Question

我试图为我的网站创建一种Twitter样式。每当下面的脚本在数据库中添加“ 阵列 ”, 而不是你试图跟踪的人的用户代号。 Dunno 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();
}

?>

爆炸(arry,2)

任何感谢的帮助

/////// 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

我也认为问题在于将$mem变量加到$folfyArry1 阵列中。

试试这个代码...

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

} else { 

    $followArry1 = $mem1; 
}

Answer 2

$followArry1是一个数组,转换成字符串显示“Array ”。

见此证明(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(例如:http://ideone.com/h5ilz ):

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

友情链接