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User content from sql table - igoogle type thing


Alexhoward

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Hello everybody! (Dr. nick)

 

I’m attempting to build a rather complicated website, and have a limited knowledge of php.

 

Saying that, I’ve come a long way from when I stated…

 

Here's where I'm stuck:

 

What I’d like to do is have a page of links,

 

For example.

 

This displays a link for ebay

 

<a href="http://rover.ebay.com/rover/1/710-5232-2978-17/1?aid=5826369&pid=2804964&id=12" target="_blank" onmouseover="window.status='http://www.ebay.co.uk';return true;" onmouseout="window.status=' ';return true;"><img src="http://www.lduhtrp.net/image-2804964-5826369" width="120" height="60" alt="Click Here to shop at eBay.co.uk" border="0"/></a>

 

Sorry, it's shown the link and i don't know how to show it just as the text...?!

 

Then I’d like people to be able to choose this link and it will add it to their own login page (dynamically not a separate url)

 

I’ve managed to drag the link out of the table via the username from a cookie and display the link of their page. But directly and not how I’d like

 

However…

 

This is how I’d like it work –

 

Select link, this will save into a table – username, link_id, lets call it “memberslinks”

 

Table of links = link_id, link, call this one “links” (this Is where I’ll save all the links)

 

Next pull the link out of the table and display on page

 

Something like:

 

//$mysite_username is the username cookie

 

$test = @mysql_query("select link_id FROM memberslinks WHERE member ='$mysite_username'");

$result = @mysql_query("select link FROM links WHERE member ='$test'");

 

Display the links in a nice little 9 x 9 grid, with paging script (?)

 

Hope this is enough to explain what I’m up to

 

Thanks in advance

 

Alex Howard

 

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Also,

 

just to add.

 

Doing it directly, so have a table with

 

username, link

 

i can pull an array and show all the results all at once (not in a nice grid however, just row by row)

 

but doing the lookup between tables i can't

 

anyone know a way round all this

 

Thanks alot!

 

ALex Howard

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Hi Thorpe.

 

Apologies...

 

it's quite hard to expain....

 

say there was two tables

 

1. Username, link_id.  and it had johnsmith, 1

 

then.

 

2. link_id, link.  and it had 1, www.phpfreaks.com

 

i could use the previous example to pull this link out of the two tables and display it on the page.

 

however it would only show one, and not all using array...

 

secondly,

 

i'm unsure how to display the array as a nice grid on the page.

 

i kind of want it to work as igoogle does with adding links (don't need to pull them about thou)

 

and be arranged as www.gamershood.com is

 

does this make it any clearer...?

 

Thanks again

 

Alex Howard

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Hi,

 

I've got a little further and now i'm stuck again

 

I've used some code that i found here, and adapted it to meet the needs of my site

 

The problem is that the submit button does not work properly, and the username cookie and link are not inserted into the table.

 

there are 2 pages used to acheive this, although i'd rather it just did it when you hit the buttin, rather than go to another page, then have to return to add more links

 

The Code - ignore the page names

 

add link page - http://www.everyonlinestore.co.uk/change_password.php

 

 

<?php

 

echo $_COOKIE["mysite_username"];

 

?>

 

 

<?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 "<FORM action=addlinks.php method=post><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><p align=center>$user2</p><p align=center>rating</p><p align=center><input type=submit></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></form>"; // end the table

 

?>

 

Sumit page query

 

// 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 (!$_POST['$mysite_username']) {

echo "You did not enter a Username.<br>";

echo $_COOKIE["mysite_username"];

echo "<a href=change_password.php>Please Try again</a>";

exit;

}

 

// check link entered

if (!$_POST['$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['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>";

}

 

?>

 

Thanks in advance

 

if you're gettign an error on the change_password page, it because you're not logged in.

 

click the banner and sign up if you like

 

then log in, and go back to the change_password page. refresh browser

 

Thankyou!

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ok,

 

so i've made them all their own seperate forms:

 

 

code

 

<?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

 

?>

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Hi redarrow,

 

if you goto - http://www.everyonlinestore.co.uk/change_password.php

 

you'll see.

 

It drags all of the links out of my sql table, and displays them in a nice little grid on the page

 

(i've just put a few in to test)

 

(i'd still like the grid to be spread out a bit more, but don't know how to do that either...)

 

(also eventually there will be hundreds of links, so i would like to include some kind of paging script as well)

 

the "rating script here" will eventualy be a rating script

 

The problem is the submit query button.

 

I'd like it to save the username (from the cookie) and the link into another table,

 

ready to be retrived from the users login page

 

 

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