Question

我正在尝试使用将所有对象插入我的数据库的循环解码json，以下是我的PHP代码，但我有错误：致命错误：无法在phptest中使用stdClass类型的对象作为数组。第21行的PHP

$postdata = file_get_contents("php://input");
$request = json_decode($postdata);


$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";


$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}else{
    foreach($request as $key => $value) {
    $equip = $value['equipment'];
    $sql = "INSERT INTO contratos (equipo) VALUES ('$equip')";

}
}

我使用angular发布了json，下面的代码是我的controller.js

$scope.continue = function(choices)
 {
     var data = $scope.choices;
     $http.post('php/phptest.php', data)
     .then(function(response) {
      console.log(response);


   });
  };

和json

[  
   {  
      "id":"choice1",
      "quantity":1,
      "equipment":"BLONG - F25",
      "teamvox":true,
      "plandatos":"D50"
  },
  {  
      "id":"choice2",
      "quantity":1,
      "equipment":"OUTERFONE - S17",
      "evidence":true,
      "mobictrl":true,
      "plandatos":"D100"
   }
]

Answer 1

您尝试在foreach循环中使用像数组一样的对象。但是json_decode成了一个对象。您可以通过将true传递给第二个参数来更改此行为。

$postdata = file_get_contents("php://input");
$request = json_decode($postdata, true);


$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";


$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}else{
    foreach($request as $key => $value) {
    $equip = $value['equipment'];
    $sql = "INSERT INTO contratos (equipo) VALUES ('$equip')";

}
}

Answer 2

您可以使用$value->equipment代替$value[equipment]，因为返回的值不是数组而是stdClass对象。

Answer 3

解决房间里的另一头大象; SQL注入漏洞

$postdata = file_get_contents("php://input");
$request = json_decode($postdata);


$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";

// make MySQLi throw exceptions
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new mysqli($servername, $username, $password, $dbname);

$stmt = $conn->prepare('INSERT INTO `contratos` (`equipo`) VALUES (?)');
$stmt->bind_param('s', $equip);

foreach ($request as $data) {
    $equip = $data->equipment; // $data is a stdclass
    $stmt->execute();
}

解码json数组错误

3 个答案: