Question

如果我对此查询结果有误，可以有人帮助我。

<?php

include 'dbconfig.php';
$sql = 'SELECT SUM(total_fund.total_funds), total_fund .name from total_fund join project 
        WHERE total_fund.funds_id = project.project_id AND project_id=1200 GROUP BY project.project_id';
$retval = mysqli_query($dbconfig, $sql);

while ($data = mysqli_fetch_array($retval)) {

    echo $data['total_funds'] . '</br>';
    echo $data['name'];
}
?>

Answer 1

您的查询中有一个拼写错误：

total_fund .name

应写成：

total_fund.name

并且您可能需要一个AS别名，无论如何，以这种方式编写查询会更好：

select
  total_fund.name
  SUM(total_fund.total_funds) as total_funds
from
  total_fund join project on total_fund.funds_id = project.project_id
where
  project_id=1200
group by
  total_fund.name

Answer 2

这是完整的查询

<?php

include 'dbconfig.php';
$sql = 'SELECT SUM(total_fund.total_funds)AS total, total_fund .name, project.Amount_to_be_raised from total_fund join project 
        where total_fund.funds_id = project.project_id AND project_id=1200';
$retval = mysqli_query($dbconfig, $sql);

while ($data = mysqli_fetch_array($retval)) {

    echo $data['total'] . '</br>';
    echo $data['name'];
}
?>

加入查询错误未知索引

2 个答案: