php csv问题

Question

我有这个功能

function csv_to_array($filename='', $delimiter=',')
{
    if(!file_exists($filename) || !is_readable($filename))
    {
    echo "Not a valid file name";
        return FALSE;

     }
    $header = NULL;
    $data = array();



    $fname = file($filename);
    $a1 = array('username','fullname','email','dept');  

    if (($handle = fopen($filename, 'r')) !== FALSE)
    {       
        while (($row = fgetcsv($handle, 1000, $delimiter)) !== FALSE)
        {    
            if(!$header)
            {
                 $header = $row; 

                // print_r(array_diff_key($header,$array)) ;
                }
            else { 
                $data[] = array_combine($header, $row);  
                } 
        }     
        fclose($handle);
    }

    return $data;

}

它返回关联数组。 我的csv文件包含标题 用户名全名电子邮件部门 但我要求标题不要提及csv文件如何去做它

Answer 1

怎么样：

$seen_header = false; // initialize to false.
$data = array();

// keep reading from the file till you have lines.
while (($row = fgetcsv($handle, 1000, $delimiter)) !== FALSE) { 

        // $seen_header will be false for the fist time.
        if(!$seen_header) {
                // don't add header to $data and make $seen_header true.
                $seen_header = true;
        }else{   
                // non header..add it.
                $data[] = $row;
        }       
}