fetching from DB and store them in a new array PHP

Question

I am fetching data from my DB and they look like this:

[{"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","3":"1","key-4":"1"}]

where the key-1 are the name of the column. (I only have one entry so).

I want to extract only the column values and save them into a new array that will output like this:

{"key-1":"1","key-2":"1","key-3":"1","key-4":"1"}

I want it to look exactly like this and not : [{"key-1":"1","key-2":"1","key-3":"1","key-4":"1"}]

I tried this:

$cart["key-1"]=$output["key-1"];

where $output is the outcome of the DB that shown first (the one with []).

and the $cart is the new array I want.

Both are declared as:

$cart=array();
$output=array();

and $output[]=$row where row is the result of the DB fetch. How to do it?

Dale · Accepted Answer · 2012-12-15 11:29:24Z

1

Here's one way to do it, I've substituted the database row for a string here, and made use of json_decode() and json_encode()

$data = '[{"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","3":"1","key-4":"1"}]';

// convert to an array
$data = json_decode($data, true);

// create new array here
$cart = array();

for ($i = 0; $i < count($data); $i++)
{
    foreach ($data[$i] as $k => $v)
    {
        if (strpos($k, 'key') !== FALSE)
        {
            $cart[$k] = $v;
        }
    }
}

echo $cart['key-1'] . '<br/>';

echo json_encode($cart);

Output:

1
{"key-1":"1","key-2":"1","key-3":"1","key-4":"1"}

answered Dec 15, 2012 at 11:29

Dale

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Comments

Smuuf · Accepted Answer · 2012-12-15 11:27:49Z

0

This is a very chaotically asked question. From what I gathered you want maybe this..?

$cart=array();
foreach ($output as $index=>$value){
  if stripos($index,"key-"){
    cart[$index]=$value;
  }
}

answered Dec 15, 2012 at 11:27

Smuuf

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2 Comments

ghostrider Over a year ago

It is not chaotically at all. I want to get only the values from the column and store them in a new array. What is complicated in this?? I will try your answer.

Smuuf Over a year ago

Ok, my bad. I didn't realize it was JSON and assumed it was just an array. Dale's answer seems to be correct.

0xmtn · Accepted Answer · 2012-12-15 11:28:46Z

0

Use mysql_fetch_assoc() to get only column names ;)

while ($row = mysql_fetch_assoc($result_of_query))
{
    echo $row['key-1'];
    echo $row['key-2'];
}

answered Dec 15, 2012 at 11:28

0xmtn

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2 Comments

ghostrider Over a year ago

Ok thanks with that. I also want my ooutput not to have this [] in front of it. because i need that output to be in that format. ( start with {}). I know almost nothing about php so I am unaware of what is the difference between [] and {} but I need it in that format({}).

0xmtn Over a year ago

You return a json from db? I mean, I understood your question like this: You tried to fetch some rows from DB, and they came in an array with both numbered(indexed) arrays and column-named key-value paired arrays. So my code is a solution to the fetching "problem" like this... If you have a json, then you just have to call json_encode() on $output_array[0] and then loop through its returned value, and get only the column names.

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