PHP 解码、更改和编码 JSON

Question

我猜这个问题已经在某个地方得到了回答，但我找不到解决方案，我把头发扯掉了。我有一个存储在 MySQL 中的 JSON 数组，如下所示：

[{"ip":"8.8.8.8","name":"Bob"},{"ip":"","name":""},{"ip":"","name":""},{"ip":"","name":""}]

我想替换特定对象的“ip”和“name”。所以我将$slot_num 设置为0 并尝试更改值和UPDATE 数据库。下面的 SELECT 子句应该没问题，因为它在其他地方多次使用。

//Recieves POST info such as ip=1.1.1.1&group=204&slot=3&name=help
$ip = $_POST['ip'];
$group_id = $_POST['group'];
$slot_num = $_POST['slot'] -1; //PHP receives slot num increased by 1. IE- $slot_num 1 would be array[0]
$name = $_POST['name'];

if($result = $mysqli->query("SELECT * FROM `open_groups` WHERE `group_id` = $group_id")) {
    $row = mysqli_fetch_array($result);
    $slot_ar = json_decode($row['players'], true);
    //Check if array has correct number slots
    if($slot_num => count($slot_ar) || !is_int($slot_num)){
        die('Injection attempt');
    }

    $slot_ar[$slot_num]['ip'] = $ip;
    $slot_ar[$slot_num]['name'] = $name;
    $players = json_encode($slot_ar);
    $players = $mysqli->real_escape_string($players);
    if(!$mysqli->query("UPDATE `open_group` SET players = '$players' WHERE group_id = $group_id")) {
        echo $mysqli->error;
        exit;
    }
    if(!$mysqli->query("INSERT INTO `occupied`(`ip`, `group`) VALUES ('$ip', '$group_id')")) {
        echo $mysqli->error;
        exit;
    }
    echo "Success";
}
else echo $mysqli->error;

我是在错误地访问数组还是什么？

固定代码

$ip = $_POST['ip'];
$group_id = $_POST['group'];
$slot_num = $_POST['slot']; //PHP receives slot num increased by 1. IE- $slot_num 1 would be array[0]
$name = $_POST['name'];

if($result = $mysqli->query("SELECT * FROM `open_groups` WHERE `group_id` = $group_id")) {
$row = mysqli_fetch_array($result);
$slot_ar = json_decode($row['players'], true);
//Check if array has correct number slots
if($slot_num-1 >= count($slot_ar) || !is_numeric($slot_num)){
    echo "Injection attempt";
    exit;
}
$slot_ar[$slot_num-1]['ip'] = "$ip";
$slot_ar[$slot_num-1]['name'] = "$name";
$players = json_encode($slot_ar);
$players = $mysqli->real_escape_string($players);
if(!$mysqli->query("UPDATE `open_groups` SET players = '$players' WHERE `group_id` = $group_id")) {
    echo "Update error";
    exit;
}
if(!$mysqli->query("INSERT INTO `occupied`(`ip`, `group`) VALUES ('$ip', '$group_id')")) {
    echo "Occupied error";
    exit;
}
echo "Success";
}
else echo "Fail";

Answer 1

你需要转义 $players，因为 json 数组中充满了引号，所以当你使用查询将其插入数据库时​​，mysql 将无法执行和理解它。
也尝试在转义后将变量放在单引号内。试试这个：

$players = $mysqli->real_escape_string($players);
$mysqli->query("UPDATE `open_group` SET players = '$players' WHERE group_id = '$group_id' ");

Answer 2

正如 AlexanderO'Mara 指出的那样，您真的应该选择 bind_param() 而不是 mysqli_escape_string()，这在某些情况下已被证明是不可靠的。我使用 PDO 语句，所以这是未经测试的。

来源：mysqli real escape string, should I use it?

$slot_num = $_POST['num'];
$name = $_POST['name'];

// Initial select
$sql = "SELECT * FROM `open_groups` WHERE `group_id` = ?";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('d', $group_id);
$stmt->execute();

if ($stmt->errno) {
  die("Error : " . $stmt->error);
}

$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
    $slot_ar = json_decode($row['players'], true); //Get JSON array

    // Make sure $slot_num isn't bigger than desired
    if($slot_num > count($slot_ar) || !is_int($slot_num)){
        die('Injection attempt');
    }
    // Store new variables in the array
    $slot_ar[$slot_num]['ip'] = "$ip";
    $slot_ar[$slot_num]['name'] = "$name";

    // encode the new $players
    $players = json_encode($slot_ar);

    // new SQL query
    $sql = "UPDATE `open_group` SET players = ? WHERE group_id = ?";
    $stmt = $mysqli->prepare($sql);
    $stmt->bind_param('sd', $players, $group_id);
    $stmt->execute();
    if ($stmt->errno) {
      die("Error : " . $stmt->error);
    }

}