求教 关于PHP中操作CSV文件问题

Question

源码如下： 
<?php
         if(isset($_POST['upcsv'])){
              if("application/vnd.ms-excel"!=$_FILES['upfile']['type']){
                    echo "<script>alert('文件格式不确')</script>"; return  false;
                                            }
             else{
                $csvfile=fopen($_FILES['upfile'],"r");
                 while($data=fgetcsv($csvfile)){
                      print_r($data);
                            }
                 fclose($csvfile);
             }
        }
        ?>
        <form method="post" action="?" enctype="multipart/form-data">
            <input type="file" name="upfile" id="upfile">
            <input type="submit"  name="upcsv" value="上传">
        </form>
流览报错
Warning: fopen() expects parameter 1 to be string, array given in D:\AppServ\www\srctec\test_csv.php on line 18Warning: fgetcsv() expects parameter 1 to be resource, boolean given in D:\AppServ\www\srctec\test_csv.php on line 19Warning: fclose(): supplied argument is not a valid stream resource in D:\AppServ\www\srctec\test_csv.php on line 22请高人指教 

Answer 1

$csvfile = fopen($_FILES['upfile']['tmp_name'],"r");