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?

delayedinsanity · 08-02-2008, 04:31 AM

PHP Code:


			
$data['options']['name'] = 'some array name here';
$data['options']['array'] = array(0 => "opt1", 
                                  1 => "opt2", 
                                  2 => "opt3");

sjaq · 08-02-2008, 08:56 AM

PHP Code:


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

or something like that

xenon · 08-02-2008, 09:32 AM

or, you could simply put the $data['options'] variable into the single quotes...I guess...assuming that you're evaluating the code after.

Kalle · 08-02-2008, 08:58 PM

Use serialization insted, much easier, but abit slower:

PHP Code:


			
$fp[] = "foreach (@unserialize('" . str_replace('\'', '\\\', serialize($data['options'])) . "') as $option) {

Or something similar
=)

benton · 08-05-2008, 02:54 AM

I appreciate all of the suggestions though I wasn't able to get any to work.

1

Quote:

$data['options']['array'] = array(0 => "opt1",
1 => "opt2",
2 => "opt3");

I'm probably missing the point but if I can't get the code to run through a regular array, how will changing it to a multiarray help?

2

Quote:

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

Changed to the following to get it to not fail:
$fp[] = "foreach (array('" . implode(',\'', $data['options']) . "') as \$option) {" . "\n";

This almost worked. The output was

PHP Code:


			
 foreach (array('opt1,'opt2') as $option) {

I only used two options to test the above. Is there a way to surround each member of the array in single quotes since that seems to be the problem with it?

3

Quote:

put the $data['options'] variable into the single quotes

I'm not sure what was meant by this but if you mean something like

PHP Code:


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

If I do that, the actual name shows in the output as

PHP Code:


			
foreach ($data['options'] as $field ) {

But that won't work since that array is not known in the created file.

4

Quote:

$fp[] = "foreach (@unserialize('" . str_replace('\'', '\\\', serialize($data['options'])) . "') as $option) {

I couldn't get this to run due to parsing errors. I changed it to the following but it still fails.

PHP Code:


			
$fp[] = "foreach (@unserialize(str_replace('\'', '\\', serialize(" . $data['options'] . ")) . ") as $option) {" . "\n";

I would apreciate any other suggestions.