Question

我正在使用一个简单的PHP脚本，该脚本将以简单HTML形式填充的数据解析为JSON文件。问题是我无法使所有添加的对象都显示在同一数组内，如下所示：

{
  "arraynamefoo": [
    {
      "name": "Testing¹",
      "latitude": "32.75039432",
      "longitude": "-117.01482831",
      "description": "Testing, first description...",
      "category": "park",
      "photourl": "http://www.availableideas.com/wp-content/uploads/2016/10/Colorful-Polygonal-Render-iPhone-6-Plus-HD-Wallpaper.jpg"
    },
    {
      "name": "Testing²",
      "latitude": "32.7896552",
      "longitude": "-117.1686369",
      "description": "Testing, second description. . .",
      "category": "park",
      "photourl": "http://1.bp.blogspot.com/-olGHQWTmk3A/T0tZVSU3CDI/AAAAAAAABbA/EgjurSou6_Q/s1600/12.jpg"
    },
    {
      "name": "Testing³",
      "latitude": "32.89119111",
      "longitude": "-116.86513959",
      "description": "Testing, third description.",
      "category": "park",
      "photourl": "http://getwallpapers.com/wallpaper/full/c/f/0/674534.jpg"
    }
  ]
}

我什至无法在PHP脚本中设置给定的数组名称，所以我也无法使所有写入的数据都属于同一个JSON数组...所以，有人可以帮我指出我需要在PHP代码中进行哪些修改才能使其实现？请先谢谢。

write.php文件：

<?php

    $filetxt = 'file.json';

        if(isset($_POST['name']) && isset($_POST['latitude']) && isset($_POST['longitude']) && isset($_POST['description']) && isset($_POST['category']) && isset($_POST['photourl'])) {
        if(empty($_POST['name']) || empty($_POST['latitude']) || empty($_POST['longitude']) || empty($_POST['description']) || empty($_POST['category']) || empty($_POST['photourl'])) {
            echo 'You need to fill all the fields';
        }
        else {
        $data = array(
          'name'=> $_POST['name'],
          'latitude'=> $_POST['latitude'],
          'longitude'=> $_POST['longitude'],
          'description'=> $_POST['description'],
          'category'=> $_POST['category'],
          'photourl'=> $_POST['photourl'],
        );

        $filetxt = 'file.json';

        $arr_data = array();

        if(file_exists($filetxt)) {
          $jsondata = file_get_contents($filetxt);

          $arr_data = json_decode($jsondata, true);
        }

        $arr_data[] = $data;

        $jsondata = json_encode($arr_data, JSON_PRETTY_PRINT);

        if(file_put_contents('file.json', $jsondata)) echo 'Successfully sent data!';
        else echo 'Error while attempting to write to JSON';
      }
    }
        else echo 'Error writing data to JSON!';
    ?>

此外，write.html文件（也许不是问题所必需的，但还是在这里）：

<form action="write.php" method="post">
Name: <input type="text" name="name" id="name" /></br>
Latitude: <input type="text" name="latitude" id="latitude" /></br>
Longitude: <input type="text" name="longitude" id="longitude" /></br>
Description: <input type="text" name="description" id="description" /></br>
Category: <input type="text" name="category" id="category" /></br>
Photo URL: <input type="text" name="photourl" id="photourl" /></br>
<input type="submit" id="submit" value="Submit" />

Answer 1

通常，StackOverflow喜欢更精确的问题。签出guide on how to ask good questions。我在下面重写了您的代码，并添加了一些注释以解释我在做什么。希望这能回答您的问题。

<?php

// Set constants as such
const FILEPATH = 'file.json';
const FIELDS   = [
  'name',
  'latitude',
  'longitude',
  'description',
  'category',
  'photourl'
];

// Loop over the fields you require rather than typing them out (so much less work!)
foreach (FIELDS as $fieldName) {
  if (!array_key_exists($fieldName, $_POST) || strlen($_POST[$fieldName]) <= 0) {
    echo 'You need to fill all the fields.';
    die();
  }
}

$arr_data = [
  'arraynamefoo' => []
];

if (file_exists(FILEPATH)) {
  $jsonString = file_get_contents(FILEPATH);
  $arr_data = json_decode($jsonString, true);
}

// arraynamefoo is the name you showed in your pastebin. 
// Adding the trailing [] will append the contents of $_POST to 'arraynamefoo'
$arr_data['arraynamefoo'][] = $_POST;

$jsonString = json_encode($arr_data, JSON_PRETTY_PRINT);

if (file_put_contents('file.json', $jsonString)) {
  echo 'Successfully sent data!';
  die();
}

echo 'Error while attempting to write to JSON';

我无法将数据写入JSON

1 个答案: