从PHP多维数组中检索值

Question

I have a multi-dimensional array that I want to print statistics from for a website admin. My problem is that once I get the data into an array in the format I want it, I can't figure out how to retrieve it. I have to know the key values in order to get it eg $uniqueBrowser["Firefox"]["14.0.1"]['times'];

I want to be able to say for example - Firefox 70%, IE 30% as a start (aggregate all version 'times used' for the browsers). I then want to drill down further to specific versions eg FF3.6 14%, FF14 51% etc. (simply printing out the times value associated with the version).

EDIT : My Question is, how do I retrieve my data in the way described above? My head is fried and I've been on this for too long. Please someone put me out of my misery. If there is a better way to do this, please let me know also.

The code I have so far is below. This code is run in a for each loop which iterates through data returned by the database. The $browser and $version variables are set to the browser values and version values returned from the database at each iteration of the foreach loop.

if(in_array($browser, $uniqueBrowser)) {        
    if(in_array($version, $uniqueBrowser[$browser])) {
        $uniqueBrowser[$browser][$version]['times'] = $uniqueBrowser[$browser][$version]['times'] + 1;
    } else {            
        $uniqueBrowser[$browser]['version'] = $version;
        $uniqueBrowser[$browser][$version]['times'] = 1;
    }       
} else {
    $uniqueBrowser[] = $browser;
    $uniqueBrowser[$browser]['version'] = $version;
    $uniqueBrowser[$browser][$version]['times'] = 1;
}

When I run this code, it gives me this (via var_dump):

Browser data: array(4) {
  [0]=>
  string(7) "Firefox"
  ["Firefox"]=>
  array(3) {
    ["version"]=>
    string(6) "14.0.1"
    [15]=>
    array(1) {
      ["times"]=>
      int(2)
    }
    ["14.0.1"]=>
    array(1) {
      ["times"]=>
      int(15)
    }
  }
  [1]=>
  string(17) "Internet Explorer"
  ["Internet Explorer"]=>
  array(2) {
    ["version"]=>
    string(9) "8.0.0.253"
    ["8.0.0.253"]=>
    array(1) {
      ["times"]=>
      int(1)
    }
  }
}

Answer 1

Instead of getting data out of the database, and put it in a complex array, you should let the database do the work for you, and retrieve the values from a query in the format that you want.

Let the database do the calculations, and do not use PHP to do that. It makes it over complex.

Answer 2

I know JvdBerg has already given an answer like this, but I added my comment first. So I don't feel like I'm copying him.

As an example I've created the following table structure. Hopefully it's like yours.

+---------+------------------+------+-----+---------+----------------+
| Field   | Type             | Null | Key | Default | Extra          |
+---------+------------------+------+-----+---------+----------------+
| id      | int(11) unsigned | NO   | PRI | NULL    | auto_increment |
| browser | varchar(120)     | YES  |     | NULL    |                |
| version | varchar(50)      | YES  |     | NULL    |                |
+---------+------------------+------+-----+---------+----------------+

Then I inserted the following rows, again, hopefully like yours.

+----+---------+---------+
| id | browser | version |
+----+---------+---------+
|  1 | Firefox | 3.6     |
|  2 | Firefox | 4.1     |
|  3 | Firefox | 3.6     |
|  4 | Safari  | 5.1.7   |
|  5 | Safari  | 6       |
|  6 | Firefox | 14.0.1  |
|  7 | IE      | 7       |
|  8 | IE      | 8       |
|  9 | IE      | 7       |
| 10 | Firefox | 14.0.1  |
| 11 | Opera   | 12.0.1  |
| 12 | Safari  | 5.1.7   |
+----+---------+---------+

Then the following query produced these results.

mysql> SELECT browser, version, COUNT(version) AS times FROM demo.browsers GROUP BY browser;
+---------+---------+-------+
| browser | version | times |
+---------+---------+-------+
| Firefox | 3.6     |     5 |
| IE      | 7       |     3 |
| Opera   | 12.0.1  |     1 |
| Safari  | 5.1.7   |     3 |
+---------+---------+-------+

If you want it to be more precise you can use GROUP BY browser, version instead of just by the browser.

mysql> SELECT browser, version, COUNT(version) AS times FROM demo.browsers GROUP BY browser, version;
+---------+---------+-------+
| browser | version | times |
+---------+---------+-------+
| Firefox | 14.0.1  |     2 |
| Firefox | 3.6     |     2 |
| Firefox | 4.1     |     1 |
| IE      | 7       |     2 |
| IE      | 8       |     1 |
| Opera   | 12.0.1  |     1 |
| Safari  | 5.1.7   |     2 |
| Safari  | 6       |     1 |
+---------+---------+-------+

Answer 3

I'm not sure I fully understand your question, but have you tried iterating over the array using an extended for loop?

eg:

foreach ($uniqueBrowser as $browserName=>$values){

    echo $browserName .' contains '. print_r($values,true);

}