javascript - 创建具有正确格式的索引 JSON

Question

我需要在 PHP 中创建一个 JSON 数组，如下所示。该数组需要使用 PHP While 循环创建。

[{
    "latitude":"124",
    "longitude":"",
    "altitude":"87654",
    "title":"test3",
    "description":"This is a test 3",
    "urlidle":"0",
    "urlselected":"0"
}, {
    "latitude":"1ert",
    "longitude":"67",
    "altitude":"9",
    "title":"tes3456",
    "description":"This is a test 123",
    "urlidle":"www.demo.blah",
    "urlselected":"0demo.blah"
}]

我实现的PHP循环如下。

$poi=array();
while ($row = $output->fetch_array(MYSQLI_ASSOC)) {

    $poi[$x]['latitude'] = $row['lat'];
    $poi[$x]['longitude'] = $row['long'];
    $poi[$x]['type'] = $row['type'];
    $poi[$x]['title'] = $row['title'];
    $poi[$x]['description'] = $row['desc'];
    $poi[$x]['urlidel'] = $row['url1'];
    $poi[$x]['urlselected'] = $row['url2'];
    $x++;
}

并且可以通过 JS for 循环方法读取

var jqxhr = $.getJSON(serverUrl, function(data) {

    loadPoisFromJsonData(data);

}

但似乎不起作用

其中 ServerUrl 是与 JSON 相呼应的 PHP 文件。

我需要按如下方式读取 JSON。

function loadPoisFromJsonDataFn(poiData) {

    for (var i = 0; i < poiData.length; i++) {
        var singlePoi = {
            "id": poiData[i].id,
            "latitude": parseFloat(poiData[i].latitude),
            "longitude": parseFloat(poiData[i].longitude),
            "altitude": parseFloat(poiData[i].altitude),
            "title": poiData[currentPlaceNr].name,
            "description": poiData[i].description,
            "urlidle": poiData[i].urlidle,
            "urlselected": poiData[i].urlselected
        };

    }
}

Answer 1

您应该使用json_encode()将数据添加到数组后

$poi=array();
$x = 0;
while ($row = $output->fetch_array(MYSQLI_ASSOC)) {

    $poi[$x]['latitude'] = $row['lat'];
    $poi[$x]['longitude'] = $row['long'];
    $poi[$x]['type'] = $row['type'];
    $poi[$x]['title'] = $row['title'];
    $poi[$x]['description'] = $row['desc'];
    $poi[$x]['urlidel'] = $row['url1'];
    $poi[$x]['urlselected'] = $row['url2'];
    $x++;
}
print_r($poi);
echo json_encode($poi);

不需要在JS中再次循环。结果(poiData)已经是一个 JSON

var jqxhr = $.getJSON(serverUrl, function(data) {

  console.log(data);
  //See it in Firebug, it's already a JSON.

  alert( JSON.stringify(data) );
  //If you want to see it as string

}