警告：連接（）[function.join]：用C通過

Question

你好，大家好我想**使用多個複選框並插入其價值到數據庫的參數無效，但我收到此錯誤說：'Warning: join() [function.join]: Invalid arguments passed in C';

下面

是代碼：使用

<?php 
require_once('db_conn.php'); $cat=$_POST['cat']; 

if(isset($_FILES['file_upload']) && isset($cat)) 
{ 
    $shuff=str_shuffle("ABD6565LSLFKDSAJFD"); 

    $food = join(', ', $_POST['food']);  

    mkdir("upload/$shuff"); 

    $files=$_FILES['file_upload']; 

    for($x = 0; $x <count($files['name']); $x++)  
    { 
     $name=$files['name'][$x];  
     $tmp_name=$files['tmp_name'][$x]; 

     if(move_uploaded_file($tmp_name, "upload/$shuff/".$name)) 
     { 
      $query="INSERT INTO image(mid, cid, name, food, path) VALUES('', '$cat', '$name', '$food', 'upload/$shuff/$name')"; 
      mysql_query($query); 
      echo 'The file '.$name. ' uploaded successfully'. '<br \>'; 
     } 
     else 
     { 
      echo 'uploading failed'; 
     } 
    } 
} 
?> 

Answer 1

嘗試破滅，而不是加入：

if(!empty($_POST['food']) AND is_array($_POST['food'])) {  
    $food = implode(', ', $_POST['food']); 
} 
else { 
    $food = ""; 
}