Filling an array with an array

benton · 08-02-2008, 03:13 AM

I'm writing a script where the user can input certain items, which are then converted to php code and written to a file. A cron job will then be used to run that file. For the most part, it is all working fine but I am having problems get an array written correctly. Here is an example:

In the main code that displays for the user:

PHP Code:


			
//collect data from form

$data = array();

$data['name'] = 'some name';

$data['email'] = 'email address';

$data['options'] = array(0 => "opt1",

                         1 => "opt2",

                         2 => "opt3");

$fp = array();

PrepareData($fp, $data);

WriteFile($fp);

For the PrepareFile function

PHP Code:


			
PrepareFile(&$fp, $data) {

$fp[] = "\$name = " . $data['name'] . "; \n";

$fp[] = "\$email = " . $data['email'] . "; \n";



$fp[] = "\$db_query = mysql_query(\"select * from table\") or die(mysql_error());" . "\n";

$fp[] = "while(\$db = mysql_fetch_array(\$db_query, MYSQL_ASSOC)) {" . "\n";

$fp[] = "foreach (" . $data['options'] . " as $option) {" . "\n";

$fp[] = "\$message .= \$db[\$option] .'</tr></td>';" . "\n";

}

}

The problem is with the line

PHP Code:


			
$fp[] = "foreach (" . $data['options'] . " as $option) {

When the written file is displayed, for that line it shows

PHP Code:


			
foreach (Array as $option) {

The rest of the code is converted correctly. If I remove the foreach loop, the created file can be ran and the output is created as expected, using the complete code, of course. Can anyone see a problem with how I am doing this?