为google图表生成json -添加空

Question

如果mysql返回的行中没有结果，我需要在json中生成一个空值。所谓“不存在”，我的意思是在$comps数组中没有一个或多个值的结果。

$data = array(
    'rows' => array()
);

// these values will change, based on submitted form
$comps = array( 'activity','purchase','groups'); 

$comps_count = count( $comps ); 

// mimic return from mysql
$sql_results = array (
array ( 'year' => '2013', 'month' =>  '12', 'day' => '7', 'component' => 'activity', 'num' => '10'), 

array ('year' => '2013', 'month' => '12', 'day' => '13', 'component' => 'purchase', 'num' => '11'), 
array ('year' => '2013', 'month' => '12', 'day' => '13', 'component' => 'groups', 'num' => '12'), 

array ('year' => '2013', 'month' => '12', 'day' => '30', 'component' => 'activity', 'num' => '13'), 
array ('year' => '2013', 'month' => '12', 'day' => '30', 'component' => 'groups', 'num' => '14'),

array ('year' => '2013', 'month' => '12', 'day' => '31', 'component' => 'dummy', 'num' => '0'), 
);

$results = array();


foreach ( $sql_results as $row ) {

    $results[] = (object)$row;

}

$j = 0; 

foreach ( $results as $result ) {

    $num = (int) $result->num;

    $year = (int) $result->year;
    $month = (int) $result->month - 1;  // for javascript months
    $day = (int) $result->day;

    $this_date = $year . $month . $day;

    $component = $result->component; 
    $key = array_search($component, $comps);

    if ( $temp_date != $this_date ) {

        if ( !empty( $temp_arr ) ) {

            $diff = abs( ( $comps_count + 1 ) - count( $temp_arr ) );

            if ( $diff != 0 ) {
                for ( $i=0; $i<$diff; $i++ )
                    $temp_arr[] = array('v' => null);
            }

            $data['rows'][] = array('c' => $temp_arr);

            $temp_arr = array();
            $j = 0;
        }

        $temp_arr[] = array('v' => "Date($year, $month, $day)");

        $j++; 

        for ( $i=0; $i<$key; $i++ ) {

            $temp_arr[] = array('v' => null);
        }

        $temp_arr[] = array('v' => $num);

    }
    else {
        $j++; 

        $xcount = $comps_count - $key;

        if ( $xcount != $j ) { 
            for ( $i=$xcount; $i<$key; $i++ ) {

                $temp_arr[] = array('v' => null);
            }   
        }

        $temp_arr[] = array('v' => $num);

    }

    $temp_date = $this_date;
}

$data['rows'][] = array('c' => $temp_arr);

$json = json_encode($data, JSON_NUMERIC_CHECK);

echo $json;

json输出：

{"rows":[
{"c":[{"v":"Date(2013, 11, 7)"},{"v":10},{"v":null},{"v":null}]},
{"c":[{"v":"Date(2013, 11, 13)"},{"v":null},{"v":11},{"v":null},{"v":12},{"v":null}]},
{"c":[{"v":"Date(2013, 11, 30)"},{"v":13},{"v":null},{"v":14}]},
{"c":[{"v":"Date(2013, 11, 31)"},{"v":0}]}
]}

(我不担心最后一排假人)。问题在第二行。它应该是：

{"c":[{"v":"Date(2013, 11, 13)"},{"v":null},{"v":11},{"v":12}]},

如何避免插入这些额外的空号？

必须有更简单的方法来解决这件事，但我现在陷入了困境。

Answer 1

我离得很近。这很管用..。

foreach ( $results as $result ) {

    $num = (int) $result->num;

    $year = (int) $result->year;
    $month = (int) $result->month - 1;  // for javascript months
    $day = (int) $result->day;

    $this_date = $year . $month . $day;

    $component = $result->component; 
    $key = array_search($component, $comps);


    if ( $temp_date != $this_date ) {

        if ( !empty( $temp_arr ) ) {

            $diff = ( $comps_count + 1 ) - count( $temp_arr );

            if ( $diff != 0 ) {
                for ( $i=0; $i<$diff; $i++ ) {
                    $temp_arr[] = array('v' => null);
                }
            }

            $data['rows'][] = array('c' => $temp_arr);

            $temp_arr = array();

        }

        $temp_arr[] = array('v' => "Date($year, $month, $day)");


        for ( $i=0; $i<$key; $i++ ) {

            $temp_arr[] = array('v' => null);
        }

        $temp_arr[] = array('v' => $num);


    }
    else {

        $diff = ( $key + 1  ) - count( $temp_arr );

        if ( $diff != 0 ) 
                $temp_arr[] = array('v' => null);

        $temp_arr[] = array('v' => $num);

    }

    $temp_date = $this_date;

}

Answer 2

要从0到11获取google日期和月份，请尝试使用：

$date =new DateTime('2014-12-21');
$date->format('Y,n-1,d');