將json導入mysql

Question

我有這個json文件

{
    "5927803e892d4c459f2992a3a1ac2ead.16": {
        "lat": 57.532696,
        "lng": 9.949331
    },
    "e25fe65e4fe4468db1361213ba1a376b.16": {
        "lat": 57.501783,
        "lng": 9.947415,
        "name": "Vidstrup Train Station",
        "disabled": false
    },
    "dcd8b44bd0f3478cb2d99c8d5115ba50.16": {
        "lat": 57.480323,
        "lng": 9.968769,
        "name": "Vellingshøj Station",
        "disabled": false
    }
}

我有一個工作的PHP文件來導入它但我真的想要ID也是我的問題是在json文件中沒有像“id”這樣的名稱：“”

 <?php  
 //  
 $connect = mysqli_connect("localhost", "xxxxx", "xxxxxx", "Rails");  
 $filename = "location-export.json";  
 $data = file_get_contents($filename);  
 $array = json_decode($data, true);  
 foreach($array as $row)  
 {  
      $sql = "INSERT INTO location(location_lat, location_lng, location_name, location_mission) VALUES ('".$row["lat"]."', '".$row["lng"]."','".$row["name"]."', '".$row["mission"]."')";       
      mysqli_query($connect, $sql);       
 }  
 echo "Data Inserted";  
 ?> 

Answer 1

更改您的代碼如下

 foreach($array as $id => $row)  
 {  
      $sql = "INSERT INTO location(id, location_lat, location_lng, location_name, location_mission) VALUES ('{$id}','".$row["lat"]."', '".$row["lng"]."','".$row["name"]."', '".$row["mission"]."')";       
      mysqli_query($connect, $sql);       
 }  
 echo "Data Inserted";  
 ?> 

但是當然這比使用預備語句完全不安全（並且可能更慢）。

你應該試試

$stmt = mysql_prepare($link, "INSERT INTO location(id, location_lat, location_lng, location_name, location_mission) VALUES(?,?,?,?)");

 foreach($array as $id => $row)  
 {  
        mysqli_stmt_bind_param($stmt,"ssss", $id,  $row["lat"], ....);
        mysqli_stmt_execute($stmt);


 }