我正在尝试使用 PHP 将数据从 JSON 文件发送到 MySQL 数据库。

我的 99% 都能正常工作，但遇到了一个我无法弄清楚的小问题。这是我的代码:

$con = mysqli_connect($host, $username, $password, $dbname) or die('Error in Connecting: ' . mysqli_error($con));

$st = mysqli_prepare($con, 'INSERT INTO url_feed(url, results, current_date, networks, identifier) VALUES (?, ?, ?, ?, ?)');

mysqli_stmt_bind_param($st, 'sssss', $url, $results, $current_date, $networks, $identifier);

$filename = 'https://www.example.com/random.json';
$json = file_get_contents($filename);   

$data = json_decode($json, true);

foreach ($data as $row) {
$url = $row['url'];
$identifier = $row['identifier'];
$current_date = $row['current_date'];
$results = $row['results'];
$networks = $row['networks'];

    mysqli_stmt_execute($st);
}

mysqli_close($con);

这是一个包含 3 个对象的 JSON 副本:

[
   {
      "url":"http://example1.com",
      "identifier":495755330,
      "current_date":"2015-12-30 17:05:45",
      "results":3,
      "networks":{
         "FaceBook":{"detected":true,"result":"no-result"},
         "Twitter Inc":{"detected":false,"result":"no-result"},
         "Pinterest.com":{"detected":true,"result":"no-result"},
         "Other Sites":{"detected":true,"result":"some-result"}
      }
   },
   {
      "url":"http://example2.com",
      "identifier":495755331,
      "current_date":"2015-12-30 17:05:46",
      "results":0,
      "networks":{
         "FaceBook":{"detected":false,"result":"what-result"},
         "Twitter Inc":{"detected":false,"result":"some-result"},
         "Pinterest.com":{"detected":false,"result":"some-result"},
         "Other Sites":{"detected":false,"result":"what-result"}
      }
   },
   {
      "url":"http://example3.com",
      "identifier":495755332,
      "current_date":"2015-12-30 17:05:47",
      "results":1,
      "networks":{
         "FaceBook":{"detected":false,"result":"some-result"},
         "Twitter Inc":{"detected":true,"result":"some-result"},
         "Pinterest.com":{"detected":false,"result":"some-result"},
         "Other Sites":{"detected":false,"result":"some-result"}
      }
   }
]

现在如果我运行脚本，它会像这样将它插入到数据库中:

id | url                 | results   | current_date        | networks | identifier  | status
1  | http://example1.com | 3         | 2015-12-30 17:05:45 | Array    | 495755330   | queued
2  | http://example2.com | 0         | 2015-12-30 17:05:46 | Array    | 495755331   | queued
3  | http://example3.com | 1         | 2015-12-30 17:05:47 | Array    | 495755332   | queued

但这就是我希望输入的方式:

id | url                 | results   | current_date        | networks                           | identifier | status
1  | http://example1.com | 3         | 2015-12-30 17:05:45 | FaceBook,Pinterest.com,Other Sites | 495755330  | queued
2  | http://example3.com | 1         | 2015-12-30 17:05:47 | Twitter Inc                        | 495755332  | queued

这是我想不通的部分:

它试图以数组形式输入，这显然不起作用 - 它只是在 networks 列下插入文本“Array”。如果 detected 设置为 true，我只想插入数据库。如果不是，我不想在数据库中列出该社交网络。

如果某个对象没有任何社交网络设置为 true，我根本不想输入该对象。这就是为什么您会看到我上面的示例中缺少某些网络以及第二行不存在的原因。

最佳答案

代替 $networks = $row['networks']; 使用这个:

$a = json_decode($row['networks'], true);
$b = array_filter($a, function($el) {
  if ($el['detected'] == true) {
    return true;
  }
});
$c = implode(', ', array_keys($b));
$networks = $c;

关于php - 使用 PHP 将 JSON 信息发送到数据库？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34540733/