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Break4y

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Здравейте някой може ли дами напише или дами даде уплоад значи искам дае ето така: http://mp-tri.com/?p=add
Ползвайте за потребител: break4y - парола: 555267
Добавете някоя песен и вижте след като я добавите как става но искам дае без плеяра направо да добавя песента :) +1 Който ми помогне
 
add.php

Код:
<div class="center">
<div class="boxtext"><b>Добавяне на песен</b><br /> 
</div> 
<form id="form2" name="form2" method="post" action=""> 
<table width="250" border="0"  cellpadding="1" cellspacing="1" id="table"> 
<tr> 
<td>Изпълнител</td> 
<td><strong>:</strong></td> 
<td align="center"><input name="artist" type="text" class="txt" id="artist" size="30" /></td> 
</tr> 
<tr> 
<td align="left">Песен</td> 
<td align="center"><strong>:</strong></td> 
<td align="center"><input name="title" type="text" class="txt" id="title" size="30"></td> 
</tr> 
<tr> 
<td align="left">Линк</td> 
<td align="center"><strong>:</strong></td> 
<td align="center"><input name="path" type="text" class="txt" id="path" size="30"></td> 
</tr> 
<tr> 
<td colspan="3" align="center"><select name="cat" class="txt" id="cat"> 
<option>Изберете категория</option> 
<?php 
include("conn.php"); 
$r = mysql_query("SELECT * FROM [color=red]songs_cat[/color]"); 
while($row = mysql_fetch_array($r)){ 
$cat .= "<option value=\"$row[id]\" class=\"bottom\">$row[value]</option>"; 
} 
echo $cat; 
?> 
</select></td> 
</tr> 
<tr> 
<td colspan="3" align="center"><input name="submit" type="submit" class="btn" id="button" value="Изпрати" /> 
<input name="button2" type="reset" class="btn" value="Изчисти" /></td> 
</tr> 
</table> 
</form> 
<?php 

if($_POST['submit']){ 
if($_POST['title'] == ""){ 
echo "<script type=\"text/javascript\"> 
alert('Не сте написали изпълнител или песен.Моля направете го.'); 
</script>";} 
elseif($_POST['cat'] == "Изберете категория" || $_POST['pic'] == "http://"){ 
echo "<script type=\"text/javascript\"> 
alert('Не сте избрали категория или линк на песента .Моля направете го.'); 
</script>";} 
else { 
$date = date("d.m.Y"); 
echo "Успешно добавена песен!<br><a href='index.php'><<< Начална страница.</a>"; 
$sql = "INSERT INTO `songs` (`id`, `author`, `artist`, `title`, `path`, `cat`, `date`) VALUES (NULL, '$_POST[author]', '$_POST[artist]', '$_POST[title]', '$_POST[path]', '$_POST[cat]', '$date');"; 
$result = mysql_query($sql) or die(mysql_error()); 
} 
} 
?> 
</td> 
</tr> 
</table> 

</diV>
	</div>

sql.sql

Код:
CREATE TABLE `songs_cat` ( 
`id` int(4) NOT NULL auto_increment, 
`value` varchar(255) NOT NULL, 
PRIMARY KEY (`id`) 
) ENGINE=MyISAM DEFAULT CHARSET=cp1251; 

INSERT INTO `songs_cat` VALUES (2, 'Club'); 
INSERT INTO `songs_cat` VALUES (3, 'Dance'); 
INSERT INTO `songs_cat` VALUES (4, 'Electronic'); 
INSERT INTO `songs_cat` VALUES (5, 'Gothic'); 
INSERT INTO `songs_cat` VALUES (6, 'Hip-Hop'); 
INSERT INTO `songs_cat` VALUES (7, 'JazZ'); 
INSERT INTO `songs_cat` VALUES (8, 'Latin'); 
INSERT INTO `songs_cat` VALUES (9, 'Pop'); 


CREATE TABLE `songs` ( 
`id` int(4) NOT NULL auto_increment, 
`author` varchar(255) NOT NULL, 
`artist` varchar(255) NOT NULL, 
`title` varchar(255) NOT NULL, 
`path` varchar(255) NOT NULL, 
`year` varchar(255) NOT NULL, 
`album` varchar(255) NOT NULL, 
`size` varchar(255) NOT NULL, 
`cat` varchar(255) NOT NULL, 
`date` varchar(255) NOT NULL, 
`broken` varchar(255) NOT NULL default '0', 
`views` varchar(255) NOT NULL default '0', 
PRIMARY KEY (`id`) 
) ENGINE=MyISAM DEFAULT CHARSET=cp1251;
 
add.php
Код:
Стил на песента: 
<form method="POST" action="added.php">
<select name="style">
<option value="Jazz">Jazz</option>
<option value="Metal">Metal</option>
</select> 
Изпълнител: <input type=text name=pevec><br>
Име на песента: <input type=text name=pesen><br>
Линк към песента: <input type=text name=link>
</form>
<input type=submit value=Добави>
added.php
Код:
<?
$pesen = $_POST['pesen'];
$style = $_POST['style'];
$pevec = $_POST['pevec'];
$link = $_POST['link'];
$file = "$pesen.html";
$handle = fopen($file, 'a') or die("Не мога да отворя файла");
$data = "Песента, която разглеждате е $pesen и се изпълнява от $pevec.<br>Стилът и е $style<br><embed src=$link>";
fwrite($handle, $data);
fclose($handle);
echo "Песента бе добавена успешно.<br><a href=$pesen.html>Прослушайте я</a>";
?>

:roll:
п.п. Edited
Пробвах я сега системата и си работи безпроблемно ;)
 
sql.sql
Код:
CREATE TABLE `pesni` ( 
`id` int(50) NOT NULL auto_increment, 
`id_cat` varchar(10) NOT NULL, 
`pesen` varchar(255) NOT NULL, 
`link` text NOT NULL, 
`izpalnitel` varchar(255) NOT NULL, 
PRIMARY KEY (`id`) 
); 


CREATE TABLE `categorii` ( 
`id` int(50) NOT NULL auto_increment, 
`cat` varchar(255) NOT NULL, 
PRIMARY KEY (`id`) 
);

config.php
Код:
<?php 
$server = "localhost"; 
$dbusername = "ime"; 
$dbpassword = "parola"; 
$db_name = "bazaime"; 

mysql_connect($server, $dbusername, $dbpassword)or die("Сайтът не може да се свърже към базата данни"); 
@mysql_select_db($db_name) or die("Грешна база данни"); 
mysql_query("SET CHARACTER SET cp1251");

?>

all.php
Код:
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
<?php
include "config.php";
$query = " SELECT * FROM pesni"; 
$result = mysql_query($query) or die('Error, query failed'); 

while($row = mysql_fetch_array($result)) 
{ 
$id_cat=$row['id_cat'];
$query1 = " SELECT * FROM categorii WHERE id='$id_cat'"; 
$result1 = mysql_query($query1) or die('Error, query failed'); 
$row1 = mysql_fetch_array($result1);

echo $row1['cat'].' <a href="view.php?id='.$row['id'].'">'.$row['pesen'].'</a><br>'; 
} 
?>

addcat.php
Код:
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
<form action="" method="post">
<input name="cat" type="text"></br>
<input type="submit">
</form>

<?php
if($_POST)
{
include "config.php";
$catname=$_POST['cat'];
$add="INSERT INTO categorii (`cat`) VALUE ('$catname')";
$ok = mysql_query($add) or die('Error, query failed'); 
if($ok)
{
echo "<META HTTP-EQUIV=\"refresh\" CONTENT=\"0; URL=addcat.php\">"; 
}
}
?>

addsong.php
Код:
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
<form action="" method="post">
pesen <input name="pesen" type="text"></br>
<select name="cat">
<?php
include "config.php";
$query1 = " SELECT * FROM categorii"; 
$result1 = mysql_query($query1) or die('Error, query failed'); 
while($row1 = mysql_fetch_array($result1))
{
echo '<option value="'.$row1['id'].'">'.$row1['cat'].'</option>'
}
?>
</select><br>
izpalnitel <input name="izpalnitel" type="text"></br>
link <input name="link" type="text"></br>
<input type="submit">
</form>
<?php
if($_POST)
{
$pesen=$_POST['pesen'];
$categ=$_POST['cat'];
$izpal=$_POST['izpalnitel'];
$link=$_POST['link'];

$add="INSERT INTO pesni (`id_cat`,`pesen`,`link`,`izpalnitel`) VALUE ('$categ','$pesen','$link','$izpal')";
$ok = mysql_query($add) or die('Error, query failed'); 
if($ok)
{
echo "<META HTTP-EQUIV=\"refresh\" CONTENT=\"0; URL=view.php?id=".mysql_insert_id()."\">"; 
}

}
?>

view.php
Код:
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
<?php
$id=$_GET['id'];
include "config.php";
$query = " SELECT * FROM pesni WHERE id='$id'"; 
$result = mysql_query($query) or die('Error, query failed'); 

while($row = mysql_fetch_array($result)) 
{ 
$id_cat=$row['id_cat'];
$query1 = " SELECT * FROM categorii WHERE id='$id_cat'"; 
$result1 = mysql_query($query1) or die('Error, query failed'); 
$row1 = mysql_fetch_array($result1);

echo 'Категория: '.$row1['cat'].'<br>';
echo 'Pesen: '.$row['pesen'].'<br>'; 
echo 'Izpalnitel: '.$row['izpalnitel'].'<br>';
echo 'link: '.$row['link'].'<br>';  
} 
?>
 
Ами няма да е зле ако можем да видим какъв е ъоплоуд скрипта който ни даваш за пример ! Показва го само на регистрирани!
 
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
<form action="" method="post">
pesen <input name="pesen" type="text"></br>
<select name="cat">
<?php
include "config.php";
$query1 = " SELECT * FROM categorii";
$result1 = mysql_query($query1) or die('Error, query failed');
while($row1 = mysql_fetch_array($result1))
{
echo '<option value="'.$row1['id'].'">'.$row1['cat'].'</option>';
}
?>
</select><br>
izpalnitel <input name="izpalnitel" type="text"></br>
link <input name="link" type="text"></br>
<input type="submit">
</form>
<?php
if($_POST)
{
$pesen=$_POST['pesen'];
$categ=$_POST['cat'];
$izpal=$_POST['izpalnitel'];
$link=$_POST['link'];

$add="INSERT INTO pesni (`id_cat`,`pesen`,`link`,`izpalnitel`) VALUE ('$categ','$pesen','$link','$izpal')";
$ok = mysql_query($add) or die('Error, query failed');
if($ok)
{
echo "<META HTTP-EQUIV=\"refresh\" CONTENT=\"0; URL=view.php?id=".mysql_insert_id()."\">";
}

}
?>
 
Мерси сега работи отлично ;)

EDIT: може ли даго направиш view.php с плеяр песните да могат дасе слушат :o
 

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