Comments box

**keyboard** · 08-07-2011, 08:06 AM

I think I changed the right bit

PHP Code:


<?php 



 mysql_connect("localhost", "root", "") or die(mysql_error()); 

 mysql_select_db("keyboard_test") 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><br />   $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>

but it's still coming up with the same 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 58

**keyboard** · 08-07-2011, 08:13 AM

I changed one of the bits of code

PHP Code:


$cheese =  "<b>$username</b><br />   $comment  ";

So that it displays their name then on the next line their comment. Like this

Name
Their comment
Name
Their comment
Name
Their comment
Name
Their comment

When I would like it to look like this

Name
Their comment

Name
Their comment

Name
Their comment

Name
Their comment

Any help?

**JShor** · 08-07-2011, 04:01 PM

Originally Posted by keyboard1333

I think I changed the right bit

but it's still coming up with the same 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 58

This time you had a period missing in that same line of code.

Use this:

PHP Code:


<?php 



 mysql_connect("localhost", "root", "") or die(mysql_error()); 

 mysql_select_db("keyboard_test") 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><br />   $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>

**JShor** · 08-07-2011, 04:02 PM

Originally Posted by keyboard1333

I changed one of the bits of code

PHP Code:


$cheese =  "<b>$username</b><br />   $comment  ";

So that it displays their name then on the next line their comment. Like this

Name
Their comment
Name
Their comment
Name
Their comment
Name
Their comment

When I would like it to look like this

Name
Their comment

Name
Their comment

Name
Their comment

Name
Their comment

Any help?

Can't you just add another line-breaker tag?

PHP Code:


$cheese =  "<b>$username</b><br /><br />   $comment  ";

That should work for you.

**keyboard** · 08-08-2011, 06:39 AM

Yep, that all works now. Thanks for all the help.

**keyboard** · 08-08-2011, 06:48 AM

I ran the code and into the date column it inserted this

2007-08-11

PHP Code:


$date = date("d/m/y");

That's what it should be inserting but the date is wrong. it should be 08/08/2011.
Any help?

**JShor** · 08-08-2011, 06:35 PM

What do you mean by inserting? Inserting into MySQL? YYYY-MM-DD is the typical date format for a field that has the type set to DATE.

If that's the case, try changing it to VARCHAR or TEXT or something. You could also insert the date in UNIX time using BIGINT, which is actually a more elegant solution.

**keyboard** · 08-08-2011, 10:55 PM

How do I use unix time?

**JShor** · 08-09-2011, 12:09 AM

http://php.net/time

UNIX time is the number of seconds since the beginning of the UNIX epoch (January 1st, 1970 at midnight Greenwich time).

The time() function returns the time in UNIX time. Store that large number in a BIGINT column in MySQL and when you go to fetch the date, you can convert it with the date() function.

Let's suppose you have the number 1312848370. That is the UNIX time for August 8th, 2011 8:06:24 PM EST. The second argument of the date() function accepts a UNIX timestamp and converts it into a string format.

For example, the following would return 08/08/11

PHP Code: