2016-01-24 59 views
0

我想從第一次讀取excel文件,我有一些問題。我使用PhpExcel庫,我使用下面的代碼,現在Excel文件的文件夾中上傳的,但它不是顯示Excel文件中的任何數據,它給錯誤PhpExcel庫給出錯誤

"Error is catchedCould not open C:\xampp\tmp\phpF85B.tmp for reading! File does not exist."

及通告像

Notice: Undefined variable: obj in C:\xampp\htdocs\PhpExcel\index.php on line 24

Notice: Trying to get property of non-object in C:\xampp\htdocs\PhpExcel\index.php on line 24

最後

Fatal error: Call to a member function toArray() on a non-object in C:\xampp\htdocs\PhpExcel\index.php on line 24

<?php 

if(isset($_POST['sub']) && !empty($_FILES['ex_file']['name'])){ 

$file=$_FILES['ex_file']['name']; 
$type=explode(".",$file); 
if(end($type)!= "xls" && end($type)!= "xlsx"){ 
    echo "The File is not and excel file"; 
    } 

else{ 
    include "PhpExcelLib/Classes/PhpExcel/IOFactory.php"; 
    $dir="uploads/"; 
    $name=$_FILES['ex_file']['name']; 
    $tmp_name=$_FILES['ex_file']['tmp_name']; 
    if(move_uploaded_file($tmp_name,$dir.$name)){ 

     try{ 
      $obj=PhpExcel_IOFactory::load($tmp_name); 
     } 
     catch(Exception $e){ 
      echo "Error is catched". $e->getMessage(); 
     } 
     $data=$obj->getActiveSheet->toArray(null,true,true,true); 
     print_r($data); 
    } 
    else{ 
     echo "File is not uploaded"; 

} 
} 
} 
?> 
<html> 
<body> 
<form method="POST" enctype="multipart/form-data"> 
<label>Choose File To View Data</label> 
<input type="file" name="ex_file" id="ex_file" /> 
<input type="submit" name="sub" value="Submit" /> 
</form> 
</body> 
</html> 

回答

1

您的文件無法上傳。

檢查這個,看看是什麼原因造成的錯誤:

// Check $_FILES['upfile']['error'] value. 
switch ($_FILES['upfile']['error']) { 
    case UPLOAD_ERR_OK: 
     break; 
    case UPLOAD_ERR_NO_FILE: 
     throw new RuntimeException('No file sent.'); 
    case UPLOAD_ERR_INI_SIZE: 
    case UPLOAD_ERR_FORM_SIZE: 
     throw new RuntimeException('Exceeded filesize limit.'); 
    default: 
     throw new RuntimeException('Unknown errors.'); 
} 

這應該給你錯誤的想法