Its expecting a ;

futurewii · June 24, 2008

Hey, I get this message..

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /mounted-storage/home36c/sub005/sc28991-YIVP/futurewii.com/pages/videos/view.php on line 10

With this script..

<div id="content"><br />
<br />
<?php

include "includes/conn.php";

$result = mysql_query("SELECT * FROM videos WHERE id = '".$_GET['id']."'")  
or die(mysql_error());  
while($row = mysql_fetch_array( $result )) {
echo "<h2><span lang="en-gb">$row['title']</span></h2>";
	echo "<br />";
echo $row['script'];
	echo "<br />";
echo "<h7>$row['download']</h7>";
	echo "<br />";
	echo "<br />";
echo "<p><span lang="en-gb">$row['description']</span></p>";
echo "<br />";
   echo "<br />";
   echo "<br />";
echo "<br />";
   echo "<br />";
   echo "<br />";

$page_id = "$row['id']"; // for example
include("comments/comments_show.php");
include("comments/comments_form.php");
?>

vbnullchar · June 24, 2008

do it like this..

echo "<h2><span lang='en-gb'>$row[title]</span></h2>";

futurewii · June 24, 2008

ok thanks a lot

DarkWater · June 24, 2008

Wow. So many people code so sloppily these days. When using arrays, the keys should be enclosed in ' ' or " " (for variable keys). Then, when echo'd in a string, they must be enclosed in { }.

echo "{$row['title']}";

As an example.

The only reason that it works the way you have it is because PHP doesn't WANT to through errors at you, so it tries to "fix" it. First, it looks for a constant named title, and if it doesn't find one, it ASSUMES you wanted 'title'. You should really clean that up.

whizard · June 24, 2008

and... there's no such thing as an <h7> tag...

DarkWater · June 24, 2008

Ahaha, I just noticed that.

This line:

echo "<h7>$row['download']</h7>";

There's no such thing as an h7 tag. =) What did you mean to put there? xD

futurewii · June 24, 2008

The h7 tag refers to a CSS tag because the page is actually an include

whizard · June 24, 2008

It doesn't matter if its an include page or line one of index.html in root. It doesn't matter if its HTML, XHTML, CSS, XML, or anything else. There's still no such thing as an <h7> tag.

kenrbnsn · June 24, 2008

Just because there isn't a <h7> tag defined in the standards, doesn't mean you can't use it and define the style via CSS. See:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">

<html>
<head>
<title></title>
<style type="text/css" media="screen">
h7 {
	color: red;
	font-size: 150%;
}
</style>
</head>

<body>
<h7>This is a h7 area</h7>


</body>
</html>

True, it won't validate, but it does work (at least in Firefox3).

Ken

futurewii · June 24, 2008

Thanks

Just one more thing and i'm finished bugging you for tonight :-[

<tr style='font-size:12px;' bgcolor="#c0c0c0">
	<th>Game Title</th>
	<th>Video Type</th>
<?php

include "includes/conn.php";

$query_count = "SELECT * FROM videos WHERE console='wii'"; 

		print "<p><tr><a href='?page=video&id=".$row['id']."'>".$row['title'] . "</a></tr>";
		print "<td>".$row['type']."</td>";
		print "</tr>";
		print "</tr>";
		print "</table>";

	 ?>

I'm wanting the table to only show entries which have their 'CONSOLE' value as 'wii' but nothing shows up in the table ??? any suggestions?

kenrbnsn · June 24, 2008

You need to fetch the results from the query:

<?php
$query_count = "SELECT * FROM videos WHERE console='wii'"; 
$row = mysql_fetch_assoc($query_count));
print "<p><tr><a href='?page=video&id=".$row['id']."'>".$row['title'] . "</a></tr>";
print "<td>".$row['type']."</td>";
print "</tr>";
print "</tr>";
print "</table>";
?>

Ken

peranha · June 24, 2008

EDIT : Beat to it by ken

futurewii · June 24, 2008

Hey, Thanks a lot for your help. The entry is called wii not 'wii' and when I try your code Ken i get this error..

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /mounted-storage/home36c/sub005/sc28991-YIVP/futurewii.com/pages/videos/index.php on line 15

kenrbnsn · June 24, 2008

Sorry, forgot to put the query in:

<?php
$query_count = "SELECT * FROM videos WHERE console='wii'"; 
$rs = mysql_query($query_count) or die("Problem with the query: $query_count<br>" . mysql_error());
$row = mysql_fetch_assoc($rs);
print "<p><tr><a href='?page=video&id=".$row['id']."'>".$row['title'] . "</a></tr>";
print "<td>".$row['type']."</td>";
print "</tr>";
print "</tr>";
print "</table>";
?>

You need to enclose strings in single quotes for mysql.

Ken

futurewii · June 24, 2008

Now I get this ???

Warning: mysql_query() [function.mysql-query]: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) in /mounted-storage/home36c/sub005/sc28991-YIVP/futurewii.com/pages/videos/index.php on line 15

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /mounted-storage/home36c/sub005/sc28991-YIVP/futurewii.com/pages/videos/index.php on line 15

Problem with the query: SELECT * FROM videos WHERE console=wii

Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2)

Im really sorry for the constant questions, everyone here is extremely helpful and its the only way i'll learn

kenrbnsn · June 24, 2008

Those errors are saying you're having problem connecting to your MySQL database.

Ken

futurewii · June 24, 2008

Oh dear..

All my other databases seem to still be active though, but as long as there's no scripting errors By the way can i just check that

WHERE console='wii'";

Is the correct phrase to make the table only show fields which have wii as the entry under 'console'?

Thanks muchly

futurewii · June 24, 2008

Servage has a bad reputation of randomly cutting off my MySql databases anyway for about 20 seconds on end, but its still not right >:( ???

futurewii · June 24, 2008

Ok its working now But how do I get it to show all the entries in the database? at the moment it is just showing one..

http://futurewii.com/?page=videos/index

Thanks

futurewii · June 24, 2008

Its also skipping the tables down by one too :-\

kenrbnsn · June 24, 2008

Please repost the code you have now.

Ken

futurewii · June 24, 2008

ï»¿<div id="content">
<h2><span lang="en-gb">Video Database..</span></h2>

		<br /><br />
<h2>Nintendo Wii</h2>
<br />
<table width="400">
<tr style='font-size:12px;' bgcolor="#c0c0c0">
	<th>Game Title</th>
	<th>Video Type</th>
<?php
include "includes/conn.php";
$query_count = "SELECT * FROM videos WHERE console='wii'"; 
$rs = mysql_query($query_count) or die("Problem with the query: $query_count<br>" . mysql_error());
$row = mysql_fetch_assoc($rs); {
		print "<tr style='font-size:12px;'>";
		print "<td><a href='?page=videos/view&id=".$row['id']."'>";
		echo $row['title'];
		print "</a></td>";
		print"<td>".$row['type']."</td>";
		print "</tr>";

	}

?>

<br /><br />
<h2>Gamecube</h2>
<br />
<table width="400">
<tr style='font-size:12px;' bgcolor="#c0c0c0">
	<th>Game Title</th>
	<th>Video Type</th>
<?php
include "includes/conn.php";
$query_count = "SELECT * FROM videos WHERE console='ngc'"; 
$rs = mysql_query($query_count) or die("Problem with the query: $query_count<br>" . mysql_error());
$row = mysql_fetch_assoc($rs); {
		print "<tr style='font-size:12px;'>";
		print "<td><a href='?page=videos/view&id=".$row['id']."'>";
		echo $row['title'];
		print "</a></td>";
		print"<td>".$row['type']."</td>";
		print "</tr>";

	}

?>
<br /><br />
<h2>Virtual Console</h2>
<br />
<table width="400">
<tr style='font-size:12px;' bgcolor="#c0c0c0">
	<th>Game Title</th>
	<th>Video Type</th>
<?php
include "includes/conn.php";
$query_count = "SELECT * FROM videos WHERE console='vcg'"; 
$rs = mysql_query($query_count) or die("Problem with the query: $query_count<br>" . mysql_error());
$row = mysql_fetch_assoc($rs); {
		print "<tr style='font-size:12px;'>";
		print "<td><a href='?page=videos/view&id=".$row['id']."'>";
		echo $row['title'];
		print "</a></td>";
		print"<td>".$row['type']."</td>";
		print "</tr>";

	}

?>
<br /><br />
<h2>Other Videos</h2>
<br />
<table width="400">
<tr style='font-size:12px;' bgcolor="#c0c0c0">
	<th>Game Title</th>
	<th>Video Type</th>
<?php
include "includes/conn.php";
$query_count = "SELECT * FROM videos WHERE console='oth'"; 
$rs = mysql_query($query_count) or die("Problem with the query: $query_count<br>" . mysql_error());
$row = mysql_fetch_assoc($rs); {
		print "<tr style='font-size:12px;'>";
		print "<td><a href='?page=videos/view&id=".$row['id']."'>";
		echo $row['title'];
		print "</a></td>";
		print"<td>".$row['type']."</td>";
		print "</tr>";

	}

?>

<br />

Thanks

futurewii · June 24, 2008

There it is ^

http://www.futurewii.com/?page=videos/index

Sign In

Its expecting a ;

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