$datetime2 = new DateTime($turnOff);
$interval = $datetime1->diff($datetime2);
$datedifference=$interval->format('%Y-%m-%d %H:%i:%s');
$startReading=$result['startReading'];
$endReading=$result['endReading'];
$dailyConsumption=$endReading-$startReading;
  array_push($postData,array(
    "turnOn" => $turnOn,
    "turnOff" => $turnOff,
    "runningTime"=>$datedifference,
    "startReading"=>$startReading,
    "endReading"=>$endReading,
    "dailyConsumption"=>$dailyConsumption
    ));
  }