Question

我试图从json文件向表中插入数据，但行得到0.而不是来自json的值

JSON代码：

{
"posts": [{
    "dr_DeviceID": "323",
    "dr_UserLocalLat": "38.7482572",
    "dr_UserLocalLong": " -9.1847516"
}]
}

$connection = mysql_connect("localhost", "***", "!*****!");
if (!$connection)
{
    die('PHP Mysql database connection could not connect : ' . mysql_error());
} 
$db_selected = mysql_select_db("*****", $connection);


$result=mysql_query("SELECT * FROM $tbl_name wHERE ad_IMEI=ad_IMEI ");
$i=0;
while($row=mysql_fetch_array($result)) { 
$response[$i]['dr_DeviceID']  = $row['ad_IDDevice']; 
$response[$i]['dr_UserLocalLat']= $row['user_location_lat'];
$response[$i]['dr_UserLocalLong']= $row['user_location_long'];
$data['posts'][$i] = $response[$i];
$i=$i+2;}
$json_string = json_encode($data);
$file = 'select.json';
file_put_contents($file, $json_string);

$jsondata = file_get_contents('select.json');
$obj = json_decode($jsondata, true);
$id = $obj['posts']['dr_DeviceID'];
$dr_UserLocalLat = $obj['posts']['dr_UserLocalLat'];
$dr_UserLocalLong = $obj['posts']['dr_UserLocalLong'];
$sqlj = "INSERT INTO $tbl_name1 (dr_DeviceID, dr_UserLocalLat, dr_UserLocalLong) VALUES('$dr_DeviceID', '$dr_UserLocalLat', '$dr_UserLocalLong')";
$result=mysql_query($sqlj,$connection);

Answer 1

问题是你正在尝试访问一个对象数组，就像它是一个对象一样。这条线就在这里 $data['posts'][$i] = $response[$i]; 你在$ data ['posts']数组中添加了一个项目。如果您的结果有多行，那么您上面留下的json示例将是

{
 "posts": [{
   "dr_DeviceID": "323",
   "dr_UserLocalLat": "38.7482572",
   "dr_UserLocalLong": " -9.1847516"
  },
  {
   "dr_DeviceID": "324",
   "dr_UserLocalLat": "39.7482572",
   "dr_UserLocalLong": " -19.1847516"
  }]
}

因此，当您在之后解码json时，会得到一个对象数组。要访问数组中的每个项目，您需要一些循环周期。否则，要从json获取第一个项目，您需要执行 $obj['posts'][0]['dr_UserLocalLat'], instead of $obj['posts']['dr_UserLocalLat']。

从json将数据从json插入到mysql表中

1 个答案: