php - 将 JSON 数据输入 MySQL

Question

我正在尝试获取 JSON 数据并将其输入到我的数据库中。

这是我当前的代码:

while($row= mysqli_fetch_assoc($query)){
    $id = $row["id"];
    $domain = $row["domain"];
    $time = date('Y-m-d H:i:s');

    $ch = curl_init();
    $url = "https://www.example.com/";
    curl_setopt($ch, CURLOPT_URL,$url);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, "apikey=123&test=$domain");
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    $output = curl_exec ($ch);
    print $output;

    // CODE SHOULD BE ADDED HERE I THINK

    curl_close ($ch);

    $sql_to_update = "UPDATE monitoring SET socials='$socials', update_time='$time' WHERE id='$id'";
    mysqli_query($conn,$sql_to_update);
}

这是我得到的 JSON 结果的示例:

{
   "resource":"random data",
   "tests":100,
   "total":250,
   "networks":{
      "FaceBook":{
         "detected":true,
         "result":"ignore"
      },
      "Twitter Inc":{
         "detected":false,
         "result":"ignore"
      },
      "MySpace":{
         "detected":true,
         "result":"ignore"
      },
      "Pinterest":{
         "detected":true,
         "result":"ignore"
      },
      "Instagram":{
         "detected":false,
         "result":"ignore"
      }
   }
}

我需要做的就是获取 networks有 detected设置为 true并使用变量 $socials 将它们插入到我的数据库中.

例如，对于上面列出的 JSON 示例，我希望将以下内容输入到 socials 中数据库列:FaceBook, MySpace, Pinterest

如果没有networks有 detected设置为 true , 那么我要输入 NONE FOUND改为进入该列。

有人知道我如何使用 JSON 输出将其放入我的数据库吗？

Answer 1

Check the code to get the desired data that you want to update:-

<?php
$link = '{
   "resource":"random data",
   "tests":100,
   "total":250,
   "networks":{
      "FaceBook":{
         "detected":true,
         "result":"ignore"
      },
      "Twitter Inc":{
         "detected":false,
         "result":"ignore"
      },
      "MySpace":{
         "detected":true,
         "result":"ignore"
      },
      "Pinterest":{
         "detected":true,
         "result":"ignore"
      },
      "Instagram":{
         "detected":false,
         "result":"ignore"
      }
   }
}'; // suppose json variable is $link
$dataArr = json_decode($link);
echo "<pre/>";print_r($dataArr);// print the variable to see how json data looks when it converted to php array
$socials = '';
foreach($dataArr->networks as $key=>$dataA){
    if($dataA->detected == 1){
        $socials .= $key.',';
    }

}
$socials = trim($socials,',');
echo $socials;
?>

输出:- https://eval.in/506693

现在输入您的更新代码即可。