php - PHP爆炸功能导致问题

Question

我正在尝试为我的网站创建一个 twitter 风格的跟踪系统。当下面的脚本运行时，会在数据库中添加一个“数组”，而不仅仅是您要关注的人的用户 ID。不知道如何修复它。

<?
session_start();
# Connect to the mysql database
include_once "library/connect_to_mysql.php"; 

if(isSet($_POST['mem'])){

#filter everything but numbers for security
$mem1 = preg_replace('#[^0-9]#i', '', $_POST['mem']); 
$mem2 = $mem1;

#Decode the Session IDX variable and extract the user's ID from it
$decryptedID = base64_decode($_SESSION['idx']);
$id_array = explode("p3h9xfn8sq03hs2234", $decryptedID);
$my_id = $id_array[1];

$sql = mysql_query("SELECT following_array FROM Members WHERE id='$my_id' LIMIT 1"); 

while($row = mysql_fetch_array($sql)) { 
$following = $row["following_array"]; 
}

$followArry1 = explode(',', $following);

if (in_array($mem1, $followArry1)) {

exit(); 
}

if ($followArry1 != "") { 

$followArry2 = "$followArry1,$mem2"; 

} else { 

$followArry2 = "$mem2"; 
}

$UpdateArray = mysql_query("UPDATE Members SET following_array ='$followArry2' WHERE id='$my_id'") or die (mysql_error());

exit();

}else{

exit();
}

?>

任何帮助表示赞赏

/////// Updated Code ////////////////////

$followArry1 = explode(",", $following);

if (in_array($mem1, $followArry1)) { exit(); }

if ($followArry1 != "") { 

    $followArry1 = implode(',', $followArry1 + array($mem1)); 

} else { 

    $followArry1 = $mem1; 
}

$UpdateArray = mysql_query("UPDATE myMembers SET following_array ='$followArry1' WHERE id='$my_id'") or die (mysql_error());

Answer 1

$followArry1是一个数组，转换为字符串显示“ Array”。

在此处查看证明（现场演示：http: //ideone.com/tMndH）：

$followArry1 = array('a', 'b', 'c'); // just an array
$result = "$followArry1"; // it is now string containing "Array"

尝试以下操作（示例：http: //ideone.com/h5ilz）：

$followArry1 = array('a', 'b', 'c');
$mem = 'd';
$followArry2 = implode(',', $followArry1 + array($mem));

Answer 2

我还认为问题在于将 $mem 变量添加到 $followyArry1 数组。

试试这个代码...

if ($followArry1 != "") { 
    $followArry1[] = $mem1;
    $followArry1 = implode(',', $followArry1); 

} else { 

    $followArry1 = $mem1; 
}