Upload скрипт малко редакция

B

bojko91

Registered
Здравейте имам ей този ъплоад код. Иксам да помоля да ми разясните как може да се получи следното

Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form> 

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded <br />


";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
} 


?>

Иска ми се като се качи файла да ми се покаже ембед код


Код: 
<object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/ името на качения файл &displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object>

Мъчих се да го направя по следния начин но не се получи.

Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form> 

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded <br />
<object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/ ". basename( $_FILES['uploadedfile']['name']). " &displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object>

";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
} 

if ($uploaded_type =="text/php")
{
echo "No PHP files<br>";
$ok=0;
} 


?>

Надявам се да ми помогнете. Поздрави!
 
Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "<object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/".basename( $_FILES['uploaded']['name'])."&displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object>


";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}
?>
 
asiudfoisaznhfo каза:
Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "<object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/".basename( $_FILES['uploaded']['name'])."&displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object>


";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}
?>
Увеличете...

супер стана, но може ли да не се показваплеара а само кода, за даможе да се копира?
 
Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "<textarea name=\"textfield\" rows=\"5\" id=\"textfield\"><object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/".basename( $_FILES['uploaded']['name'])."&displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object></textarea>


";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}
?>
 
asiudfoisaznhfo каза:
Код: 
<form enctype="multipart/form-data" action="uploadvd.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>

<?php
$target = "videos/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "<textarea name="textfield" rows="5" id="textfield"><object width='425' height='344'><param name='movie' value='sajta.com/player/embed_player.swf'></param><param name='allowFullScreen' value='false'></param><embed src='sajta.com/player/embed_player.swf' width='425' height='344' type='application/x-shockwave-flash' pluginspage='http://www.macromedia.com/go/getflashplayer' flashvars='file=sajta.com/videos/".basename( $_FILES['uploaded']['name'])."&displayheight=344&logo=sajta.com/player/mini_logo.png&lightcolor=0xCC0000&title=test&usefullscreen=false'  allowfullscreen='false' width='425' height='344'/></object></textarea>


";
}
else {
echo "Sorry, there was a problem uploading your file.";
}

if ($uploaded_size > 350000)
{
echo "Your file is too large.<br>";
$ok=0;
}
?>
Увеличете...

Супер . най - накрая пробачка. Трудно ли се добавя, някакъв Loading бар, който да ми показва до къде е стигнало качванто на файла?
 
за прогрес бар
php кода който проверява на колко процента е качен файла е 5 реда макс 10
но ajax-а който да стартира прогрес бара е друг въпрос
ако намериш някой да ти напише на ajax следната функция ще ти напиша код за прогрес бар

функцията
когато се натисне бутона "Upload"
в div по доло на всеки 5 секунди примерно
да се зарежда файла progress.php с get информация name= информацията от полето "uploaded"
 
asiudfoisaznhfo каза:
за прогрес бар
php кода който проверява на колко процента е качен файла е 5 реда макс 10
но ajax-а който да стартира прогрес бара е друг въпрос
ако намериш някой да ти напише на ajax следната функция ще ти напиша код за прогрес бар

функцията
когато се натисне бутона "Upload"
в div по доло на всеки 5 секунди примерно
да се зарежда файла progress.php с get информация name= информацията от полето "uploaded"
Увеличете...

Ако някой може да помогне с Ajax функцията, че абсолютно ми е тъмна индия там много ще съм благодарен! Поздрави :)

Благодаря ви предварително за отзивчивостта :wink:
 
Трябва да си влезнал или регистриран, за да отговориш тук.

Back
Горе