Php 使用count从两个表中选择行总数

我有两张MySQL表

学生=150（代表学生总数的行总数）

雇主=230（代表雇主总数的行总数）

完整代码

它显示结果=2

我也尝试使用：

SELECT  (SELECT COUNT(*) FROM   student) AS count1,(SELECT COUNT(*) FROM   employer) AS count2 FROM    dual*

它显示结果=1

我想我的代码可能有错， 我如何从学生和雇主处获得总计，以便查看数据并在图表中显示

---

150230

我在MySQL中用这样一个查询对两个表求和

SELECT  ( (SELECT COUNT(*) FROM scheme.student) + 
          (SELECT COUNT(*) FROM scheme.employer) 
         ) AS 'Column' ;

在我的例子中，我使用了相同的表2次，该表有280行（）

---

希望这对您有所帮助

您的第二个查询应该可以工作

    SELECT  (SELECT COUNT(*) FROM   student) AS count1,(SELECT COUNT(*) 
    FROM employer) AS count2 FROM    dual* <-- make sure you remove (*)
// so it should be like this 
    SELECT  (SELECT COUNT(*) FROM   student) AS count1,(SELECT COUNT(*) 
    FROM employer) AS count2 FROM    dual

---

作为代码的第一行是行不通的，因为您试图显示尚未执行的内容，而您似乎正在下面执行。谢谢，我已经修复了它，但它仍然显示总计1谢谢，它帮助我显示数据。现在我必须弄清楚我想如何在piechart中显示它。嗨，你的代码对我很有帮助，如果我在mysql中查询它，它确实显示了正确的总数，但是为什么当我编写done php时，网站中的总数看起来是1？你也会这样吗？还是我遗漏了什么？

SELECT  ( (SELECT COUNT(*) FROM scheme.student) + 
          (SELECT COUNT(*) FROM scheme.employer) 
         ) AS 'Column' ;

    SELECT  (SELECT COUNT(*) FROM   student) AS count1,(SELECT COUNT(*) 
    FROM employer) AS count2 FROM    dual* <-- make sure you remove (*)
// so it should be like this 
    SELECT  (SELECT COUNT(*) FROM   student) AS count1,(SELECT COUNT(*) 
    FROM employer) AS count2 FROM    dual

$sql = "SELECT  (SELECT COUNT(*) FROM   playlists) AS count1,(SELECT COUNT(*) FROM   artists) AS count2 FROM    dual";

if ($result=mysqli_query($con,$sql))
  {
      $row = mysqli_fetch_array($result);
      var_dump($row); // just checking the raw array...
      echo "<br />";
      echo $row[0]. ", ". $row[1];  // got your results -> 150, 230
  }
mysqli_close($con);

$sql = "SELECT  (SELECT COUNT(*) FROM students), (SELECT COUNT(*) FROM employer) FROM    dual";