PHP for each loop

Question

This is what I'm trying to achieve. If All is selected from the drop down menu then I select all clients from my database. For each client it will then run the script creating the html page and then the pdf. This is where I have got to but I cant get it to generate a html page and pdf for each client.

<?php
$client_id=$_POST["client_id"];
$date_start=$_POST["date_start"];
$date_end=$_POST["date_end"];

if ($client_id == 'ALL') 
{
  $con = mysql_connect("localhost","user","password");
  if (!$con)
  {
    die('Could not connect: ' . mysql_error());
  }

  mysql_select_db("mydatabase", $con);  

  $query = "select client_id from ca_client_account";

  $result = mysql_query($query) or die(mysql_error());

  while($row = mysql_fetch_array($result))
  {
    $command="php $result.php $result $date_start $date_end > $result.html";
    exec($command, $output, $status);
    echo $command;
    if ($status!=0) {print_r($output); die("wget failed with status $status"); }

    $command="wkhtmltopdf-i386 --margin-left 5mm --margin-right 5mm $result.html $result.pdf";
    exec($command, $output, $status);
    if ($status!=0) die("htmltopdf failed");
  }

}
else
{
  $command="php $client.php $client_id $date_start $date_end > $client.html";
  exec($command, $output, $status);
  if ($status!=0) {print_r($output); die("wget failed with status $status"); }

  $command="wkhtmltopdf-i386 --margin-left 5mm --margin-right 5mm $client.html $client.pdf";
  exec($command, $output, $status);
  if ($status!=0) die("htmltopdf failed");
}
?>

For a single client everything is perfect. When I try and generate all client statements I cant get it to work.

What am i doing wrong?

Many thanks

the $result variable does not output each client_id as intended. It outputs resource id 3. — Sam Corbet
– Sam Corbet, Commented Feb 23, 2012 at 2:40
curious, why do you exec a php file instead of just includeing the file and running the functions like any other? Seems like less hassle to me. — jb.
– jb., Commented Feb 23, 2012 at 2:48

Kai Qing · Accepted Answer · 2012-02-23 02:41:57Z

2

$result will not deliver what you expect here. You need to specify the exact field you intend to echo out. In this case $row['client_id'] would work.

answered Feb 23, 2012 at 2:41

Kai Qing

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3 Comments

Sam Corbet Over a year ago

Thank you, its almost there but I get Undefined offset:1 now. I used $row[1] instead of $row['client_id'] - im sure this isnt the cause tho

Sam Corbet Over a year ago

@TheManiac you are right. I did not mean to change them they are a static file myfile.php - I changed them without thinking when copying in here to make the post generic

The Maniac Over a year ago

Great -- one more thing: if you want to use $row['client_id'] instead of $row[1] (the former being much more descriptive and a better practice in general) replace your call to mysql_fetch_array with mysql_fetch_assoc

Marc B · Accepted Answer · 2012-02-23 02:41:16Z

You have a fundamental misunderstanding of how database calls work. In your $command, you use $result, which is a database requery RESULT HANDLE. It is not a value from the query, it not useable by anything OTHER than mysql_*() functions. A standard query sequence looks something like this:

$sql = "SELECT ...";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);

echo $row['some_field'];

$result contains a statement handle representing your query results. You use this statement handle to FETCH a row of data. Within that $row of data you've fetched, will be the individual data fields you specified in your query. It is THOSE bits of data which you would pass on to your external script.

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