php - 如何删除 header 错误

Question

我的代码在上传和显示图像时遇到问题。我打算在上传过程完成后重定向页面，所以我使用了 header 函数，但给出了警告和错误，不幸的是上传失败了。怎么办我可以删除它吗？这是代码..

<?php 

//connect to the database//
$con = mysql_connect("localhost","root", "");
if(!$con)
{
 die('Could not connect to the database:' . mysql_error());
 echo "ERROR IN CONNECTION";
}

 $sel = mysql_select_db("imagedatabase");
if(!$sel)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
//file properties//

$file = $_FILES['image']['tmp_name']; 

echo '<br />';

 /*if(!isset($file))
    echo "Please select your images";

else
{
 */for($count = 0; $count < count($_FILES['image']); $count++)
{
//$image = file_get_contents($_FILES['image']['tmp_name']);
$image_desc[$count] = addslashes($_POST['imageDescription'][$count]);
$image_name[$count] = addslashes($_FILES['image]']['name'][$count]); echo '<br \>';
$image_size[$count] = @getimagesize($_FILES['image']['tmp_name'][$count]);
$error[$count] = $_FILES['image']['error'][$count];

if($image_size[$count] === FALSE  || ($image_size[$count]) == 0)
    echo "That's not an image";
else
{

// Temporary file name stored on the server
 $tmpName[$count]  = $_FILES['image']['tmp_name'][$count];

  // Read the file
   $fp[$count]   = fopen($tmpName[$count], 'r');
   $data[$count] = fread($fp[$count], filesize($tmpName[$count]));
   $data[$count] = addslashes($data[$count]);
   fclose($fp[$count]);


 // Create the query and insert
 // into our database.

 $results = mysql_query("INSERT INTO images( description, image) VALUES                 ('$image_desc[$count]','$data[$count]')", $con);

    if(!$results)
    echo "Problem uploding the image. Please check your database";  
   //else 
   //{
      echo "";
    //$last_id = mysql_insert_id();
    //echo "Image Uploaded. <p /> <p /><img src=display.php?    id=$last_id>";
    //header('Lcation: display2.php?id=$last_id');
        }
     //}
}


mysql_close($con);
header('Location: fGallery.php');
?>

header 函数应该将我引导到另一个可以制作画廊的页面..这是代码..

<?php

//connect to the database//
    mysql_connect("localhost","root", "") or die(mysql_error());
    mysql_select_db("imagedatabase") or die(mysql_error());

    //requesting image id




    $image = mysql_query("SELECT * FROM images ORDER BY id DESC");


    while($row = mysql_fetch_assoc($image))
    {
         foreach ($row as $img) echo '<img src="img.php?id='.$img["id"].'">';
    }

    mysql_close();


?>

我的画廊也有问题.. 一些帮助会很棒!谢谢! :D

Answer 1

必须在产生输出的任何其他 echo 或 die 调用之前调用 header() 函数。

如果您需要输出，您可以缓冲您的输出，但在您的情况下没有任何区别，因为输出永远不会显示给用户。浏览器将读取重定向并导航到第二页。

---

<?php 

//connect to the database//
$con = mysql_connect("localhost","root", "");
if(!$con) {
 // this output is okay the redirect will never be reached.
 die('Could not connect to the database:' . mysql_error());
 // remember after a die this message will never be shown!
 echo "ERROR IN CONNECTION";
}

 $sel = mysql_select_db("imagedatabase");
if(!$sel) {
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION"; // same here with the die!
}
//file properties//

$file = $_FILES['image']['tmp_name']; 

// OUTPUT
// echo '<br />';

// removed out commented code

for($count = 0; $count < count($_FILES['image']); $count++)
{
$image_desc[$count] = addslashes($_POST['imageDescription'][$count]);
$image_name[$count] = addslashes($_FILES['image]']['name'][$count]);
// OUTPUT
// echo '<br \>';
$image_size[$count] = @getimagesize($_FILES['image']['tmp_name'][$count]);
$error[$count] = $_FILES['image']['error'][$count];

if($image_size[$count] === FALSE  || ($image_size[$count]) == 0)
// you may better use a die if you want to prevent the redirection
    echo "That's not an image";
else
{

// Temporary file name stored on the server
 $tmpName[$count]  = $_FILES['image']['tmp_name'][$count];

  // Read the file
   $fp[$count]   = fopen($tmpName[$count], 'r');
   $data[$count] = fread($fp[$count], filesize($tmpName[$count]));
   $data[$count] = addslashes($data[$count]);
   fclose($fp[$count]);


 // Create the query and insert
 // into our database.

 $results = mysql_query("INSERT INTO images( description, image) VALUES                 ('$image_desc[$count]','$data[$count]')", $con);

    if(!$results) // use die
    echo "Problem uploding the image. Please check your database";  
// OUTPUT
//      echo "";
        }
}


mysql_close($con);
header('Location: fGallery.php');
?>

上面我为您标记了每个输出，还删除了所有注释行。