php & mysql editing returned values

[Munch]

Hi guys,

Im new to php and have been working on a image that returns infomation from a mysql database,
In mysql i have a list of jobs which are numbered 111, 211, 311 etc
What i want to do is be able to convert those numbers into the names of the actual job and display the name in the image.

Note: Editing the numbers in the database is not a option

Thank you for any help that you can provide.
Chris

[thedarkwinter]

Hi

Am i missing something?

Is your table struc something like?

jobid  |  jobname    |    data
111    |  something  |    random
211    |  blahblah      |      data
311    |  etc              |      here

then just select jobname instead of jobid?

if it isn't that is probably what you should be doing...

cheers,
tdw

[HuggieBear]

I'm assuming that you have the names of the job in your database e.g.

[b]Jobs[/b]
[table]
[tr][td]ID[/td][td]Number[/td][td]Name[/td][/tr]
[tr][td]1[/td][td]111[/td][td]Clean Windows[/td][/tr]
[tr][td]2[/td][td]211[/td][td]Wash Clothes[/td][/tr]
[tr][td]3[/td][td]311[/td][td]Empty Bins[/td][/tr]
[/table]

If so then yes, it's quite simple.

Regards
Huggie

[Munch]

The only infomation that is in the database except personal infomation is the job numbers

In the DB..
Name| Job | Phone
john | 111 | +44
dave| 211 | +44
rich  | 311 | +44

Not in the DB
111 = Plumber
211 = Electrician
311 = Joiner

I want to be able to call the infomaton from the database by http://domain.com/job.php?name=John
then have the business card shown with name job title andphone number

I hope this clears things up :)
Chris

[HuggieBear]

OK, if you don't have the information in the database then you'll need something like an array with the data in.

[code]
<?php
$profession = array(111 => "Plumber", 211 => "Electrician", 311 => "Joiner");

$sql = "SELECT job FROM tablename";
$result = mysql_query($sql);
while ($roles = $mysql_fetch_array($result, MYSQL_ASSOC)){
  extract($roles);
  echo "{$job} => {$profession[$job]}";
}
?>
[/code]

Regards
Huggie

[Munch]

hi huggie,

how would i display that script in a imagecreatefrompng image please?

sorry to be a pain
chris

[HuggieBear]

You wouldn't... You just put this code around any image creation code and then use the variables with the image functions, something like this:

[code]<?php
$profession = array(111 => "Plumber", 211 => "Electrician", 311 => "Joiner");

$sql = "SELECT job FROM tablename";
$result = mysql_query($sql);
while ($roles = $mysql_fetch_array($result, MYSQL_ASSOC)){
  extract($roles);
  $text = $profession[$job];
  // All your image code goes here.  You should use imagettftext() to add the text to the image
  // and you'll want to provide the last argument, which is string as $text
}
?>[/code]

Regards
Huggie

[Munch]

This is what i have so far...

[b]staff.php[/b]
[code]

<?php

  error_reporting(0);
  include ("staff.inc.php");

  $name = $_GET['name'];
  if ($name == null || $name == '') {
    $name = "";
  }

$link = mysql_connect($MySQLSrv, $MySQLUser, $MySQLPass);
if (!$link) {
  die('Could not connect to MySQL! : ' . mysql_error());
}

$db_selected = mysql_select_db($MySQLDB, $link);
if (!$db_selected) {
  die ('Could not find the DB : ' . mysql_error());
}

if ($name != "") {
$result = mysql_query("SELECT name, job, phone FROM staff WHERE name = '{$name}'");
if (!result) {
  echo 'Could not run query: ' . mysql_error();
  exit;
}
$row = mysql_fetch_row($result);

  header("Content-type: image/png");
  $im    = imagecreatefrompng("$pngimage");
  $cBlack = imagecolorallocate($im, 0, 0, 0);

  if ($name != "") {
    imagestring($im, 5, 10, 10, 'Name: ' . $row[0], $cBlack);
    imagestring($im, 5, 10, 30, 'Job: ' . $row[1], $cBlack);
    imagestring($im, 5, 10, 50, 'Phone: ' . $row[2], $cBlack);
    imagepng($im);
    imagedestroy($im);
}
  }
?>[/code]

[b]staff.inc.php[/b]

[code]<?php

$pngimage = "staff.png";

$MySQLSrv="localhost";
$MySQLDB="test";
$MySQLUser="root";
$MySQLPass="";

?>[/code]

i was typin this out while you were posting.. ill ready ur post now :)

[HuggieBear]

OK, change this:

[code]
<?php
if ($name != "") {
  imagestring($im, 5, 10, 10, 'Name: ' . $row[0], $cBlack);
  imagestring($im, 5, 10, 30, 'Job: ' . $row[1], $cBlack);
  imagestring($im, 5, 10, 50, 'Phone: ' . $row[2], $cBlack);
  imagepng($im);
  imagedestroy($im);
}
?>
[/code]

To this:

[code]
<?php
if ($name != "") {
  imagestring($im, 5, 10, 10, 'Name: ' . $row[0], $cBlack);
  $profession = array(111 => "Plumber", 211 => "Electrician", 311 => "Joiner");
  $job_id = $row[1];
  $job = $profession[$job_id];
  imagestring($im, 5, 10, 30, 'Job: ' . $job, $cBlack);
  imagestring($im, 5, 10, 50, 'Phone: ' . $row[2], $cBlack);
  imagepng($im);
  imagedestroy($im);
}
?>
[/code]

Regards
Huggie

[Munch]

thank you very much :) its working how i wanted it to now :D