Question

我有一个简单的查询，可以找到名为permanent_employee的表的平均工资：

SELECT  AVG(E.salary) 
FROM employee as E,permanent_employee as P
WHERE E.empID=P.empID

我想将其绑定到一个PHP变量。如果返回薪水，我将像这样echo "<td>{salary}</td>";将其绑定，一切都会正常运行。但是echo "<td>{$AVG(E.salary)}</td>";却给了我错误。我可以使该查询返回一个变量，该变量以后可以覆盖到PHP形式吗？

更新：解决方案是使用AVG（E.salary）AS sth

Answer 1

您需要一个适当的别名

SELECT  AVG(E.salary)  my_avg 
FROM employee as E,permanent_employee as P
WHERE E.empID=P.empID


echo "<td>{$my_avg}</td>";

Answer 2

有多种方法可以将其提取到php变量

$result = mysqli_query($con,"SELECT  AVG(E.salary)  my_avg 
FROM employee as E,permanent_employee as P
WHERE E.empID=P.empID");
$row = mysqli_fetch_array($result));
$avg = $row['my_avg'];
echo "<td>".$avg."</td>";

您也可以使用PDO，请参见Get results from from MySQL using PDO

查询结果与PHP变量的绑定

2 个答案: