使用count從兩個表中選擇行總數

Question

我有兩個MySQL表，

學生= 150（總行數代表學生總數）

雇主= 230（總行數代表雇主總數）

完整代碼

<?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);

結果：1

我怎樣才能解決這個問題？

我嘗試使用

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

它顯示結果= 2

我也嘗試使用：

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

它顯示結果= 1，

我想我的代碼可能有誤，如何從學生和雇主那里得到總計，以便我可以查看數據並以餅圖形式顯示？

150、230

Answer 1

我用這樣的查詢在MySQL中完成了兩個表的總和

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

在我的情況下，我使用同一張表2次，該表有280行（ 280 x 2 = 560 ）。

希望這對您有幫助

Answer 2

您是第二個查詢應該工作。

    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

然后，您只需要獲取數組即可顯示結果

$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，如果您將+替換為，則您的第一個查詢將起作用

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