使用while循环对PHP MYSQL进行排名答案

【问题标题】：Ranking PHP MYSQL Using while loop使用while循环对PHP MYSQL进行排名
【发布时间】：2015-10-11 15:56:10
【问题描述】：

我有问题。我正在根据申请人的考试成绩制定排名系统。代码如下：

$testScore = "SELECT Overall_Score, First_Name
                      FROM applicant_details, person, person_details
                      where person.ID_No like person_details.ID_No and Position_Applied = 'Work137' and Person_Type = 'Applicant' and applicant_details.ID_No like person.ID_No";

$result = mysql_query($testScore);

$rank = 0;
$lastScore = false;
$rows = 0;

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

       $name = $row['First_Name'];
       $overall = $row['Overall_Score'];
       $rows++;

       if( $lastScore != $overall){
           $lastScore = $overall;
           $rank = $rows;
        }
        echo"Name:$name rank:$rank score: $overall </br>";
}

此代码的输出是第一个被查询的人是数字1。示例输出：

姓名：犹他州排名：1 得分：85

姓名：立夏等级：2 得分：90

我想要一个输出：

姓名：犹他州排名：2 得分：85

姓名：立夏等级1得分：90

【问题讨论】：

修复您的查询以使用正确的join 语法并在表之间使用正确的连接条件。

标签： php mysql

【解决方案1】：

只需使用 ORDER BY [列名排序] [DESC/ASC] 按值排序...

SELECT Overall_Score, First_Name
                  FROM applicant_details, person, person_details
                  where person.ID_No like person_details.ID_No and Position_Applied = 'Work137' and Person_Type = 'Applicant' and applicant_details.ID_No like person.ID_No ORDER BY Overall_Score DESC

如果你想保持顺序，那么应该使用 mysql_num_rows() 来获取答案的总数，并使用该数字对它们进行排名：

$rows = mysql_num_rows($result);
while( $row = mysql_fetch_array( $result ) ){               

   $name = $row['First_Name'];
   $overall = $row['Overall_Score'];

   if( $lastScore != $overall){
       $lastScore = $overall;
       $rank = $rows;
    }
    echo"Name:$name rank:$rank score: $overall </br>";
   $rows--;

}

【讨论】：

从来不知道解决方案非常简单。谢谢！

【解决方案2】：

您的查询中需要一个 ORDER BY 子句：

$testScore = "SELECT Overall_Score, First_Name
              FROM applicant_details, person, person_details
              WHERE person.ID_No like person_details.ID_No
              AND Position_Applied = 'Work137'
              AND Person_Type = 'Applicant'
              AND applicant_details.ID_No like person.ID_No
              ORDER BY Overall_Score DESC
";

正如 Gordon Linoff 所暗示的，您可以通过使用正确的 JOIN 语法来改进此查询：

$testScore = "SELECT Overall_Score, First_Name
              FROM applicant_details
              JOIN person ON applicant_details.ID_No like person.ID_No
              JOIN person_details ON person.ID_No like person_details.ID_No
              WHERE Position_Applied = 'Work137'
              AND Person_Type = 'Applicant'
              ORDER BY Overall_Score DESC
";

【讨论】：