[RESOLVED] displaying adresses Using Mysql , javacsript n php

Thikho

Hi All

I just bought a new book PHP5 in easy step but i have managed to connect to my database.

I have a database which is holding branches adresses it uses a drop-down to select one adress and diplay its adress name so im battling with writing a query that will select one adress from the db and show the details pls help

How do i select one adress and show the details of each branch using php a for loop

The javascript file work well

this is my contact.php

<html>
<head><TITLE>Main Centre Adresses</TITLE></head>

<body>

<h2>

<?php

$conn = @mysql_connect ("localhost", "thikho", "password")  or die ("Sorry - unable to connect to Mysql");
$rs = @mysql_select_db("MainIntranet", $conn ) or die ("Err:Db");

$sql  = "SELECT * FROM Maincentre_adress WHERE `id` = id ORDER BY `id` ASC";
$rs = mysql_query($sql,$conn);
while($row=mysql_fetch_array($rs))

{

?>

<?php

}

?>

<table width="142" border="0" cellspacing="0" cellpadding="0" height="450">
<form method="post" action="">
<input type="hidden" name="currSel" value="">
<tr>
<td rowspan="2" valign="top"></td>
<td rowspan="2" valign="top" ></td>
<td valign="top" height="60" width="138"><br>
 
<select name="select2" onChange="DoOnChange();">
<option selected>Select Centre</option>
<option value="blm">Bloemfontein</option>
<option value="ct">Cape Town</option>
<option value="dur">Durban</option>
<option value="el">East London</option>
<option value="geo">George</option>
<option value="jhb">Johannesburg</option>
<option value="nel">Nelspruit</option>
<option value="pta">Pretoria</option>
<option value="pe">Port Elizabeth</option>
<option value="ptg">Polokwane</option>
<option value="val">Vaal</option>

</select>
</td>
</tr>
<tr>
<td valign="top" height="380"><br>
</td>
</tr>
</form>
</table>

</h2>

</body>
</html>

hoffmeister

what error are you getting?
if you are just getting a plank screen try putting

$conn = @mysql_connect ("localhost", "thikho", "password")

$host="localhost";
$user="thikho";
$pass="password";
$conn = mysql_connect ($host, $user, $pass) or die("Connect Error: ".mysql_error());

I had this problem but that was when i devoloped on PHP5, then put it on a PHP 4.3 server.

also what is

&nbsp;

doing????
try using notpad++ it has sytax highlighting

Thikho

Tx Hoffmeister

I can connect to the db and i can view my branches , but i want to have sql query that will be able to choose one branch and show the details
I have 14 branches and i want to select one branch from the drop down and it displays the branch info, hope im clear

Thikho

Tx Hoffmeister

hoffmeister

is the branch's in a seprate table? if so is what you want to do is refresh the page with the info for the brance displayed at the bottom?

Thikho

there is only one table

mysql> describe Maincentre_adress;
+------------+-------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+-------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | | |
| adresline1 | varchar(50) | NO | | | |
| adresline2 | varchar(50) | NO | | | |
| adresline3 | varchar(50) | NO | | | |
| tel | varchar(15) | NO | | | |
| fax | varchar(15) | NO | | | |
+------------+-------------+------+-----+---------+----------------+
7 rows in set (0.03 sec)

hoffmeister

is the id in your query the branch name?if so take a look at this
http://www.w3schools.com/PHP/php_ajax_database.asp

arty56

Why don't you use the search function built in phpmyadmin and use the query it writes for you.

It's amazing the different types of queries you can generate by using phpmyadmin and it also is a great way to learn to code your own queries.

hoffmeister

<select name="select2" onChange="DoOnChange();">

change to

<select name="ID" onChange="DoOnChange();">

and

<form method="post" action="">

<form method="post" action="contact.php">

then at the top of the page

if(isset($_POST["id"])) $id=$_POST;
$conn = @mysql_connect ("localhost", "thikho", "password") or die ("Sorry - unable to connect to Mysql");
$rs = @mysql_select_db("MainIntranet", $conn ) or die ("Errb");

$sql = "SELECT * FROM Maincentre_adress WHERE `id` = \"$id\" ORDER BY `id` ASC";

Thikho

Tnks a lot for the good example this is what i wanted 😉