选择没有列名的列值

Question

I have a table with two columns: title and content.

I want to pick the values in title columns so I execute

"SELECT title FROM table"

and it returns me this

[{"title":"Just a test"},{"title":"Just a test 2"},{"title":"Just a test 3"}]

Now my question is simple: how can I pick values Just a test, Just a test 2 and Just a test 3 without the column's name?

I need the results to be sent to an app using android code so I need them to be like this in order to parse them and fill a listview (otherwise I have to manipulate the results in the android code but I don't know how to do that).

UPDATE: this is my PHP code:

<?php
require 'jsonwrapper.php';
mysql_connect("+++++++","++++++++","+++++++++");
mysql_select_db("my_tripleleon");
$q=mysql_query("SELECT titolo FROM articoli");
while($raw=mysql_fetch_assoc($q))
{       $output[]=$raw;
}
print(json_encode($output));
mysql_close();
?>

Using json_decode($q) doesn't print anything...

Answer 1

Assign the result of your query to a $variable passed by json_decode(), as this is the format you're getting.

$myvar = json_decode($your_mysql_result, true);

This will return you an associative array like this:

Array
(
    [0] => stdClass Object
        (
            [title] => Just a test
        )

    [1] => stdClass Object
        (
            [title] => Just a test 2
        )

    [2] => stdClass Object
        (
            [title] => Just a test 3
        )

)

So for getting the values, you'd do:

foreach ($myvar as $item)
  echo "My item is: " . $item->title . "
";

---- EDIT ----

Replace this code:

while($raw=mysql_fetch_assoc($q))
{       $output[]=$raw;
}
print(json_encode($output));

By this one:

while($raw=mysql_fetch_assoc($q))
{       $output[]=$raw['title'];
}
print(json_encode($output));