PHP How to filter array values

Question

I'm trying to show in a div 3 thumbnails stored in different folders.
In my bd determined (with value "1") which record has pictures. So this is my db:

+------+----------------+---------+--------+--------+
| ID   | BRAND          |   PHP   |  RUBY  |  JAVA  |
+------+----------------+---------+--------+--------+
| 1    | ford           |    1    |    0   |    0   |
+------+----------------+---------+--------+--------+
| 2    | seat           |    1    |    1   |    1   |
+------+----------------+---------+--------+--------+
| 3    | fiat           |    1    |    1   |    0   |
+------+----------------+---------+--------+--------+
| 4    | toyota         |    1    |    0   |    0   |
+------+----------------+---------+--------+--------+
| 5    | vw             |    1    |    0   |    1   |
+------+----------------+---------+--------+--------+

Selecting records:

$result = mysqli_query($connecDB,"SELECT * FROM brands WHERE php = '1' OR ruby = '1' OR java = '1' ORDER BY id ASC");


while($row = mysqli_fetch_assoc($result)){
    $new_array[] = $row;
}

But my problem is: How I can select the values "1" and how to know which is which?
Maybe I could use: $row['php'] $row['java'] $row['ruby'] But how to filter the values "0"?

tip: simplify your query with WHERE 1 IN (PHP, RUBY, JAVA). — Marc B
– Marc B, Commented Aug 19, 2013 at 19:33
Just loop through each one with an if($row['ruby'])... what are you wanting to do if it is a one? — cwhelms
– cwhelms, Commented Aug 19, 2013 at 19:33
@cwhelms With the value "1" say there is a folder with images that correspond to the record. Example: directory php-> folder 1 (id of ford), folder 2 (id of seat), ruby directory -> folder 3 (id of fiat) etc. So should have a "while" or "foreach" with this data and display a url depending on the values "1" in the array: folder/'.$code.'/'.$id.'/ — santyas
– santyas, Commented Aug 19, 2013 at 19:41
@Jon Perhaps there is some way to simplify many if(php) if(ruby) if(java)? thank you — santyas
– santyas, Commented Aug 19, 2013 at 19:49
@santyas: Not really unless I 'm missing something. You can of course hide the conditionals inside a method, which often is a good way to increase both the abstraction and readability of your code at the expense of a tiny performance hit that you probably don't care about (exchanging something you don't need for something you do is a common attribute of good engineering). But the logic still needs to exist somewhere. — Jon
– Jon, Commented Aug 19, 2013 at 20:31

cwhelms · Accepted Answer · 2013-08-19 22:19:39Z

1

It looks like you want to break up the array into php, java, and ruby. Below is not tested, just to give you an idea.

$resultsArray = array();
while($row = mysqli_fetch_assoc($result)){
    if($row['php'] == 1) {
        $resultsArray['php'][] = $row['id'];
    }
    if($row['java'] == 1) {
        $resultsArray['java'][] = $row['id'];
    }
    if($row['ruby'] == 1) {
        $resultsArray['ruby'][] = $row['id'];
    }
}

Then loop through that 2d array, grabbing the name from the key in resultsArray.

foreach($resultsArray as $language => $array) {
    foreach($array as $id) {
        echo "url/".$language."/".$id;
        echo "<br>";
    }
}

edited Aug 19, 2013 at 22:19

answered Aug 19, 2013 at 19:51

cwhelms

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4 Comments

santyas Over a year ago

Like this? foreach($resultsArray as $key => $val){print_r($val);}

cwhelms Over a year ago

Look at my edit. print_r will only get you the array of an individual language.

santyas Over a year ago

Hello again, works perfect. I use it with a single foreach and the id in this way: echo "url/".$language."/".$row['id']; I can ask you something else? Surely you understand the concept and I take long to reformulate a new post. How I can use opendir with directories I got? Thank you very much!

santyas Over a year ago

I could write the names of the files that are inside the folders, but I do not know the names. So I need to use readdir @cwhelms

santyas · Accepted Answer · 2013-08-20 18:28:53Z

0

foreach($resultsArray as $language => $array){
    foreach(array_slice(glob('work/'.$language.'/'.$rowtest['id'].'/*.jpg'),0,1) as $image){    
        echo $image . "<br />";
    }
}

answered Aug 20, 2013 at 18:28

santyas

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