PHP Array

[greens85]

Hi All,

 

Im relativley new to both PHP & mySQL, I want to drag out all the data in my database and have it displayed in an array:

 

Lets say for example it the data was jobs,

 

I want to display all the jobs in a list ordered by the job id and where a certain field contains the data 'yes', therefore if a certain row of the data doesnt contain the data yes the job isn't displayed.

 

I currently have the following code:

 

<?php
$connection = mysql_connect("localhost", "user", "pass") or die (mysql_error());
$db = mysql_select_db("mydb", $connection) or die (mysql_error());
?>

<?php
$query = mysql_query ("SELECT * FROM jobs WHERE CS = 'yes' ORDER BY jobid");
	while ($data1 = mysql_fetch_array ($query)) {
		echo $data1 [country];
}
?>

 

But this just returns the country for every job, where as i want to return the data in a list and it also includes the data for rows that dont have CS set to yes.

 

Hope this makes sense!

 

Can anyone point me in the right direction?  ???

[trq]

The code you have provided should do exactly what you have described. maybe you should better describe what you mean by a 'list'?

[greens85]

Appologises perhaps i didnt explain myself very well.

 

What i mean is, there may be say 3 jobs in the db with the CS param. set to yes.

 

Each jobs then has different data, i.e. each is a different jobs and needs to be displayed as a seperate entity.

 

At the moment the only output im getting is:

 

Job Title: TEST JOB

 

Where as im trying to achieve something such as:

 

Job Title: Test Job

Job Desc: Description Here

Comany: Company ABC

Address: Address Here

 

---------------------------------

 

Job Title: Test Job 2

Job Desc: Description 2 Here

Comany: Company DEF

Address: Address 2 Here

 

And so on for each row that has the CS field set to 'yes'.

 

Hope this clears up what I mean and thanks for your quick reply!

[trq]

<?php

mysql_connect("localhost", "user", "pass") or die (mysql_error());
mysql_select_db("mydb", $connection) or die (mysql_error());

$sql = "SELECT country FROM jobs WHERE CS = 'yes' ORDER BY jobid";
if ($result = mysql_query($sql)) {
  if (mysql_num_rows($result)) {
    while ($row = mysql_fetch_array ($query)) {
      echo $row['country'];
      echo "------------<br />";
    }
  }
}

?>

 

Hope that helps as an example.

[greens85]

Hi Thorpe,

 

Ive now managed to get it to return this:

 

TEST JOB

This is a test job

 

0001

 

---------------------

 

Test Job

Test Job

 

1112223

 

Using this code:

 

<?php
//Run Query's To Select The Data
$query5 = mysql_query("SELECT * FROM jobs WHERE CS = 'CS' ORDER BY jobid ASC");
while ($data1 = mysql_fetch_array($query5)){
//Show The Data In The Browser
echo $data1['position'];
echo '<br/>';
echo $data1['description'];
echo '<br/>';
echo '<br/>';
echo $data1['jobref'];
echo '<br/>';
echo $data['hour'];
echo '<br/>';
echo $data['subcounty'];
echo '<br/>';
echo $data['contract'];
echo '<br/>';
echo $data['salary'];
echo '<br/>';
echo $data['deadline'];
}
?>

 

But it stops displaying after jobref, when i want it to display all the fields under that!

 

Do you have any idea why this might be occuring?

 

If I could get it to display all the fields i specify in the code, the problem would be solved 

[trq]

After $data1['jobref'] you start using an undefined $data array, you should still be using $data1.

 

You should always have this at the top of your code when developing, it will point out allot of simple errors.

 

<?php error_reporting(E_ALL) ; ini_set('display_errors','1'); ?>