爆炸（）預計參數2爲字符串

-1

UPDATE anggota SET id = '1000',name = 'Danti',Address = 'Jalan Mawar' WHERE Kode_anggota='10000'

我用$_GET爲我的表名，所以我可以用一個查詢所有的表，這是我的代碼：

<?php 
    function updateData($value) { 
     $table = $_GET['tujuan']; 
     $id = $_GET['id']; 
     $query1 = mysql_query("SELECT * FROM $table ORDER BY Kode_$table"); 
     $fieldNum = mysql_num_fields($query1); 
     for($q=0;$q<$fieldNum;$q++){ 
      $fieldName = mysql_field_name($query1, $q); 
      $name [] = $fieldName; 
     } 
     $output1 = explode (" ",$name); 

     foreach ($value as $arrVal) { 
      $total [] = $arrVal; 
     } 
     $output = explode("", $total); 
     echo "UPDATE $table SET"; 
     for($a=0;$a<$fieldNum;$a++){ 
     $akhir = "$output1".$a." = $output".$a.""; 
     $akhir1 [] = $akhir; 
     } 
     $akhir2 = implode (",",$akhir1); 
     echo $akhir2; 
     echo " WHERE Kode_$table='$id'"; 
    } 
?>

但是，當我跑步時我的節目，我得到了這樣的通知，2 - >

Warning：explode（）期望參數2是字符串，數組在C：\ xampp \ htdocs \ Proyek 2 \ cobalagi \ koneksi_class.php上在線88

警告：explode（）期望參數2是字符串，陣列中的C中給出：\ XAMPP \ htdocs中\ Proyek 2 \ cobalagi \ koneksi_class.php上線93 UPDATE anggota SET0 = 0,1 = 1,2 = 2 WHERE Kode_anggota = '10000'

我應做？

來源

2015-10-20 Danti Iswandhari

因爲變量$ name是數組不是字符串 –

首先，也是最重要的，你要修復SQL注入漏洞。永遠不要在sql語句中使用未過濾和未轉義的請求數據。看看http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php –