查询-php / json的问题

Question

我有这段代码，输出是：{"round (avg (salary), 0)":"1750"}

function tableOne() {

    $query = mysql_query("select round (avg (salary), 0) from worker, formation_area where id_formation_area=1") or die(mysql_error());
    $row = mysql_fetch_assoc($query);
    $result = $row;

    echo json_encode($result);
}

tableOne(); 

?>

但我需要这样的事：{1750}。

任何帮助？

Answer 1

试试这个：

function tableOne() {

    $query = mysql_query("select round(avg (salary), 0) as `round_avg` from worker, formation_area where id_formation_area=1") or die(mysql_error());
    $row = mysql_fetch_assoc($query);
    $result = $row['round_avg'];

    echo json_encode($result);
}

tableOne(); 

记住json需要一个something: something_else所以你可能会在这里得到一个错误，因为现在没有编码的数组