Dynamic links and content

[The_Squirrel]

Ok, I have a site that I would like to keep just one file that needs to be edited if things need to be changed.

I have my template.php page set up with the following code:

<?

error_reporting (E_ALL ^ E_NOTICE);


// Connect to the database server
$dbcnx = @mysql_connect('x', 'x', 'x');
if (!$dbcnx) {
  die( '<p>Unable to connect to the ' .
       'database server at this time.</p>' );
}

// Select the content database
if (! @mysql_select_db('fibercoating') ) {
  die( '<p>Unable to locate the Fibercoating content ' .
       'database at this time.</p>' );
}

// If index page has been called,
// display it.
if (isset($_GET['index'])) {
  $itemid = $_GET['index'];
  $sql = "SELECT * FROM `index`";
 if (@mysql_query($sql)) {
	echo $result;
  } else {
    echo('<p>Error loading page: ' .
         mysql_error() . '</p>');
  }
}

// If history page has been called,
// display it.
if (isset($_GET['history'])) {
	$itemid = $_GET['history'];
	$sql = "SELECT * FROM `history`";
   if (@mysql_query($sql)) {
	   echo $result;
   } else {
	   echo('<p>Error loading page: ' .
		   mysql_error() . '</p>');
   }
}

   // Request the ID and text of index page
    $result = @mysql_query('SELECT ID, text FROM `index`');
    if (!$result) {
      die('<p>Error performing query: ' .
          mysql_error() . '</p>');
    }

// Request the ID and text of history page
$result = @mysql_query('SELECT ID, text FROM `history`');
if (!$result) {
	die('<p>Error performing query: ' .
		mysql_error() . '</p>');
}

// Display the text of each item in a paragraph
while ( $row = mysql_fetch_array($result) ) {
  //$itemid = $row['ID'];
  $itemtext = $row['text'];
}

?>

with my links I have them set up like so:
<a href="template.php?history=true"

But it just simple loads the page and changes the URL to www.mydomain.com/template.php?hisotry=true instead of loading the content from that portion of the database.

How do I get this to work properly?

Thanks for any help!

[mtimdog]

your link says hisotry, not history. Could this be the error?

[The_Squirrel]

oops, I typed that portion of the code into this forum. On the code I use in the script it is just fine and correctly spelled. Just an error I made here. 🙂

[leatherback]

Hi,

By the way your queries are set-up, there are a lot of errors in the statements. All the 'íf statements' are non-exclusive: No matter what happens, your routine will run every query.

Also, when you start expanding your site, you need to add a new query for every page you build.

More convenient, and easier to get to work would be including the section in your query:

www.mydom.com/template.php?section=history&page=10

You would then say:

$thisSection = $_GET[section];
$ThisPage = $_GET[[page];

//and use this directly as a queryinput. (Of course you want some checking on your variables for hostile scripts being parsed to your page).

$query = select * from table where section=$ThisSection and page=$ThisPage

$result = mysql_query($query);

while $row = mysql_fetch_array($result)
  {
  echo $row[columName];
  }

You would need to add a check to make sure the right page is loaded if no variables are pased.

In your original page, you have a lot of statements where you call the variable $result., without actually creating it. Have a look at the way I get the results from the database.

[mtimdog]

if (@($sql)) {
echo $result;

you're not setting the query to anything ($result)....And even if you did, echoing out result will echo out something like ResourceID#14. You need to loop out the result with one of the mysql_fetch_* functions.

[The_Squirrel]

Originally posted by leatherback


More convenient, and easier to get to work would be including the section in your query:

www.mydom.com/template.php?section=history&page=10

You would then say:

$thisSection = $_GET[section];
$ThisPage = $_GET[[page];

//and use this directly as a queryinput. (Of course you want some checking on your variables for hostile scripts being parsed to your page).

$query = select * from table where section=$ThisSection and page=$ThisPage

$result = mysql_query($query);

while $row = mysql_fetch_array($result)
  {
  echo $row[columName];
  }

[/B]

So basically I need to change my database tables to have the fields of section -- which would be for example, history on the history table -- and then a page field that would be... what exactly?

You will have to forgive me, I am not quite grasping the concept you are showing me. But from what I can understand it does look better then what I had, I just need some more help understanding it fuller so I know exactly what is going on. 🙂

[leatherback]

Hi,

Of course I cannot see how you structure your database, and how you plan on going about on expanding your site -or not-.
So exactly how to go on with this is a bit tricky.

What is an other option is to pass the table in the url and the id of te record you want.

You would then for instance get something like:
select * from $table where id=$record

Basically all I was trying to show you, is the idea that you can work most database queries from one basic section, instead of creating a whole bunch of different queries, for different sections.

IF you decide you want to stick with the way you set it upt (Which can work just fine in a lot of cases, where you do not expand your setup, and only have a few different sections) make sure you have either nested if statements:

if(-- some table is set)
  {
  do something
  }
   else
   {
    if(-- some other table is set --)
     {
      do something
      }
    }

Or use switches. (Have a look in the manual for how htis works. Basically you test a variable for a value, and based on the value you go to a specific set of routine:

switch($section)
  {
  case "history":
   // Do the stuff for history
  break;

   case "index":
   // Do stuff for index
   break;
   }

[The_Squirrel]

Ok, I am starting to understand this.

If I do my code as follows:

<?php

error_reporting (E_ALL ^ E_NOTICE);


// Connect to the database server
$dbcnx = @mysql_connect('x', 'x', 'x');
if (!$dbcnx) {
  die( '<p>Unable to connect to the ' .
       'database server at this time.</p>' );
}

// Select the content database
if (! @mysql_select_db('fibercoating') ) {
  die( '<p>Unable to locate the Fibercoating content ' .
       'database at this time.</p>' );
}

switch($result)
{
	case "history":
		$historysql = "my SQL query...";
		break;
	case "index":
		$indexsql = "my SQL query...";
		break;
	case "etc...":
		$etcsql = "my SQL query...";
		break;
}

?>

Now I have a question...

How do I get the $result variable filled? In a book that I have it uses a select (drop down box) form named for example, result. Then it runs the switch statements with the $result variable being used. I looked at the manual but I cannot glean anything from there to help answer this particular portion of the problem.

Thanks for all of your help so far!

I forgot to mention that I will only have about 7 pages to call on this site. It is a small company and with very little content to their site. It may reach 10 pages but that is all it will ever have.

[The_Squirrel]

Greetings,

Here is some more code that I threw together that I thought might work.

<?php
error_reporting (E_ALL ^ E_NOTICE);


// Connect to the database server
$dbcnx = @mysql_connect('localhost', 'root', 'l2irx8am');
if (!$dbcnx) {
  die( '<p>Unable to connect to the ' .
       'database server at this time.</p>' );
}

// Select the content database
if (! @mysql_select_db('fibercoating') ) {
  die( '<p>Unable to locate the Fibercoating content ' .
       'database at this time.</p>' );
}

if (isset($_GET['history'])) {
   $historysql = @mysql_query('SELECT * FROM `history`');
   if (!$result) {
	   die('<p>Error performing query: ' .
		   mysql_error() . '</p>');
   }
} elseif (isset($_GET['whatisflock'])) {
   $whatisflocksql = @mysql_query('SELECT * FROM `whatisflock`');
   if (!$result) {
	   die('<p>Error performing query: ' .
		   mysql_error() . '</p>');
   }
} elseif (isset($_GET['index'])) {
   $indexsql = @mysql_query('SELECT * FROM `index`');
   if (!$result) {
	   die('<p>Error performing query: ' .
		   mysql_error() . '</p>');
   }
}

?>

But it does not... I click on a link and the page goes blank and just says "There was an error performing your query:"

It does not print what the error is or anything. I can run the query in phpmyadmin with out problems.

[laserlight]

Try to have verbose error messages when debugging, e.g.

<?php
error_reporting(E_ALL); //for debugging

// Connect to the database server
$dbcnx = mysql_connect('localhost', 'root', 'l2irx8am')
	OR die('<p>Unable to connect to the database server at this time.' . mysql_error() . '</p>');

// Select the content database
mysql_select_db('fibercoating')
	OR die('<p>Unable to locate the Fibercoating content database at this time: ' . mysql_error() . '</p>');

if (isset($_GET['history'])) {
	$historysql = mysql_query('SELECT * FROM `history`')
		OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} elseif (isset($_GET['whatisflock'])) {
	$whatisflocksql = mysql_query('SELECT * FROM `whatisflock`')
		OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} elseif (isset($_GET['index'])) {
	$indexsql = mysql_query('SELECT * FROM `index`')
		OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} else {
	echo '<p>No information provided</p>';
}
?>

[The_Squirrel]

Ok, thanks for the help. That fixed the problem I was having.

But now I am pretty much where I was at the start, I click a link and it changes the URL but it does not run the query.

error_reporting(E_ALL); //for debugging

// Connect to the database server
$dbcnx = mysql_connect('localhost', 'root', 'l2irx8am')
    OR die('<p>Unable to connect to the database server at this time.' . mysql_error() . '</p>');

// Select the content database
mysql_select_db('fibercoating')
    OR die('<p>Unable to locate the Fibercoating content database at this time: ' . mysql_error() . '</p>');

if (isset($_GET['history'])) {
    $historysql = mysql_query('SELECT * FROM `history`')
        OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} elseif (isset($_GET['whatisflock'])) {
    $whatisflocksql = mysql_query('SELECT * FROM `whatisflock`')
        OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} elseif (isset($_GET['index'])) {
    $indexsql = mysql_query('SELECT * FROM `index`')
        OR die('<p>Error performing query: ' . mysql_error() . '</p>');
} else {
    echo '<p>No information provided</p>';
}

That is my PHP script. Then on each link I just link to "xyz.php?history=true" and it should run the history query but it does not.

Just trying somethings I decided to echo one of the sql queries, so in my content area I put "echo $historysql;" when I first load my page I get an undefined variable notice, which is to be expected. So I press on the History link and it echos "Resource ID 3" and that is it. How can I get it to echo the information that is stored in the history table? I echoed the other links in the same way and I get the same results.

Any help is appreciated!
Thanks,

[laserlight]

You should use something along the lines of [man]mysql_fetch_array/man.

[The_Squirrel]

I feel like a dumby on that one. :rolleyes: I knew about mysql_fetch_array() but it totally did not even register for this.

So I tried using this code where I want it to display:

    while ( $row = mysql_fetch_array($indexsql) ) {
      $itemid = $row['ID'];
      $itemtext = $row['text'];
	}

But now it does not return anything. I am trying some variations to get it to work but having no success.
🙁

[The_Squirrel]

Greetings again,

I just thougtht I would update you all to the progress being made in my endeavors. I almost have this working 100% but I still need a little bit of help.

I redid my entire code. At the beginning of my page I have this code:

error_reporting(E_ALL); // for debugging


$page = 1; //this is the first section

$dbhost = 'xx';
$dbuser = 'xx';
$dbpass = 'xx';
$dbname = 'xx';

Then my code has the HTML part of the page and everything for my menu and so forth.

In my content area I have this code:

$db = mysql_connect("$dbhost","$dbuser","$dbpass");
	mysql_select_db($dbname,$db);
if ($page == 1)
{
	echo 'Main index<br />';
	//displays the title on top of page

$requete = "SELECT * FROM Main WHERE Section='index'";
}
else if ($page == 2) {
	echo 'History<br />';
	//displays the title on top of page

$requete = "SELECT * FROM Main WHERE Section='history'";
}
else if ($page == 3) {
	echo 'What is Flock?<br />';
	//displays the title on top of page

$requete = "SELECT * FROM Main WHERE Section='what is flock'";
}

$result = mysql_query ($requete,$db);

$row = mysql_fetch_array($result);

echo ($row['phpfile']);

Currently it works fine if I manually set the $page variable to 1, 2, or 3 at the top of the page in the first section of code I posted. If I change that to 2 or 3 it will change what is displayed.

But what I am trying to accomplish now is getting that to change when I click on a link. My links right now are in the format of: "template3.php?page=1"

How do I get PHP to change the $page variable from 1 to 2 or 3 by clicking on the link?

Thanks for any help provided.

[The_Squirrel]

Got it fixed. Someone else helped me from another forum.

The solution is changing my $page variable to be something like this

$page = isset($_REQUEST['page']) && ctype_digit($_REQUEST['page']) ? $_REQUEST['page'] : 1;

Thanks for trying anyways.