Question

我试图弄清楚如何从MySQL查询中获得正确的结果，该查询包括当输出编码为json时的求和函数。

当没有sum函数时，结果会正确返回（这对我来说也可能是一个问题，但是我还没有尝试过）但是当MySQL查询中有一个sum函数时它只有返回一个结果/对象（我假设它是总和的总和）。

这是我第一次使用mysql_fetch_array（）来检索数组结果，以及如何在网上实现json_encode的所有示例，我发现使用此函数。 mysql_fetch_array（）能否与sum和GROUP BY函数一起使用？

{


    $sql = mysql_query("SELECT portfolio,agency,program,objectives_text,last,current FROM     `table`  WHERE MATCH(program,objectives_text) AGAINST('health'IN BOOLEAN MODE)  ");

    $results = array();
    while($row = mysql_fetch_array($sql))
   {
   $results[] = array(

      'portfolio' => $row['portfolio'],
      'agency' => $row['agency'],
      'program' => $row['program'],
        'objectives_text' => $row['objectives_text'],
      'component' => $row['component'],
      'last' => $row['last'],
      'current' => $row['current']
   );
}
$json = json_encode($results);

  echo
"$json";  

}

//the above code produces correct results in json output with integer value for each row correct

    {


    $sql = mysql_query("SELECT portfolio,agency,program,objectives_text,last,current,sum(last),sum(current) FROM `table`  WHERE MATCH(program,objectives_text) AGAINST('health'IN BOOLEAN MODE)  ");

    $results = array();
    while($row = mysql_fetch_array($sql))
    {
   $results[] = array(

      'portfolio' => $row['portfolio'],
      'agency' => $row['agency'],
      'program' => $row['program'],
        'objectives_text' => $row['objectives_text'],
      'component' => $row['component'],
      'last' => $row['sum(last)'],
      'current' => $row['sum(current)']
   );
    }
    $json = json_encode($results);

     echo
    "$json";  

    }
//produces one result/object in the json output with the integer results being a sum of the individual rows

使用mysql sum函数和json_encode

0 个答案: