Need a help to show related file to show from mysql DB

[ayesham]

i'm working on an autoindex mp3 downloading script, GET_($id) @ Download page shows relevant id information. from $artist variable I need to search my DB_table and view Artists other songs in this page.

table is like

table name 'mp3'

id - songurl - Song Title - Artist - Music - Lyrics - Size - Category - Date


please help me

[something else]

$result = mysqli_query($con, "SELECT songurl, SongTitle FROM mp3 WHERE Artist='".$artist."'");
$out = 'More songs by same artist: <br/><br/>';
while($row = mysqli_fetch_assoc($result))
{
$out .= '<a href="'.$row[0].'">'.$row[1].'</a><br/>';
}
echo $out; 

[ayesham]

it shows blank bro, tried so meny timesand I don't know why?

But I just copied songurl, SongTitle FROM mp3 WHERE Artist="Gayan" and run on Phpmyadmin sql it is showing results .

[ayesham]

<?php
include(db_con.php);

$result = mysqli_query($con, "SELECT `songurl`, `Songtitle`, `artist`, `cover` FROM `mp3` WHERE `artist`='".$artist."'");

while($row = mysqli_fetch_array($result))
{
?>

<?php echo $row['cover'];?><br/>
<?php echo $row['artist'];?><br/>
<?php echo $row[' SongTitle'];?>

<?php

}

mysqli_close($con);
?>

[something else]

<?php
include(db_con.php);
$artist = mysqli_real_escape_string($con, $_REQUEST['artist']);  //use mysqli_real_escape_string on any user inputs to help avoid mysql injections
//variable $artist needs to be set to something or collected from url or a form ($_GET or $_POST or for both $_REQUEST)


$result = mysqli_query($con, "SELECT `songurl`, `Songtitle`, `artist`, `cover` FROM `mp3` WHERE `artist`='".$artist."'");

while($row = mysqli_fetch_array($result))
{
echo $row['cover'].'<br/>'
.$row['artist'].'<br/>'
.$row['SongTitle'].'<br/><br/>'; //note you had a space in: $row['_SongTitle'];
//try not to open and close php too much as it will cause your script to slow down, also avoid using too many echo's 
}
mysqli_close($con);
?>

[something else]

to use php parsing in the forums use:
[php]Your code here[/php]

=

Your code here 

or for other codes:
[code]Your code here[/code]

=

Your code here

[s2k]

Hi @ayesham,

Have you tried to catch any errors from the query? perhaps try:

<?php
include("db_con.php");

$result = mysqli_query($con, "SELECT `songurl`, `Songtitle`, `artist`, `cover` FROM `mp3` WHERE `artist`='".$artist."'");

// check that a result was found?
if (!$result) {
printf("Error: %s\n", mysqli_error($link));
}

// perhaps for debugging maybe just dump the result if the above doesn't work?
// this should give you a peak inside the result variable
die(var_dump($result));

while ($row = mysqli_fetch_array($result)) {
echo "{$row['cover']}<br/>";
echo "{$row['artist']}</br>";
echo "{$row['SongTitle']}";
}

mysqli_close($con);
?>

Also I would recommend escaping your parameters like something else suggested, hope this helps

[something else]

Slight adjustment of s2k 's code:

// check that a result was found?
if (!$result) {
printf("Error: %s\n", mysqli_error($con));
} 

[s2k]

Thanks something else took it from the PHP docs page :P

[ayesham]

tried my best but it shows blank only brothers

[something else]

With it as your code:

<?php
include(db_con.php); //guessing mysqli has been connected here and stored in variable $con

$result = mysqli_query($con, "SELECT * FROM mp3 WHERE artist='".$_REQUEST['artist']."'");

while($row = mysqli_fetch_array($result))
{
?>

<?php echo $row['cover'];?><br/>
<?php echo $row['artist'];?><br/>
<?php echo $row['SongTitle'];?>

<?php

}

mysqli_close($con);
?>

Note: This is unprotected against mysqli injections

[ayesham]

$result = mysqli_query($con, "SELECT * FROM files WHERE artist='".$row['artist']."'");

while($row = mysqli_fetch_array($result))
{ 

this was working fine thanks genius?