[SOLVED] Dynamic links

[porta325]

Hey, i'm sure it looks childish for you but i'm in the beginning and i couldn't find something i could understand on this topic anywhere.The ideea is that i want to make some kind of directory. I have the code that lists entries from a database but i don't know how to make all those entries links that point to other pages.

Basicaly, i need every entry(category) to point to a new page that will contain data about that category.

 

<?php

//connect

mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("directory") or die(mysql_error());

 

//select table

$result = mysql_query ("SELECT * FROM countries");

 

//get entries

$row = mysql_fetch_array ($result);

 

//print result

while ($row = mysql_fetch_array ($result)){

echo $row ['Country']." <br>";

}

?>

 

 

This is where i am and where i am stuck.All this part works just fine.I have an Id assigned to every entry so if you can help me move on i would apreciate. I just need to know what other files i need to create and how the href tag should look like.

[chigley]

<?php
//connect
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("directory") or die(mysql_error());

//select table
$result = mysql_query ("SELECT * FROM countries");

//get entries
$row = mysql_fetch_array ($result);

//print result
while ($row = mysql_fetch_array ($result)){
echo "<a href=\"showlinks.php?id={$row["id"]}\">{$row["name"]}</a><br />";
";
}
?>

 

Assuming you have a "name" field. Then to show them in showlinks.php, use $_GET["id"] in your query

[porta325]

Assuming you have a "name" field. Then to show them in showlinks.php, use $_GET["id"] in your query Smiley

 

The name field is "Country". I'm not sure though how to use $_GET["id"] in my query and i'm not sure what showlinks.php should contain.

[chigley]

showlinks.php demo:

 

<?php

if(is_numeric($_GET["id"]) {
$id = $_GET["id"];
}

if(!isset($id)) {
die("Invalid ID");
}

$query = mysql_query("SELECT * FROM table WHERE categoryid = '{$id}'") or die(mysql_error());

while($row = mysql_fetch_assoc($query)) {
// Output...
}

?>

[porta325]

After 8 hours diggin it's hard to understand. I would apreciate the simplest working model you can provide me with the simplest database you can immagine.

[chigley]

You have it all there for you! All you need to do is shove a few echos in the while() loop and change the table names in the query and Bob's your uncle!

[porta325]

<?php

 

//connect

mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("directory") or die(mysql_error());

 

if(is_numeric($_GET["id"]) {

$id = $_GET["id"];

}

 

if(!isset($id)) {

die("Invalid ID");

}

 

$query = mysql_query("SELECT * FROM countries WHERE Country = '{$id}'") or die(mysql_error());

 

while($row = mysql_fetch_assoc($query)) {

// Output...

}

 

?>

 

Index works great, i still have a problem with

if(is_numeric($_GET["id"]) {

$id = $_GET["id"];

}

 

if(!isset($id)) {

die("Invalid ID");

}

I get an error and i don't know why, i'm not familiar with that function.

[chigley]

Which error do you get?

[porta325]

Parse error: syntax error, unexpected '{' in C:\wamp\www\showlinks.php on line 13

[porta325]

Index.php

<?php

//connect

mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("directory") or die(mysql_error());

 

//select table

$result = mysql_query ("SELECT * FROM countries");

 

//get entries

$row = mysql_fetch_array ($result);

 

//print result

while ($row = mysql_fetch_array ($result)){

$id=$row["Id"];

$status=$row["Status"];

$country=$row["Country"];

 

echo "<a href=\"showlinks.php?id={$id}\">{$country}</a><br />";

 

}

?>

 

 

Showlinks.php

 

 

<?php

 

//connect

mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("directory") or die(mysql_error());

 

 

if(!isset($id)) {

die("Invalid ID");

}

 

$id = $_GET["id"];

 

$query = mysql_query("SELECT * FROM links WHERE Id = '{$id}'") or die(mysql_error());

 

while($row = mysql_fetch_assoc($query)) {

// Output...

$id1=$row["id"];

echo "$id1";

}

 

?>

 

I'm not sure why it says INVALID ID, as i see it the id's from both tables are connected right. Any ideeas ?

[sasa]

Index.php

<?php
//connect
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("directory") or die(mysql_error());

//select table
$result = mysql_query ("SELECT * FROM countries");

//get entries
//$row = mysql_fetch_array ($result); <--- remove this line

//print result
while ($row = mysql_fetch_array ($result)){
$id=$row["Id"];
$status=$row["Status"];
$country=$row["Country"];

echo "<a href=\"showlinks.php?id={$id}\">{$country}[/url]
";

}
?>

Showlinks.php

<?php

//connect
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("directory") or die(mysql_error());
$id = $_GET['id']; // <---- add this line

if(!isset($id)) {
die("Invalid ID");
}

$id = $_GET["id"];

$query = mysql_query("SELECT * FROM links WHERE Id = '{$id}'") or die(mysql_error());

while($row = mysql_fetch_assoc($query)) {
// Output...
$id1=$row["id"];
echo "$id1";
}

?>

[porta325]

Thx alot, solved this one, now i'm fighting with another.I can't asociate a column from first table with one from the second.Meaning that every category in first column has an id associated, id that i have to associate with entries from the second table that corespond to the categories from the first table but witch are positioned random.