PHP script for SQL querying

Question

I'm still learning php so please take it easy on me. This might sound a silly question for you guys.

Right so. I have categories lets say Blogs, eCommerce, Portfolios etc. Files according too. Blog.php etc.

I also have functions.php where all my functions are.

MYSQL database is where I store information from them files. Website information, descriptions etc.

I want to have 1 function that queries data from a website just about 1 category so then I can display it in Blogs.php, eCommerce.php etc. My function to query data from mysql looks like this.

This is an example:

function querying_category($category){
    $db = DB::getInstance();
    $all = $db->query('SELECT * FROM website WHERE category = {$category} ORDER BY id DESC');

    if($all->count()){
       foreach($all->results() as $website){
              $web_data[] = $website;
       }
    }
    return $web_data;
}

and then let's say in my blog.php would go like:

$category = 'blog';
$website = querying_category($category);

Could you please tell me what am I doing wrong? I want to declare a variable in my blog.php or ecommerce.php etc without rewriting the following query:

'SELECT * FROM website WHERE category = {$category} ORDER BY id DESC

Can I achieve it with passing in $category as an argument in querying_category() but declaring $category in my blog.php or ecommerce.php?

Sidenote: Shouldn't that be return $website;? I stand at being wrong ;-) — Funk Forty Niner
– Funk Forty Niner, Commented Mar 27, 2014 at 19:23
Are you receiving an error? If I had to guess you need to wrap {$category} in single quotes in your query but without an error message i'm not sure. — Joe Meyer
– Joe Meyer, Commented Mar 27, 2014 at 19:29

Funk Forty Niner · Accepted Answer · 2014-03-28 13:13:26Z

1

It seems as if the editing I made to the question fixed the OP's problem.

There were a few lines of code that were not properly indented along with a few spelling mistakes.

As per the OP's request, this answer has been given in order to close the question.

However, this line return $web_data; should have been return $website;

answered Mar 28, 2014 at 13:13

Funk Forty Niner

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