Read PHP Files From A Directory And Push A Common Variable In Each File To An Array

Question

I want to write a function which reads all the files in a dir and pushes a common variable value in each file to an array.

The idea is kinda to have a wordpress like feature in a way... You add a php file to a plugins folder with certain characteristics. For example every file you add must have a $fileName variable. My goal here is to grab each of the $fileName from each file in the dir and push them to an array so I can call on the array to create a navigation. The navigation will then load the php file into a content area when the link is activated with ajax.

My file path is,

/plugins/reports.php
/plugins/RandomPlugin2.php
/plugins/RandomPlugin3.php

I was trying to get this done doing something like this,

in /assets/load-plugins.php

function loadPlugins(){
$files = scandir('../plugins/');
foreach($files as $file) {
 if(($file_handle = fopen($file, "r")))
while (!feof($file_handle)) {
   $line = fgets($file_handle);
   echo $fileName;
}
fclose($file_handle);
}
}

loadPlugins();

But this is the error I get,

Warning: fopen(reports.php) [function.fopen]: failed to open stream: No such file or directory in /Applications/AMPPS/www/wubase/assets/load-plugins.php on line 12

Warning: fclose() expects parameter 1 to be resource, boolean given in /Applications/AMPPS/www/wubase/assets/load-plugins.php on line 17

It tells me there is no such file or directory but it even mentions the file currently in the plugins directory. Could this be a permission problem because I am trying to open a file from a different directory?

Also if someone has a better idea to achieve my goal I am all ears and would appreciate the constructive criticism.

Thanks,

Outsite the loop null, inside array(3) { [0]=> string(1) "." [1]=> string(2) ".." [2]=> string(11) "reports.php" } array(3) { [0]=> string(1) "." [1]=> string(2) ".." [2]=> string(11) "reports.php" } — wuno, Feb 06 '16 at 06:52
Try switching the `scandir()` function to the [`glob()`](http://php.net/manual/en/function.glob.php) function. `scandir()` only returns *filename*, not the full path. — D4V1D, Feb 06 '16 at 06:53
May be this will helpful:- http://stackoverflow.com/questions/2922954/getting-the-names-of-all-files-in-a-directory-with-php — Alive to die - Anant, Feb 06 '16 at 06:53
@D4V1D would you mind answering the question with glob. I cant get a working function and it seems like your comment is the closes to a reasonable answer. — wuno, Feb 06 '16 at 07:20

score 2 · Accepted Answer · answered Feb 06 '16 at 07:25

2

Try this:

function loadPlugins() {
    $files = glob('../plugins/');

    foreach($files as $file) {
        if(($file_handle = fopen($file, "r"))) {
            while (!feof($file_handle)) {
                $line = fgets($file_handle);
                echo $line;
            }
        }
        fclose($file_handle);
    }
}

Switch the function to glob() instead of scandir() (the former returns the full Unix path when the latter returns only the filename).
Take the habit of always use curly brackets for your if() statements, even when they are optional.
$fileName was not set, I assumed you meant $line (which is set right above).

answered Feb 06 '16 at 07:25

D4V1D

5,805
3
30
65

so if I have a variable like this in the file $fileName = "reports"; can I just do echo $fileName; inside the loop to call it and then do what I want with the variable? – wuno Feb 06 '16 at 07:32
I don't exactly get what is your prupose with `$fileName`. What does it suppose to contain? And where do you declare this variable? – D4V1D Feb 06 '16 at 07:34
in each file in the plugins folder will be a variable called $filename i want to read thane variable from each file and save it to an arrray to make a navigation with that variable being the link name and then the full path of the file being the href. So if you wanted to add a new function or feature to my app you could just write the php file with a few key parameters and then my app will properly display the php file in the content area of the ajax application i am making allowing me to easily add new functions and have them load dynamically. – wuno Feb 06 '16 at 07:37
So you want to execute PHP code that will be stored as a string in a variable. [`eval()`](http://php.net/manual/en/function.eval.php) can help you but as stated in the most up-voted comment in this manual section: *"If eval() is the answer, you're almost certainly asking the wrong question." -- Rasmus Lerdorf, BDFL of PHP*. Meanwhile, you should mark your question resolved if my piece of code helped you fetch the files in the `../plugins/` directory as asked. Feel free to ask another question regarding this new aspect of your development :). – D4V1D Feb 06 '16 at 07:40
sure thanks bro. Have a good night. now your scaring me that I am doing this a bad way... – wuno Feb 06 '16 at 07:41
You most certainly are. You may revisit the conception of your functionality. – D4V1D Feb 06 '16 at 07:50
Thank you very much. You made me rethink the whole thing. I appreciate you. – wuno Feb 06 '16 at 08:13

Armen · Answer 2 · 2016-02-06T07:00:41.913

0

Seems you have problem with file path, you can define and use this function to read your directory all files with ab

function dirToArray($dir) {

  $result = array();      
  $cdir = scandir($dir);

  foreach ($cdir as $key => $value){
    if (!in_array($value,array(".",".."))){
        if (is_dir($dir . DIRECTORY_SEPARATOR . $value)){
            $result = array_merge(
                   $result, 
                   dirToArray($dir . DIRECTORY_SEPARATOR . $value) 
            );
        }else{
            $result[] = $dir . DIRECTORY_SEPARATOR . $value;
        }
    }
  }

   return $result;
}

it will give you all files with absolute path located in that mentioned directory or subsirectories

edited Feb 06 '16 at 07:00

answered Feb 06 '16 at 06:52

Armen

4,064
2
23
40

Yes but does this read the file names to a directory? I am trying to read a variable which will hold the desired link name for the file in each file which will be names $fileName – wuno Feb 06 '16 at 06:53
yes it reaturn full path with file name /var/www/project/file.php, you can take only file name for this via basename("/var/www/project/file.php") it will give you back file.php, for experement you can run var_dump( dirToArray($path) ) and see the result – Armen Feb 06 '16 at 06:55

score 0 · Answer 3 · edited May 23 '17 at 11:59

scandir returns just name of file or directory(not full path). So, for example

.
├── 1.txt
├── 2.txt
├── 3.txt
└── public
    └── script.php

Your script in public folder, required files in ../ folder. scandir('../') will return

array (
  0 => '.',
  1 => '..',
  2 => '1.txt',
  3 => '2.txt',
  4 => '3.txt',
  5 => 'public',
);

So you need function, that returns full path to files, or make this path by yourself.

UPD: as D4V1D mentioned, glob() will solve your problem.

Read PHP Files From A Directory And Push A Common Variable In Each File To An Array

3 Answers3