我有一个简单的查询,它查找一个名为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
我想要的是将这个绑定到一个PHP variable.If,这返回了工资,我会像这个echo "<td>{salary}</td>";
一样绑定它,一切都会正常工作,OK.However这个echo "<td>{$AVG(E.salary)}</td>";
给了我errors.How,我可以让这个查询返回一个变量,这个变量可以在以后转换成PHP形式吗?
更新:解决方案是使用AVG(E.salary)作为东西
发布于 2019-06-05 03:18:02
你需要一个合适的别名
SELECT AVG(E.salary) my_avg
FROM employee as E,permanent_employee as P
WHERE E.empID=P.empID
echo "<td>{$my_avg}</td>";
发布于 2019-06-05 03:35:23
有多种方法可以将其提取到php变量中
$result = mysqli_query($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,请参阅Get results from from MySQL using PDO
https://stackoverflow.com/questions/56449968
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