[SOLVED] $_Get('currentpage') I think ...?

[Gem]

Hiya

 

Trying to build a script that will list titles and subtitles, and when you click on the title, you see title, subtitle and ARTICLE ... and its not working very well ...

 

I think im getting close though ... I have very limited knowledge of PHP so im doing alot of trial and error lol

 

This is what is happening when I load the page (Note the error at the bottom)

 

www.bradleystokejudoclub.co.uk/test.php

 

And heres the code (Line 15 is $currentpage = $_GET('currentpage'); )

 

Hope you can help =)

 

<?php
// database connection info
$conn = mysql_connect("80.94.196.33","USER_INFO") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);

$result = mysql_query ("SELECT title, subtitle, article FROM articles ORDER BY ID DESC");
while(list($title, $subtitle, $article) = mysql_fetch_row($result))
{
    echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$title'>$title</a><BR />";
echo $subtitle;
echo "<BR />";
echo "<BR />";

} 
$currentpage = $_GET('currentpage');

if ($currentpage = $title)
{
echo $title;
echo "<BR />";
echo $subtitle;
echo "<BR />";
echo nl2br($article);
echo "<BR />";
}
else
{
echo "ERROR";
}

?>

[revraz]

$_GET uses [ ] around the array, not ( )

 

Also, you need to pass something in order to $_GET it.

[Gem]

Hi Revraz ...

 

Thank you

 

What do you mean by pass something to $_GET it?? Im kind of making it up as I go, so dont really know what I'm doing ... ???

 

Cheers

 

Gem

[Gem]

^

[unrelenting]

Hi Revraz ...

 

Thank you

 

What do you mean by pass something to $_GET it?? Im kind of making it up as I go, so dont really know what I'm doing ... ???

 

Cheers

 

Gem

 

http://www.tizag.com/phpT/postget.php

[Gem]

I kind of get what that is saying, thanks.

 

Does that mean I cant do it this way?

 

Can you suggest another way?

 

Thanks

[Gem]

^

[revraz]

Don't know, you really don't say what you are actually doing.  What is current page supposed to represent?

 

I kind of get what that is saying, thanks.

 

Does that mean I cant do it this way?

 

Can you suggest another way?

 

Thanks

[Gem]

LOL sorry..

 

I'll try to explain

 

Might be easier to see for yourself though

 

www.bradleystokejudoclub.co.uk/test.php

 

On that page I have listed title and subtitle from my database....

 

I want to be able to click on title, and it take me to a new "page" where it shows title, subtitle, and article ...

 

Does that make any sense??

 

Thanks mate - I really appreciate your help

 

Gem

[revraz]

Instead of using the title to build the URL, use the ID of the Row that holds the data.  Then have it go to a new page on how you want that data to be displayed.

 

Then you need to Select that row of data to display it.

[Gem]

???  my head hurts ...

 

so, when you say have it go to a new page ... do you literally mean a new page, or ..test.php?id=1  ??

 

that would be perfect, but I dont know how to do it.

 

a clue perhaps?

 

I really appreciate you help so much, I'm enjoying learning php and what it can do!! =)

 

Gem x

 

 

 

 

[revraz]

I personally would go to a new page, but still use your id=1 with it, so you pass which row you are displaying.  Then that page, all you have to do is worry about displaying just that data.

[unrelenting]

I personally would go to a new page, but still use your id=1 with it, so you pass which row you are displaying.  Then that page, all you have to do is worry about displaying just that data.

 

Try something like this.

 

<?
// database connection info
if ((isset ($_GET['currentpage']))
{
$currentpage = $_GET('currentpage');
$conn = mysql_connect("80.94.196.33","USER_INFO") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);

$result = mysql_query ('SELECT title, subtitle, article FROM articles WHERE title = "' . $title . '"');
while(list($title, $subtitle, $article) = mysql_fetch_row($result))
{
	echo $title;
	echo "<BR />";
	echo $subtitle;
	echo "<BR />";
	echo nl2br($article);
	echo "<BR />";
}
}

$conn = mysql_connect("80.94.196.33","USER_INFO") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);

$result = mysql_query ("SELECT title, subtitle, article FROM articles ORDER BY ID DESC");
while(list($title, $subtitle, $article) = mysql_fetch_row($result))
{
    echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$title'>$title</a><BR />";
   echo $subtitle;
   echo "<BR />";
   echo "<BR />";
   
}
else
{
   echo "ERROR";
}

?>

[Gem]

Hiya

 

Thanks for your help.  I uploaded your code, but I got this error:

 

PHP Parse error: syntax error, unexpected '{' in d:\webspace\bradleystokejudoclub.co.uk\wwwroot\test.php on line 4

 

If I remove the {} then I get "Unexpected T_VAR" which is this one: $currentpage

 

What to do?

 

X

[Gem]

Sorry to bump, but im stuck without this ... x

[revraz]

That is because you have 3 ((( and only 2 )) here

 

if ((isset ($_GET['currentpage']))

 

Get rid of the first (

 

if (isset ($_GET['currentpage']))

[Gem]

Thanks Rev - This is what i've got so far

 

Test.php (This works)

<?php

$conn = mysql_connect("80.94.196.33","gem","landseer") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);

$result = mysql_query ("SELECT id, title, subtitle, article FROM articles ORDER BY ID DESC");
while(list($id, $title, $subtitle, $article) = mysql_fetch_row($result))
{
    echo " <a href='test1.php?currentpage=$id'>$title</a><BR />";
   echo $subtitle;
   echo "<BR />";
   echo "<BR />";
   
}

?>

 

Test1.php (this doesnt work)

<?php
// database connection info
if (isset ($_GET['currentpage']))
$currentpage = $_GET('currentpage');
$conn = mysql_connect("80.94.196.33","gem","landseer") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);
$result = mysql_query ('SELECT title, subtitle, article FROM articles WHERE title = "' . $title . '"');
while(list($title, $subtitle, $article) = mysql_fetch_row($result))
{
echo $title;
echo "<BR />";
echo $subtitle;
echo "<BR />";
echo nl2br($article);
echo "<BR />";
}
?>

 

Error: PHP Fatal error: Call to undefined function: array() in d:\webspace\bradleystokejudoclub.co.uk\wwwroot\test1.php on line 4

 

I'm getting close ... I can feel it in my bones =)

[Maq]

Like revraz already mentioned you need to change:

 

$currentpage = $_GET('currentpage');

 

to:

 

$currentpage = $_GET['currentpage'];

[Gem]

LOL - took me like a minute to see what the difference was there!

 

Heres what I got now... I change the query to to select where id equals $id but all i get is a blank screen ...

 

<?php
// database connection info
if (isset ($_GET['currentpage']))
$currentpage = $_GET['currentpage'];
$conn = mysql_connect("80.94.196.33","gem","landseer") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);
$result = mysql_query ('SELECT title, subtitle, article FROM articles WHERE id = "' . $id . '"');
while(list($id, $title, $subtitle, $article) = mysql_fetch_row($result))
{
echo $title;
echo "<BR />";
echo $subtitle;
echo "<BR />";
echo nl2br($article);
echo "<BR />";
}
?>

[Maq]

Put this right after your opening php tag <?php

 

ini_set ("display_errors", "1");
error_reporting(E_ALL);

[Gem]

Thank you!! 

 

Error:

 

 

Notice: Undefined variable: id in d:\webspace\bradleystokejudoclub.co.uk\wwwroot\test1.php on line 9

PHP Notice: Undefined variable: id in d:\webspace\bradleystokejudoclub.co.uk\wwwroot\test1.php on line 9

[Maq]

Well where does $id come from?  Before your clause was "WHERE title = '$title'" now you're using id, which is it?

[Gem]

well i figured that because in test.php i was linking using the $id ... I should probably use that in the query on test1.php ??

 

I am really stuck ...  ???

 

Yesterday I loved PHP, today I'm hating it!

[Maq]

well i figured that because in test.php i was linking using the $id ... I should probably use that in the query on test1.php ??

 

I am really stuck ...  ???

 

Yesterday I loved PHP, today I'm hating it!

 

Looks like you should be using '$currentpage'.

[Gem]

where to though mate??

 

<?php
ini_set ("display_errors", "1");
error_reporting(E_ALL);
// database connection info
if (isset ($_GET['currentpage']))
$currentpage = $_GET['currentpage'];
$conn = mysql_connect("80.94.196.33","gem","landseer") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db('bssql',$conn) or trigger_error("SQL", E_USER_ERROR);
$result = mysql_query ('SELECT title, subtitle, article FROM articles WHERE id = "' . $id . '"');
while(list($id, $title, $subtitle, $article) = mysql_fetch_row($result))
{
echo $title;
echo "<BR />";
echo $subtitle;
echo "<BR />";
echo nl2br($article);
echo "<BR />";
}
?>

 

All i want to is get the $id from the url of the page and then use that to tell the page which article it is I want echo'd ... all seemed so easy in my head 

 

Heres the pages, that might help

www.bradleystokejudoclub.co.uk/test.php

 

and then click on one of the links and that is the page above

 

:-\