Why doesn't my code work!!! please help

[Alexhoward]

Hi,

 

i'm so close but it doesn't work!

 

what i would like to do is press the button, and it inserts the username from their cookie, and the link (full text) into a sql table...

 

This code displays my links nicely on the page,in a grid,  in their own seperate forms.

 

see - http://www.everyonlinestore.co.uk/change_password.php   (ignore the name)

 

Code1

<?php


$conn = mysql_connect($server, $db_user, $db_pass)
or die ("could not connect to mysql");  #Connect to mysql
$rs = mysql_select_db( $database, $conn )
or die ("Could not select database"); #select database 

$sql = "Select link FROM links";  //pull the users from the table

$result= mysql_query($sql)
or die(" Could not add style facts");

echo "<table align=center border=1>";        // display the users in table
$c = 0;
while($row = mysql_fetch_array($result)) { 
  $user2 = $row['link'];
  if($c%5 == 0) echo "<tr height=\"100px\">"; // If the counter has ticked 6 times, start a new row.
echo "<td><FORM action=addlinks.php method=post><p align=center>$user2</p><p align=center>rating script here</p><p align=center><input type=submit></form></p>";

  if($c%5 == 4) echo "</tr>"; // If we're drawing the 6th pic, end this row.
  $c++;
}
if($c%5 != 4) echo "</tr>"; // If there isn't a number of pics divisible by 6, end the row
echo "</table>"; // end the table

?>

end of code1

 

Then this bit of php should insert them into the table.

 

however the variables are undefined...?

 

probably becaused they're not named in code1, but i can't name them inside the " "'s

 

I'm so stuck...!

 

code2

<?php 

// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());

// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());

// check if the link has already been added to username
$mysite_username = $_COOKIE["mysite_username"];
$check = "SELECT link FROM memberlinks WHERE username = '$mysite_username'"; 

$qry = mysql_query($check)
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry);
if ($num_rows != 0) { 
echo "you have already added that link.<br>";
echo "<a href=Change_password.php>Back</a>";
exit; 
}

// check username entered
if (!$mysite_username) {
echo "You did not enter a Username.<br>";
echo "<a href=change_password.php>Please Try again</a>";
exit;
}

// check link entered
if (!$user2) {
echo "You did not enter a link.<br>";
echo "<a href=change_password.php>Please Try again</a>";
exit;
}

else {



// insert the data
$insert = mysql_query("insert into memberlinks values ('".$_POST[$_COOKIE[mysite_username]]."', '".$_POST['$user2']."')")
or die("Could not insert data because ".mysql_error());

// print a success message
echo $_COOKIE["mysite_username"];
echo "Your link has been added!<br>"; 
echo "Now you can <a href=change_password.php>add another link</a>"; 
}

?>

end of code2

 

Thanks in advance for all your help

 

I'd love to get this sorted out, and no-one i know does PHP....

[MadTechie]

your form doesn't have any form elements except a button so $_POST['user2'] won't be passed (i assume you mean $_POST['user2'] and not $_POST['$user2']);

 

also

$_POST[$_COOKIE[mysite_username]]

should be probably be

$_COOKIE['mysite_username']

 

thats after a quick look..

[Alexhoward]

hi,

 

thats the problem.

 

i can't name the feilds with this script (or don't know how to would be more accurate), like you would a signup script for example

[lemmin]

I'm not entirely sure what you are trying to do, but try adding this to your echo statement that writes the form:

<input type=HIDDEN name='user2' value='$user2'>

[Alexhoward]

Thanks

 

well...it's not hidden

 

basically this is what i'm trying to do:

 

When you click the button i want it to insert your username from the cookie, and the link into an sql table

 

e.g.

 

username    link_id

johnsmith        link

 

 

Thanks

[lemmin]

Did you try it? It doesn't need to hide anything you have, it just sends the $user2 variable through POST since you are trying to get it from there anyway.

[Alexhoward]

http://www.everyonlinestore.co.uk/change_password.php

 

this is how it came out

 

it's odd that after you state <..... value='$user2'>

 

it's no longer enclosed in the < >.   it stops and displays the '>

 

have a look

 

cheers

[lemmin]

Something is messed up in the quote scheme. Try chaning the apostrophes to escaped quotes \" and see if it fixes it. Otherwise, post the code for that line so I can see what might be wrong.

[Alexhoward]

Hey,

 

thanks for helping me out.  i really appreciate it!

 

Here's the entire code.  it's a bit scrappy and hard to look at.

 

i've also put the username bit in with the same method

 

<?php


$conn = mysql_connect($server, $db_user, $db_pass)
or die ("could not connect to mysql");  #Connect to mysql
$rs = mysql_select_db( $database, $conn )
or die ("Could not select database"); #select database 

$sql = "Select link FROM links";  //pull the users from the table

$result= mysql_query($sql)
or die(" Could not add style facts");

echo "<table align=center border=1>";        // display the users in table
$c = 0;
while($row = mysql_fetch_array($result)) { 
  $user2 = $row['link'];
  if($c%5 == 0) echo "<tr height=\"100px\">"; // If the counter has ticked 6 times, start a new row.
echo "<td><FORM action='addlinks.php' method='post'><p><input type=HIDDEN name='user' value='mysite_username'></p><p><input type=HIDDEN name='user2' value='$user2'></p><p align=center>rating script here</p><p align=center><input type=submit></form></p>";

  if($c%5 == 4) echo "</tr>"; // If we're drawing the 6th pic, end this row.
  $c++;
}
if($c%5 != 4) echo "</tr>"; // If there isn't a number of pics divisible by 6, end the row
echo "</table>"; // end the table

?>

[lemmin]

I don't see anything wrong in the code so the variable $user2 must hold a string with an apostrophe in it. Changing the apostrophes around $user2 into escaped quotes (\") should fix that.

[Alexhoward]

hmm...

 

this is what the links look like in the table

 

<a href="http://www.dpbolvw.net/click-2804964-5590692" target="_blank" onmouseover="window.status='http://www.argos.co.uk';return true;" onmouseout="window.status=' ';return true;"><img src="http://www.awltovhc.com/image-2804964-5590692" width="120" height="60" alt="Argos.co.uk 120x60" border="0"/></a>

 

is it the

</a>

on the end...?

 

are you saying like:

 

\"$user2"\

[Alexhoward]

ok took the

<a> and </a>

off the link

 

and it works

 

but now it's now NOT inserting the entire link into the sql table

 

any ideas?