php - Mutiple file_put_content not working except the last one

I have a file {-code-1} looks like the below:

$first_data = 0;
$second_data = 0;
$third_data = 0;

I have three HTML checkbox on frontend and I would like to store the value into {-code-1}.

( The value is 1 if it's checked )

In process.php, I use $_POST['data_value'] and store the value into the variables which looks like the below:

$process_first_data = isset ( $_POST['first_checkbox'] ) ? 1 : 0;
$process_second_data = isset ( $_POST['second_checkbox'] ) ? 1 : 0;
$process_third_data = isset ( $_POST['third_checkbox'] ) ? 1 : 0;

And I use this method with preg_replace to replace the value in {-code-1} from process.php which looks like the below:

$find_first_data = '/\$first_data = \d;/';
$find_second_data = '/\$second_data = \d;/';
$find_third_data = '/\$third_data = \d;/';

$replace_first_data = '$first_data = ' . $process_first_data . ';';
$replace_second_data = '$second_data = ' . $process_second_data . ';';
$replace_third_data = '$third_data = ' . $process_third_data . ';';

$dir = 'path/to/file';
$file_content = file_get_contents ( $dir );

file_put_contents ( $dir, preg_replace ( $find_first_data, $replace_first_data, $file_content ) );
file_put_contents ( $dir, preg_replace ( $find_second_data, $replace_second_data, $file_content ) );
file_put_contents ( $dir, preg_replace ( $find_third_data, $replace_third_data, $file_content ) );

But now the problem is only the last file_put_contents is working. For example, now only the third one is working. And if I remove the third one, there are two left, then the second one is working only. And if I remove the second and the third one, there is one left, then the first one is working only.

There is no error, sorry for the long story because I want to make it in detail. May I know why only the last file_put_contents is working in this case?

Solution:

It looks like all three file_put_contents() are working, but you are overwriting the same file "$dir" three times.

file_put_contents($file, $contents, $flags);

An example of $flags is FILE_APPEND which will append to the end of the file instead of overwriting.

Solution:

One way to understand code better is to introduce some meaningful names for intermediate variables. So we could change this:

$dir = 'path/to/file';
$file_content = file_get_contents ( $dir );

file_put_contents ( $dir, preg_replace ( $find_first_data, $replace_first_data, $file_content ) );
file_put_contents ( $dir, preg_replace ( $find_second_data, $replace_second_data, $file_content ) );
file_put_contents ( $dir, preg_replace ( $find_third_data, $replace_third_data, $file_content ) );

To this:

$dir = 'path/to/file';
$original_file_content = file_get_contents ( $dir );

$first_new_content = preg_replace ( $find_first_data, $replace_first_data, $original_file_content );
file_put_contents ( $dir, $first_new_content );

$second_new_content = preg_replace($find_second_data, $replace_second_data, $original_file_content );
file_put_contents ( $dir, $second_new_content);

$third_new_content = preg_replace($find_third_data, $replace_third_data, $original_file_content );
file_put_contents ( $dir, $third_new_content);

Now, notice that the definitions for the three new variables don't reference each other, or re-read the file, they just look at $original_file_content. They could actually happen in any order, before the set of file_put_contents calls:

$dir = 'path/to/file';
$original_file_content = file_get_contents ( $dir );

$third_new_content = preg_replace($find_third_data, $replace_third_data, $original_file_content );
$second_new_content = preg_replace($find_second_data, $replace_second_data, $original_file_content );
$first_new_content = preg_replace ( $find_first_data, $replace_first_data, $original_file_content );

file_put_contents ( $dir, $first_new_content);
file_put_contents ( $dir, $second_new_content);
file_put_contents ( $dir, $third_new_content);

Since file_put_contents over-writes the whole file by default, the content of the file will always be $third_new_content.

What you probably intended is for each of the replacements to happen on the result of the previous one, then write the final result to the file at the end:

$dir = 'path/to/file';
$original_file_content = file_get_contents ( $dir );

$first_new_content = preg_replace ( $find_first_data, $replace_first_data, $original_file_content );
$second_new_content = preg_replace($find_second_data, $replace_second_data, $first_new_content );
$third_new_content = preg_replace($find_third_data, $replace_third_data, $second_new_content );

file_put_contents ( $dir, $third_new_content);

Solution:

Use FILE_APPEND option

e.g.

file_put_contents ( $dir, preg_replace ( $find_third_data, $replace_third_data, $file_content ), FILE_APPEND );