Creating a function that contains a while loop - php

Question

I am new to PHP and trying to learn it by creating a database of all my jobs (I'm a freelance designer). I have created the code below to generate a table to display all jobs where the description contains logo which works fine and generates several rows....
<?php
$logojobs = mysql_query("SELECT * FROM hlgd_projects WHERE description LIKE '%logo%'", $connection);
?>
View logo jobs
<?php
if ($report == 'logo') {
echo "<h1>Logo jobs</h1>";
echo "<table id='report'>
<tr>
<th></th>
<th>Status</th>
<th>Lead</th>
<th>Start</th>
<th colspan='2'>Codes</th>
<th>Client</th>
<th>Description</th>
<th>Fee</th>
<th>Contact</th>
<th colspan='2'>Invoice</th>
<th>Paid</th>
</tr>";
while ($logojob = mysql_fetch_array($logojobs)) {
echo "<tr>
<td class='id'>" . $logojob['id'] . "</td>
<td>" . $logojob['status_id'] . "</td>
<td>" . $logojob['lead_id'] . "</td>
<td>" . $logojob['date_start'] . "</td>
<td>" . $logojob['code_lead'] . "</td>
<td>" . $logojob['code_hlgd'] . "</td>
<td>" . $logojob['client'] . "</td>
<td>" . $logojob['description'] . "</td>
<td>£" . $logojob['fee'] . "</td>
<td>" . $logojob['contact'] . "</td>
<td>" . $logojob['invoice'] . "</td>
<td>" . $logojob['date_inv'] . "</td>
<td>" . $logojob['date_paid'] . "</td>
</tr>";
}
echo "</table>";
}
?>
However I would like to be able to generate reports for lots of different things and so would like to create a function to generate the table and pass the relevant arguments each time. I've done this as follows but it only generates the first row so I presume it's ignoring the while? Can anyone tell me where I'm going wrong or how to better put the code above into a function?
<?php
function report_bytype($title,$job_set,$job_name) {
$reporthead = "<h1>$title</h1>
<table id='report'>
<tr>
<th></th>
<th>Status</th>
<th>Lead</th>
<th>Start</th>
<th colspan='2'>Codes</th>
<th>Client</th>
<th>Description</th>
<th>Fee</th>
<th>Contact</th>
<th colspan='2'>Invoice</th>
<th>Paid</th>
</tr>";
while ($job_name = mysql_fetch_array($job_set)) {
$reportrows = "
<tr>
<td class='id'>" . $job_name['id'] . "</td>
<td>" . $job_name['status_id'] . "</td>
<td>" . $job_name['lead_id'] . "</td>
<td>" . $job_name['date_start'] . "</td>
<td>" . $job_name['code_lead'] . "</td>
<td>" . $job_name['code_hlgd'] . "</td>
<td>" . $job_name['client'] . "</td>
<td>" . $job_name['description'] . "</td>
<td>£" . $job_name['fee'] . "</td>
<td>" . $job_name['contact'] . "</td>
<td>" . $job_name['invoice'] . "</td>
<td>" . $job_name['date_inv'] . "</td>
<td>" . $job_name['date_paid'] . "</td>
</tr>";
}
$reportfoot = "</table>";
$reporttable = $reporthead . $reportrows . $reportfoot;
echo $reporttable;
return $reporttable;
}
?>
<?php
if($report == 'logo') {
report_bytype("logo",$logojobs,$logojob);
}
if($report == 'stationery') {
report_bytype("stationery",$stationeryjobs,$stationeryjob);
}
?>
Many thanks in advance,
Helen

Answer 1

The problem is this line:
while ($job_name = mysql_fetch_array($job_set)) {
$reportrows = "lots of html"
}
Every time this loops $reportrows is set to the html for that row only.
Use $reportrows .= "some html"; instead which will add each row to $reportrows rather than replace $reportrows with that row.
Edit: Replacing += with .=.

Answer 2

I can see why you have errors you are not concatenating your result
Please see http://php.net/manual/en/language.operators.string.php for more detailed explanations
Replace
$reportrows = "
With
$reportrows .= "
Full Script
function report_bytype($title, $job_set, $job_name) {
$reporthead = "<h1>$title</h1>
<table id='report'>
<tr>
<th></th>
<th>Status</th>
<th>Lead</th>
<th>Start</th>
<th colspan='2'>Codes</th>
<th>Client</th>
<th>Description</th>
<th>Fee</th>
<th>Contact</th>
<th colspan='2'>Invoice</th>
<th>Paid</th>
</tr>";
$reportrows = "";
while ( $job_name = mysql_fetch_array ( $job_set ) ) {
$reportrows .= "
<tr>
<td class='id'>" . $job_name ['id'] . "</td>
<td>" . $job_name ['status_id'] . "</td>
<td>" . $job_name ['lead_id'] . "</td>
<td>" . $job_name ['date_start'] . "</td>
<td>" . $job_name ['code_lead'] . "</td>
<td>" . $job_name ['code_hlgd'] . "</td>
<td>" . $job_name ['client'] . "</td>
<td>" . $job_name ['description'] . "</td>
<td>£" . $job_name ['fee'] . "</td>
<td>" . $job_name ['contact'] . "</td>
<td>" . $job_name ['invoice'] . "</td>
<td>" . $job_name ['date_inv'] . "</td>
<td>" . $job_name ['date_paid'] . "</td>
</tr>";
}
$reportfoot = "</table>";
$reporttable = $reporthead . $reportrows . $reportfoot;
echo $reporttable;
return $reporttable;
}
Thanks