Question

我正在尝试列出MySql数据库中最常出现的前3个名称。

这就是我正在做的事情：

        $nameQuery = "SELECT PeopleName, COUNT(*) AS totalNumber FROM finaldb ORDER BY        COUNT(PeopleName) LIMIT 5";
        $nameResult = mysql_query($nameQuery);
        while($data = mysql_fetch_array($nameResult)) {

              $name = $data['totalNumber'];
        }

 echo $name;

然而，这似乎不起作用。有什么建议吗？

我的数据库由：PeopleName，ID组成，称为finaldb。

Answer 1

您缺少分组依据，请尝试此查询：

$nameQuery = "SELECT PeopleName, COUNT(PeopleName) AS totalNumber FROM finaldb GROUP BY PeopleName ORDER BY        COUNT(PeopleName) LIMIT 5";

Answer 2

SELECT 
    PeopleName
    count(PeopleName)
FROM
    finalDB
GROUP BY 
    PeopleName
ORDER BY
    count(PeopleName) DESC
LIMIT 0,3

Answer 3

您应该使用该查询返回1个结果，但就是这样。您需要在SQL中添加GROUP BY子句：

SELECT PeopleName, COUNT(PeopleName) AS totalNumber FROM finaldb GROUP BY PeopleName ORDER BY COUNT(PeopleName) DESC LIMIT 5

您还可以通过在mysql命令行或phpmyadmin内部运行查询来检查您的查询应返回的内容。

您的循环仅将最新的totalNumber分配给$ name。如果要回显所有数据，请尝试以下操作：

    while($data = mysql_fetch_array($nameResult)) {

          echo "{$data['PeopleName']} - {$data['totalNumber']}\n";
    }

要添加数据，只需不断添加列名：

          echo "{$data['PeopleName']} - {$data['totalNumber']} - {$data['Gender']} - {$data['Age']}\n";

Answer 4

您的循环不保留您提取的名称，它只是用下一个值覆盖PREVIOUS名称。您需要构建一个值数组，或者至少在循环内部输出。 e.g。

$names = array();
while(...) {
   $names[] = array('name' => $data['PeopleName'], 'total' => $data['totalNumber']);
}
var_dump($names);

根据MYSQL的出现次数显示数据

4 个答案: