在PHP中使用Variables来创建JSONTable

我试图从表gainfinal中获取数据，并在JSON_table中显示数据，以便在google API图表中进行可视化。我特别使用了两个变量$countryone和$countrytwo。它们的使用方式如下代码所示。其他一切都很好。代码的输出就像这个

{"cols":[{"label":"year","type":"string"},{"label":".$countryone.","type":"number"},{"label":".$countrytwo.","type":"number"}],"rows":[{"c":[{"v":"1995"},{"v":"76.8561073"},{"v":"46.8550182"}]},{"c":[{"v":"1996"},{"v":"77.0366637"},{"v":"47.1409752"}]},{"c":[{"v":"1997"},{"v":"77.180129"},{"v":"46.6331669"}]}

在第二列和第三列中，我需要打印变量的实际值（我的意思是，USA和NPL），但它只是将变量打印为I字符串。我该如何更正？

    <?php
    ini_set('display_errors', 1); 
    $username = "root"; 
    $password = "";   
    $host = "localhost";
    $database="climate";
    $countryone='USA'; 
    $countrytwo='NPL';
    $index ='gainfinal'; 

    $server = mysql_connect($host, $username, $password);
    $connection = mysql_select_db($database, $server);
    $myquery = "SELECT  `year`, 
        sum(case when `countrycode` = '$countryone' then `values` else 0 end) AS `countryone`,
        sum(case when `countrycode` = '$countrytwo' then `values` else 0 end) AS `countrytwo`
FROM   `$index`
GROUP BY `year`
";
    $query = mysql_query($myquery);
    $table = array();
    $table['cols'] = array(
    /* define your DataTable columns here
     * each column gets its own array
     * syntax of the arrays is:
     * label => column label
     * type => data type of column (string, number, date, datetime, boolean)
     */
    array('label' => 'year', 'type' => 'string'),
    array('label' => '.$countryone.', 'type' => 'number'),
    array('label' => '.$countrytwo.', 'type' => 'number'),

    // etc...
);
    $rows = array();
while($r = mysql_fetch_assoc($query)) {
    $temp = array();
    // each column needs to have data inserted via the $temp array
    $temp[] = array('v' => $r['year']);
    $temp[] = array('v' => $r['countryone']);
    $temp[] = array('v' => $r['countrytwo']);

    // etc...
    // insert the temp array into $rows
    $rows[] = array('c' => $temp);
}
// populate the table with rows of data
$table['rows'] = $rows;
// encode the table as JSON
$jsonTable = json_encode($table);
// set up header; first two prevent IE from caching queries
header('Cache-Control: no-cache, must-revalidate');
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
header('Content-type: application/json');
// return the JSON data
echo $jsonTable;
?>