Hi

I am new to PHP & MySQL and have been working on learning how to set up a database. Well yesterday I set up a test database (for purposes of this discussion I named it Jokes as was the example in my book) so I named the database Jokes and then created a table called Jokes as well.

Oh before we get to far i am using PHPMyAdmin to create the database and tables.

OK so I created my table called Jokes and this table currently has 3 columns in it. That being

ID | Jokes Text | Jokes date

Now someone gave me a tip that I should connect the words with an underscore IE Jokes_Text Just so you know in case you spot this you'll know I am aware of this already. So anyways I created the table and all seems to be going fine.

I then asked how I add items and he said just to click on the insert tab in PHPMyAdmin so this is easy to do.

OK now I have items added to the database how can I call these up from my webpage.

Example lets say I have 3 jokes in the database.

I then have 3 links on my page to call up items in the database how do I call up a specific joke if the link is pressed IE

link 1 goes to Joke 1
Link 2 goes to joke 3
Link 3 goes to Joke 2

Not a typo listed numbers that way on purpose but how can I set it up to call up items when a link is pressed?

Also in a databse when I am inserting items can I have HTML code as the item? Like if I want to have a list of amazon.com books can I add HTML to the database instead of just plain text that way the links then call up the HTML and then this gets converted to the iframe window on my website when a user calls for an item from the database?

TIA

The way I would do the display of the jokes, is to write a page called say, 'display.php', then you create links in this format:

link 1 goes to Joke 1
Link 2 goes to joke 3
Link 3 goes to Joke 2

becomes

<a href="display.php?joke=1>Link 1</a>
<a href="display.php?joke=3>Link 2</a>
<a href="display.php?joke=2>Link 3</a>

Then in display.php, you pull out the joke and display it:

PHP Code:

// Make database connection

$jokeid = $_GET['joke'];

$result = mysql_query("SELECT 'Jokes Text' FROM Jokes WHERE ID=$jokeid");
$text = mysql_result($result,0,'Jokes Text');

echo $text; 


To use html, you can just put it into your database, although you must be careful with quotes. To make sure they behave properly use addslashes() on the string before inserting it into the db. Also be wary that if you are accepting user input (if you ever have a form where they can input their own jokes ofr example) then if you allow html you leave yourself wide open to all sorts of hijack attacks. Without being careful, people can insert javascript into your database that gets executed when the joke is displayed.

OK maybe I set up the database wrong but here is what I get with the code you suggested.

http://www.scottish2.com/test/test/test.html

Thoughts?

change it to

PHP Code:

<?php

// Make database connection     
$jokeid = $_GET['joke'];     

$result = mysql_query("SELECT 'Jokes Text' FROM Jokes WHERE ID=$jokeid");   
$text = mysql_result($result,0,'Jokes Text');     

echo $test; 
?>

hope that will work

Sorry. didnt look propperly. You still need a connection to the database.

replace the user and password and incase you host your mysql database on a different server than you php files then change your host too.

PHP Code:

<?php

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');

if (!$link) {
   die('Could not connect: ' . mysql_error());
}


$jokeid = $_GET['joke'];      

$result = mysql_query("SELECT 'Jokes Text' FROM Jokes WHERE ID=$jokeid");    
$text = mysql_result($result,0,'Jokes Text');      

echo $test

mysql_close($link);

?>

Not sure I edited that right Lets use this example as obviously am not going to give out my info but Lets use the following

User Name: Web
Password: Master

Obviously not the actual log in but how should I edit this as I did get an error when I tried. Below is what I did is this right?

<?php 

$link = mysql_connect('localhost', 'mysql_web', 'mysql_master'); 

if (!$link) { 
   die('Could not connect: ' . mysql_error()); 
} 


$jokeid = $_GET['joke'];       

$result = mysql_query("SELECT 'Jokes Text' FROM Jokes WHERE ID=$jokeid");     
$text = mysql_result($result,0,'Jokes Text');       

echo $test 

mysql_close($link); 

?>

no not quite

try it this way.

PHP Code:

<?php 

$link = mysql_connect('localhost', 'web', 'master'); 

if (!$link) { 
   die('Could not connect: ' . mysql_error()); 
} 


$jokeid = $_GET['joke'];       

$result = mysql_query("SELECT 'Jokes Text' FROM Jokes WHERE ID=$jokeid");     
$text = mysql_result($result,0,'Jokes Text');       

echo $test 

mysql_close($link); 

?>

if you go to http://www.php.net you can find almost everything what you need to know.

OK made the swap but am getting the following error

Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in /home/scottish/public_html/test/test/display.php on line 17

ok do this one

PHP Code:

$result = mysql_query("SELECT Jokes_Text FROM Jokes WHERE ID=$jokeid");      
$text = mysql_result($result,0,'Jokes_Text'); 


Same basic thing

Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in /home/scottish/public_html/test/test/display.php on line 18

Now quick question maybe this is database related but unsure. I noticed in my database list for my other SQL programs like chat and my forums that are set up automatically I notice each has a Perl and a PHP connection string listed yet for this new jokes database it does not have these lines. Is this maybe why? if so what have I missed?