将这两个MySQL查询合并为一个

Question

这是手头的任务：我有一个名为'reports'的数据库表，这个表中的一列是'submitted_by'。 此列包含提交报告的人员的用户名。

现在，我需要做的是显示按他们发布的报告数量排序的用户名列表。 所以这样的事情应该出来：

用户| 报告

---

马克| 25

---

约翰| 18

---

玛丽| 7

等等

我已经编写了这个PHP / MySQL代码来做到这一点，它非常简单，但我想知道这些是否可以合并为1个查询。 此外，这会在效率方面获得更多收益吗？

$users=array();

$result=mysql_query("select id,submitted_by from reports group by submitted_by");
while($row=mysql_fetch_array($result)){
    $result1=mysql_query("select id,submitted_by from reports where submitted_by='$row[submitted_by]' ");
    $users[$row[submitted_by]]=mysql_num_rows($result1);
}

arsort($users);

然后只需回显来自$users每个元素。

Answer 1

请尝试以下代码：

$users = array();

$result=mysql_query("SELECT id, submitted_by, COUNT(*) as cnt FROM reports GROUP BY submitted_by ORDER BY COUNT(*) DESC");

while($row=mysql_fetch_array($result)){
    $users[$row['submitted_by']] = $row['cnt'];
}

Answer 2

$users = array();

$result = mysql_query("
SELECT id,submitted_by, COUNT(*) as amount 
FROM reports 
GROUP BY submitted_by
ORDER BY amount DESC
");

while($row = mysql_fetch_array($result))
    $users[$row['submitted_by']] = $row['amount'];

// arsort($users);

Answer 3

尝试：

$query = "SELECT id, submitted_by, COUNT(*) as numbers FROM reports GROUP BY submitted_by ORDER BY COUNT(*) DESC";

$result = mysql_query($query);

echo "<table><tr><th>User</th><th>Reports</th></tr><tr>";

while($row = mysql_fetch_assoc($result)) {

  echo "<td>$row[submitted_by]<td>";
  echo "<td>$row[numbers]<td>";

}

echo "</tr></table>"