加載文件到服務器

Question

我有一個頁面上有兩種形式，應該給用戶posibility加載文件到服務器（從網址或從用戶的PC）

<form method="post" action="bigorder.php" name="photourl"> 
    <label for="photoorig">URL</label> 
    <input type="url" name="photoorig" placeholder=""> 
    <input type="submit" value="Load" name="photoload"> 
    <br> 
</form> 

<form method="post" action="bigorder.php" name="photofile" enctype="multipart/form-data"> 
    <label for="photoloc">Load own file</label> 
    <input type="file" name="photoloc" id="photoloc"> 
    <input type="submit" value="Load" name="photoload2"> 
</form> 

和PHP

<?php 
    $tmpname=rand().".jpg"; 
    if ($_POST['photoorig']) { 
    $file=file_get_contents($_POST['photoorig']); 
    $fp = fopen("/var/www/html/uploads/tmp/".$tmpname, "w"); 
    fwrite($fp, $file); 
    fclose($fp); 
    } 
    if ($_POST['photoloc']) { 
    $tmpFile = $_FILES['photoloc']['tmp_name']; 
    $newFile = "/var/www/html/uploads/tmp/".$_FILES['photoloc']['name']; 
    $result = move_upload_file($tmpFile, $newFile); 
    echo $_FILES['photoloc']['name']; 
     if ($result) { 
     echo ' was uploaded<br />'; 
     } else { 
     echo ' failed to upload<br />'; 
    } 
?> 

第一種形式加載文件很好，但第二個完全不起作用。我甚至沒有收到任何錯誤信息。

我在做什麼錯？或缺少什麼？

Answer 1

這裏是正確的代碼，它正在工作。在你的代碼中第二種形式缺少花括號，isset（）函數應該用來檢查發佈的數據集。

<?php 
     $tmpname=rand().".jpg"; 
     if (isset($_POST['photoload'])) { 
     echo '1st'; 
     $file=file_get_contents($_POST['photoorig']); 
     $fp = fopen("/var/www/html/uploads/tmp/".$tmpname, "w"); 
     fwrite($fp, $file); 
     fclose($fp); 
     } 
     if (isset($_POST['photoload2'])) { 
     echo '2nd'; 
     $tmpFile = $_FILES['photoloc']['tmp_name']; 
     $newFile = "/var/www/html/uploads/tmp/".$_FILES['photoloc']['name']; 
     $result = move_upload_file($tmpFile, $newFile); 
     echo $_FILES['photoloc']['name']; 
      if ($result) { 
      echo ' was uploaded<br />'; 
      } else { 
      echo ' failed to upload<br />'; 
       } 
     } 
    ?>