Question

我有两个MySQL表，

student = 150（代表学生总数的行总数）

雇主= 230（代表雇主总数的行数总数）

完整代码

<?php echo mysqli_num_rows($studempl); ?> 

$conn = new mysqli ('localhost', 'root', '', 'dashboard');
$studempl = $conn ->query ("SELECT  (SELECT COUNT(*) FROM student) + (SELECT COUNT(*) FROM employer) FROM    dual");
$tot_studempl = mysqli_num_rows($studempl);

结果：1

我该如何解决这个问题？

我尝试使用

SELECT 'student' AS stdID, COUNT(*) FROM student
UNION
SELECT 'employer' AS emplID, COUNT(*) FROM employer*

显示结果= 2，

我还尝试使用：

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

显示结果= 1，

我想我的代码可能有问题，如何从学生和雇主处获得总数，以便我可以查看数据并在饼图中显示？

150,230

Answer 1

我使用像这样的查询来完成MySQL中两个表的总和

{{1}}

在我的情况下，我使用了相同的表2次，该表有280行（280 x 2 = 560）。

希望这能帮到你

Answer 2

你的第二个查询应该有效。

    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

然后你只需要获取数组来显示结果

$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);

另外，如果你将+替换为

，那么你的第一个查询是有效的

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

使用count从两个表中选择行总数

2 个答案: