多个SELECT语句的正确mySQL语法？

我试图用一个SELECT语句选择多个列，但我这样尝试了，但没有成功

mysql> SELECT AVG(salaryoffered) "$ offer" AS "Average Salary", MIN(salaryoffered)
"$ offer" AS "Min Salary", MAX(salaryoffered) "$ offer" AS "Max Salary" FROM
interview;

然而，他们确实工作了，并单独使用SELECT打开了表格。像这样：

mysql> SELECT AVG(salaryoffered) "$ offer" AS "Average Salary" 
mysql> SELECT MIN(salaryoffered) "$ offer" AS "Min Salary" 
mysql> SELECT MAX(salaryoffered) "$ offer" AS "Max Salary" 
    -> FROM interview;

知道如何在一个命令中将三个表放在一起吗？ 假设是这样的：

+----------------+------------+------------+
| Average Salary | Min Salary | Max Salary |
+----------------+------------+------------+
|      12.080357 |      10.75 |      13.75 |
+----------------+------------+------------+

---

第一个查询也不起作用。试试这个

SELECT AVG(salaryoffered) AS "Average Salary", 
MIN(salaryoffered) AS "Min Salary", 
MAX(salaryoffered) AS "Max Salary" 
FROM interview;

如果您想添加一些文本，可以使用concat，比如

SELECT concat(AVG(salaryoffered), "$ offered") AS "Average Salary", 
concat(MIN(salaryoffered), "$ offered") AS "Min Salary", 
concat(MAX(salaryoffered), "$ offered") AS "Max Salary" 
FROM interview;

---

你确定这样行吗<代码>选择平均工资（SalaryOffer）“$offer”作为“平均工资”？是的，有效！我试过了，都是单独的，都能用，但放在一起就不行了。嗯……去掉每个表达式中的

“$offer”

。这种语法是不对的。如果没有

“$offer”

，它应该可以正常工作。不，它不应该。它将抛出语法错误

AVG（salaryoffered）“$offer”

不正确。它现在可以工作了，谢谢大家！我会记住的！