如何从PHP中的SQL表获取行位置

Question

我需要使用PHP脚本从SQL表中获取“位置（行）”。 我测试了我的SQL语句，它运行良好。 输出正确。

我的SQL语句：

SET @rownum = 0; 
SELECT position, name,cash FROM (SELECT name, cash, @rownum := @rownum + 1 AS position FROM 'my_DB'.'Cash' ORDER BY cash DESC ) AS t WHERE name = 'user44';

输出：

"position":44,"name":"user44","cash":"5600"

但是，如果我把它放在PHP中：

$query="SET @rownum = 0; SELECT position, name,cash FROM (SELECT name, cash, @rownum := @rownum + 1 AS position FROM 'my_DB'.'Cash' ORDER BY cash DESC ) AS t WHERE name = 'user44';";

它向我显示了一个errant query 。

我也尝试过这样的事情：

$query="SELECT position, name,cash FROM (SELECT name, cash, @rownum := @rownum + 1 AS position FROM 'my_DB'.'Cash' ORDER BY cash DESC ) AS t WHERE name = 'user44';";

输出：

"position":null,"name":"user44","cash":"5600"

感谢VolkerK，我用以下方法解决了该问题：

$query = "SET @rownum =0;"; $result = mysql_query($query,$link) or die('Errant query: '.$query); $query="SELECT position, name,cash FROM (SELECT name, cash, @rownum := @rownum + 1 AS position FROM 'my_DB'.'HSCash' ORDER BY cash DESC ) AS t WHERE name = 'user44'"; $result = mysql_query($query,$link) or die('Errant query: '.$query);

Answer 1

我现在更好了。

诀窍不是获得位置，而是获得比用户得分更高的人数。

SQL：

// Get how much cash the user make
SELECT cash
FROM mydb.cash
WHERE name = :name;

// Get how many people did better + 1
SELECT COUNT(*)+1
FROM mydb.cash
WHERE cash > :cash

受污染的查询更加复杂，但是不需要对整个表进行排序，看起来像

SELECT position, name, cash
FROM mydb.cash
JOIN (
    SELECT COUNT(*)+1 AS position
    FROM mydb.cash
    WHERE cash > (
        SELECT cash
        FROM mydb.cash
        WHERE name = :name
    )
)
WHERE name = :name

或者，通过在加入之前进行过滤来限制加入，例如：

SELECT position, name, cash
FROM (
    SELECT name, cash
    FROM mydb.cash
    WHERE name = :name
)
JOIN (
    SELECT COUNT(*)+1 AS position
    FROM mydb.cash
    WHERE cash > (
        SELECT cash
        FROM mydb.cash
        WHERE name = :name
    )
)