Elite Coders
You need to end the statement
order the comments by the date descendingPHP Code:$sql = mysql_query ("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')";
I'd need to see the html output to tell you what is wrong with the bold username.order by date desc
Corrections to my coding/thoughts welcome.
Administrator
Somewhere on this page its coming up with this error
( ! ) Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\Canberra Amatuer Productions\echo time.php on line 59
PHP Code:<?php
mysql_connect("localhost", "****", "****") or die(mysql_error());
mysql_select_db("****") or die(mysql_error());
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{ header("Location: login.php");
}
else
{
}
}
}
else
{
header("Location: login.php");
}
$links = "link2.php";
?>
<?php
if (isset ($_POST['submit']))
{
$comment = mysql_escape_string ( '<b>' . trim (nl2br($_POST['comment'])) . '</b>' );
$quantity = $_POST['quantity'];
$item = $_POST['item'];
$cheese = "<b>$username</b> - $comment ";
$sql = mysql_query ("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')";
echo 'Your comment has been entered successfully!';
echo '<a href="echo time.php">Please click here</a>';
}
else
{
?>
<html>
<head>
<body style="background-color:lightgreen">
</head>
<body>
<?php include("$links");
echo "Hey ".$username;
?>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b>COMMENTS</b></p>
<?php
// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
}
echo '<br /><br />
<form action="echo time.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
</body>
</html>
I'm pretty sure it's on this line of code
But not 100% sure.PHP Code:("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')";
How do I organise the comments by date?
This is the comments box which is making the entire line bold
PHP Code:
</body>
</html>Hey ----
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b>COMMENTS</b></p>
<b>----</b> - <b>----</b> <br /><b>----</b> - <b>----</b> <br /><b>----</b> - <b>----</b> <br /><br /><br />
<form action="member.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form></body>
</html>
That's the html output
Last edited by djr33; 09-19-2012 at 03:48 AM. Reason: user request
Senior Coders
Yes, it is. You didn't close the function with a ")" (I know, it can get confusing sometimes if you have multiple parentheses/brackets/curly brackets).Originally Posted by keyboard1333
PHP Code:mysql_query("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')");
Use the ORDER BY in your SQL selection query. For example:
You can change DESC (descending order) to ASC (ascending order).Code:SELECT * FROM `comments` WHERE id = 1 ORDER BY date DESC
You already ended your <body> and <html> tags. Legally, you're not supposed to write any HTML after that. You also did this earlier in your code.
Try this:
PHP Code:<?php
mysql_connect("localhost", "****", "****") or die(mysql_error());
mysql_select_db("****") or die(mysql_error());
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{ header("Location: login.php");
}
else
{
}
}
}
else
{
header("Location: login.php");
}
$links = "link2.php";
if (isset ($_POST['submit']))
{
$comment = mysql_escape_string ( '<b>' . trim (nl2br($_POST['comment'])) . '</b>' );
$quantity = $_POST['quantity'];
$item = $_POST['item'];
$cheese = "<b>$username</b> - $comment ";
$sql = mysql_query ("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')";
echo 'Your comment has been entered successfully!';
echo '<a href="echo time.php">Please click here</a>';
}
else
{
?>
<html>
<head>
</head>
<body style="background-color:lightgreen">
<?php include("$links");
echo "Hey ".$username;
?>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b>COMMENTS</b></p>
<?php
// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
}
echo '<br /><br />
<form action="echo time.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
</body>
</html>
Last edited by djr33; 09-19-2012 at 03:49 AM. Reason: user request (content not changed, just removed personal info for keyboard1333)
- Josh
Senior Coders
Also, not to be picky, but I should mention that you need a doctype. From the looks of it, you're using XHTML standards.
Add this to the very beginning of your HTML code:
Code:<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" dir="ltr" lang="en">
- Josh
Administrator
I've got multiple </body> </html> because i'm using php include and the page i'm including has the tags on it. I'll fix that up. Also, thanks for the doctype.
Senior Coders
No problem, and regardless of how many includes you have, your final output should be valid HTML, or you may experience display problems.
- Josh
Administrator
I just changed the bit which is like this
to thisPHP Code:// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
PHP Code:// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM `comments` WHERE id = 1 ORDER BY date DESC") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
I don't think I did it right because when I ran the edited code the comments didn't show.
Administrator
Just tried it and it ran this error
Try this:
PHP Code:<?php
mysql_connect("localhost", "****", "****") or die(mysql_error());
mysql_select_db("****") or die(mysql_error());
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{ header("Location: login.php");
}
else
{
}
}
}
else
{
header("Location: login.php");
}
$links = "link2.php";
if (isset ($_POST['submit']))
{
$comment = mysql_escape_string ( '<b>' . trim (nl2br($_POST['comment'])) . '</b>' );
$quantity = $_POST['quantity'];
$item = $_POST['item'];
$cheese = "<b>$username</b> - $comment ";
$sql = mysql_query ("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')";
echo 'Your comment has been entered successfully!';
echo '<a href="echo time.php">Please click here</a>';
}
else
{
?>
<html>
<head>
</head>
<body style="background-color:lightgreen">
<?php include("$links");
echo "Hey ".$username;
?>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b>COMMENTS</b></p>
<?php
// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
}
echo '<br /><br />
<form action="echo time.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
</body>
</html>
( ! ) Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\Canberra Amatuer Productions\member.php on line 51
Administrator
The comment is now not displaying in bold while their name is, so that's all working. However, now the comments are not displaing with a line inbetween each. I think I might have accidentally changed something. Here is the full code.
PHP Code:<?php
mysql_connect("localhost", "****", "****") or die(mysql_error());
mysql_select_db("****") or die(mysql_error());
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{ header("Location: login.php");
}
else
{
$links = "link2.php";
}
}
}
else
{
header("Location: login.php");
}
?>
<?php
if (isset ($_POST['submit']))
{
$comment = mysql_escape_string (trim (nl2br($_POST['comment'])));
$quantity = $_POST['quantity'];
$item = $_POST['item'];
$cheese = "<b>$username</b> - $comment ";
$sql = mysql_query ("INSERT INTO comments (id,comments) VALUES ('0','".$cheese."')");
echo '<center>Your comment has been entered successfully!</center>';
echo '<center><a href="member.php"><br />Please click here</a></center>';
}
else
{
?>
<html>
<head>
<body style="background-color:lightgreen">
</head>
<body>
<?php include("$links");
echo "Hey ".$username;
?>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b>COMMENTS</b></p>
<?php
// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo $row['comments'].'<br />';
}
echo '<br /><br />
<form action="member.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
</body>
</html>
I also have a question. When you use php echo to write something on the page the rest of the page dissapears like in this bit of code
What I was wondering, is their a way to keep the background light green while still replacing all of the old content?PHP Code:echo '<center>Your comment has been entered successfully!</center>';
echo '<center><a href="member.php"><br />Please click here</a></center>';
Senior Coders
You didn't modify the code with the piece I gave you, which is why you're still receiving that syntax error.
Use this:
PHP Code:mysql_query("INSERT INTO comments (id,comments,date,ip) VALUES ('0','".$cheese."','".$date."','" . $_SERVER['REMOTE_ADDR']"')");
I didn't find any code that formats the output of the comments... only the comments that were input into the database.However, now the comments are not displaing with a line inbetween each. I think I might have accidentally changed something
This should work for separating spaces though:
PHP Code:// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
echo str_replace("\n", "<br />", $row['comments']).'<br />';
}
What disappears?I also have a question. When you use php echo to write something on the page the rest of the page dissapears like in this bit of code
The reason that your background is no longer light green is because you defined that in the <body> tag, which you closed. Then you opened a new <body> tag with no background definition. You cannot have multiple <body>, <head> or <html> tags. Also, <body> cannot be in the <head> tags.What I was wondering, is their a way to keep the background light green while still replacing all of the old content?
Try replacing this:
With this:Code:<html> <head> <body style="background-color:lightgreen"> </head> <body>
Hope this helps.Code:<html> <head> </head> <body style="background-color:lightgreen">
Last edited by JShor; 08-06-2011 at 11:06 PM.
- Josh
Posting Permissions
Reply With Quote
Bookmarks