Contents showing outside block

Worker Joined: Dec 03, 2005 Posts: 153

Hi All

I have block that takes requests and then place the information within the database. As it stands it works great, but the contents are showing outside the block. I have tried other ways but am unable to make the content stay inside the block, here is the orignial file to view.

http://camp-battlegroup.myby.co.uk/block-Request.txt

Any advise

Cheers Scorp

Posted: Fri Mar 28, 2008 10:12 am

Blocks need to capture all output in a variable called $content. The higher level logic then echo's the $content. Also get rid of the <html>, <head>, <body>, etc tags since Nuke already outputs those for you. Google the PHP-Nuke HOW-TO, there is some advice on how to make blocks that explains this.

Posted: Fri Mar 28, 2008 11:06 am

Hi

I have already tried the $content and the echo, when I place any of them in front of the code, the block either dont show or the right hand side wont show. I have been trying this out for the last few days and googled it.

Cheers Scorp

Posted: Fri Mar 28, 2008 11:42 am

I told you what you need to do in general. Without seeing your other attempts I can't help you. Again, look at the HOW-TO. You can't just echo stuff out inside a block, it needs to get captured in $content.

Posted: Fri Mar 28, 2008 11:59 am

Hi

I have tried the below:

// User Input Area
$content .="<form action="index.php" method="POST">";
$content .="<table>";
$content .="<tr><td>Song title: </td><td><input type="text" name="song" value=""></td></tr>";
$content .="<tr><td>Artist: </td><td><input type="text" name="artist" value=""></td></tr>";
$content .="<tr><td>Your name: </td><td><input type="text" name="name" value=""></td></tr>";

$content .="</table>";
$content .="<input type="submit" name="submit" value="Send!">";
$content .="</form>";

?>

All HTML Tags have been removed.

But the block and the right side does not show. The linked script is the orginal script, which needs to be turned into a block or maybe a module at a later date. I have done a block that will show the date input fromthe orignial script, but this is my first time at trying to insert data from a block.

I have looked through the php nuke How to Do I have googled it for the last few days, so, as a last resort I thought I seek help here. Can anyone show me a example or where I am going wrong.

Cheers
Scorp

Posted: Fri Mar 28, 2008 12:10 pm

Well what you have above, should "work", except I would add a

Code:




$content = '';

before all of that.

I'm not sure sure about your form posting to index.php...but that is another issue.

Post the entire block code (if you haven't already).

Posted: Fri Mar 28, 2008 12:16 pm

Hi Gremmie

Where would I had the $content =;

LMAO ok here is the full code with the database connection information.

Code:




$reg = "yes"; 


$ip = $_SERVER['REMOTE_ADDR'];


$content .="<center>Your ip is: ".$ip."</center><br />";


//


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


if (empty($_POST['song'])) { 


$content .="Sorry, you haven't supplied the song title!<br />"; 


$reg = "no"; 


} 


if (empty($_POST['artist'])) { 


$content .="Sorry, you haven't supplied the artists name!<br />"; 


$reg = "no"; 


} 


if (empty($_POST['name'])) { 


$content .="Sorry, you haven't supplied your name<br />"; 


$reg = "no"; 


} 





  $sql = "SELECT COUNT(*) FROM request_song WHERE ip='{$ip}'"; 


  $result = mysql_query($sql); 


  if (mysql_result($result, 0) > 0) { 


$content .="Sorry, your ip has already wished for one song, you can not wish for <br />another until the DJ's have seen your request!<br />"; 


$reg = "no"; 


} 


if ($reg == "yes") { 


    $sql = "INSERT INTO request_song(song, artist, name, ip) 


            VALUES('{$_POST['song']}', '{$_POST['artist']}', '{$_POST['name']}', '{$ip}')"; 


    mysql_query($sql); 


} 


} 


// User Input Area


$content .="<form action="index.php" method="POST">"; 


$content .="<table>";


$content .="<tr><td>Song title: </td><td><input type="text" name="song" value=""></td></tr>"; 


$content .="<tr><td>Artist: </td><td><input type="text" name="artist" value=""></td></tr>"; 


$content .="<tr><td>Your name: </td><td><input type="text" name="name" value=""></td></tr>";


 


$content .="</table>"; 


$content .="<input type="submit" name="submit" value="Send!">";


$content .="</form>";





?>

Posted: Fri Mar 28, 2008 2:48 pm

I edited your post to put the code within BBCode 'code' tags
Try it on the line below $ip=$_SERVER......
I will have to also assume the forums here stripped out your opening php tag

Posted: Fri Mar 28, 2008 4:19 pm

Hi guardian

Nope block will not show and the righthand side is not showing.

I left out the top part of the code with the database connection when posting the code at first.

Scorp

Posted: Fri Mar 28, 2008 5:13 pm

Your problem, scorpious, was that you had put double quotes inside a double quoted string without properly escaping them. I switched you over to single quotes, and added the $content = ''; at the top. This displays on my site.

Code:




<?php


$content = '';


$reg = "yes";


$ip = $_SERVER['REMOTE_ADDR'];


$content .="<center>Your ip is: ".$ip."</center><br />";


//


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


if (empty($_POST['song'])) {


$content .="Sorry, you haven't supplied the song title!<br />";


$reg = "no";


}


if (empty($_POST['artist'])) {


$content .="Sorry, you haven't supplied the artists name!<br />";


$reg = "no";


}


if (empty($_POST['name'])) {


$content .="Sorry, you haven't supplied your name<br />";


$reg = "no";


}





  $sql = "SELECT COUNT(*) FROM request_song WHERE ip='{$ip}'";


  $result = mysql_query($sql);


  if (mysql_result($result, 0) > 0) {


$content .="Sorry, your ip has already wished for one song, you can not wish for <br />another until the DJ's have seen your request!<br />";


$reg = "no";


}


if ($reg == "yes") {


    $sql = "INSERT INTO request_song(song, artist, name, ip)


            VALUES('{$_POST['song']}', '{$_POST['artist']}', '{$_POST['name']}', '{$ip}')";


    mysql_query($sql);


}


}


// User Input Area


$content .='<form action="index.php" method="POST">';


$content .='<table>';


$content .='<tr><td>Song title: </td><td><input type="text" name="song" value=""></td></tr>';


$content .='<tr><td>Artist: </td><td><input type="text" name="artist" value=""></td></tr>';


$content .='<tr><td>Your name: </td><td><input type="text" name="name" value=""></td></tr>';


 


$content .='</table>';


$content .='<input type="submit" name="submit" value="Send!">';


$content .='</form>';





?>

BTW, you should set your error reporting to high and you would see that you had a syntax error.

Posted: Fri Mar 28, 2008 5:27 pm

Hi

Yes worked first time

Many thanks to you all for your advise and help

Scorp

Regular Joined: Mar 12, 2008 Posts: 74

this block work as a stand alone or does this work with shoutcast or something like that?? cpz ic database for some mod?

Posted: Sat Mar 29, 2008 3:47 am

Hi Fresh

This Block was done for members to send requests for a song, it can be used for anything, but if you want to use it for shoutCast like me here is the sql info you will need:

CREATE TABLE `request_song` (
`id` bigint NOT NULL auto_increment,
`song` longtext NOT NULL default '',
`artist` longtext NOT NULL default '',
`name` longtext NOT NULL default '',
`ip` varchar(100) NOT NULL default '',
PRIMARY KEY (`id`)
) TYPE=MyISAM;

Cope and paste the above save as request.sql

Remember if you have coped the code that Gremmie posted you will need the Connection information at the top.

<?php

if (eregi("block-Request.php",$PHP_SELF)) {
Header("Location: index.php");
die();
}

mysql_connect("Localhost","username","password");
// Change the above line
mysql_select_db("DBName");
// global $prefix, $db <<< Would not work

Copy and paste the above at the top of your file.

Did you have a look at the ShoutCast Blocks at Camp-Battlegroup.

Cheers

Scorp

Posted: Sat Mar 29, 2008 9:43 am

Well...$prefix and $db would work if you convert the rest of the mysql_ calls over to the $db abstraction layer.

Posted: Sat Mar 29, 2008 1:11 pm

Cheers Gremmie

I was wondering how to use the global $prefix

Thanks for all the advise and help

Scorp

Posted: Sat Mar 29, 2008 2:16 pm

Well it doesn't look like you used a prefix on your table request_song, so you wouldn't have to use it if you didn't want to. The prefix is used in case you want to run multiple nuke sites or add-ons together in the same database and need to avoid name clashes.

To use the $db layer, just take a look at some of the included blocks and see what they are doing. There are member functions like $db->sql_query(), $db->sql_fetchrow(), etc that simply replace the mysql_xxx() calls you have in the block code. You could keep the code you have the same; the only caveats with doing that is that you are creating an extra database connection per page load for that block code, and your code isn't totally "nuked". You can decide if those are important enough to switch the code over.