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Hi I am newish to php and I have created an update page for Content Management System. I have a file upload in this case a picture. I have other inputs that contain text and I can get them to populate my form and thats fine and works great because the user can see what has already been entered. But the file name for the photo can not have a value so if the user doesn't pick the picture from the directory again it will update minus the picture. What I think I need is a isset function that says if the file (picture) input is left blank don't update this field and use whatever what already in the database for it, that way if it was left blank when created it will still be, and if the user has changed it this time it will change; or if they want to leave it the same it won't leave their picture blank. Hope that makes sence.

Here is my coding currently for the Form: 

 <p>
              Photo:
            </p>
            <input type="hidden" name="MAX_FILE_SIZE" value="350000">
            <input type="file" name="photo"/>

Below is my php code for my update if the update button is pressed:

   $con = mysql_connect("localhost","******","********");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("*******", $con);

  // run this only, once the user has hit the "Update" button

  if (isset($_POST['update'])) {

    // assign form inputs

    $name = $_POST['nameMember'];

    $position = $_POST['bandMember'];

    $pic = $_POST['photo'];

    $about = $_POST['aboutMember'];

    $bands = $_POST['otherBands'];

      // add member to database

    $result = mysql_query("UPDATE dbProfile SET nameMember='".$name."',bandMember='".$position."',photo='".$pic."',aboutMember='".$about."',otherBands='".$bands."' WHERE id='".$id."'");

        mysql_close($con);


      Header("Location: listMember.php"); 

      exit; 

    }


  else { // read member data from database

    $result = mysql_query ("SELECT * FROM dbProfile WHERE id='".$id."'");

    while($row = mysql_fetch_array($result))


{


   $name = $row['nameMember'];

    $position = $row['bandMember'];

    $pic = $row['photo'];

    $about = $row['aboutMember'];

    $bands = $row['otherBands'];

  }
  }
  mysql_close($con);
?>

If you could help I would be very please and greatful.

4

2 回答 2

2

您必须使用该$_FILES变量来上传文件。有关详细信息，请参阅PHP 手册中的处理文件上传。

于 2009-01-29T12:03:59.153 回答
0

尝试：

if(is_uploaded_file($_FILES['photo']['tmp_name']))

从手册：

如果 filename 命名的文件是通过 HTTP POST 上传的，则返回 TRUE。这有助于确保恶意用户没有试图欺骗脚本来处理它不应该在其上工作的文件——例如，/etc/passwd。

于 2009-01-29T20:47:16.553 回答