Question

当我在mysql中测试此查询时，它很好，但当我在php中运行它时，我一直收到此错误。

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT *, (@rownum := @rownum + 1) AS rank FROM ( SELECT *, (totalWins+(total' at line 1

这是我的PHP代码。

    <?php

    $sql = "    SET @rownum = 0; ";

    $sql .= "   SELECT *,  (@rownum := @rownum + 1) AS rank FROM ( ";

    $sql .= "       SELECT *, (totalWins+(totalPushs*.5)) AS totalPoints, totalWins+totalLost+totalPushs AS totalBets FROM ( ";

    $sql .= "           SELECT *, SUM(win) AS totalWins, SUM(lost) AS totalLost, SUM(push) AS totalPushs FROM ( ";

    $sql .= "               SELECT *, (finalResult = 'Winner') AS win, (finalResult = 'Loser') AS lost, (finalResult = 'Push') AS push FROM ( ";

    $sql .= "                   SELECT " . $db_prefix . "users.userID, userName, ";
    $sql .= "                   IF (pickID=visitorID, visitorResult, homeResult) AS finalResult ";
    $sql .= "                   FROM " . $db_prefix . "users ";
    $sql .= "                   JOIN " . $db_prefix . "picks ";
    $sql .= "                   ON " . $db_prefix . "users.userID = " . $db_prefix . "picks.userID ";
    $sql .= "                   JOIN " . $db_prefix . "schedule ";
    $sql .= "                   ON " . $db_prefix . "picks.gameID = " . $db_prefix . "schedule.gameID ";

    $sql .= "               ) x ";

    $sql .= "           ) x ";
    $sql .= "           GROUP BY userID ";

    $sql .= "       ) x ";

    $sql .= "   ) x ";
    $sql .= "   ORDER BY totalPoints DESC, totalWins DESC, totalPushs DESC, totalLost ";

    $result = mysql_query($sql) or die(mysql_error());

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

        echo $row[rank] . '|' . $row[userName]. '|' . $row[totalWins] . '|' . $row[totalLost] . '|' . $row[totalPushs] . '|' . $row[totalPoints];
        echo '<br>';

    }

    ?>

我可以在没有第一行代码的情况下使用php代码

$sql = " SET @rownum = 0; ";

但它不会回显排名列。

当它在php中时，我是否需要以不同的方式对代码进行排序？

Answer 1

mysql_query不支持一次运行多个查询。你必须先运行
mysql_query("SET @rownum = 0;");，然后您可以在第二次mysql_query来电中运行其余查询。

Answer 2

请尝试使用tablename。*而不是*

尝试使用“SET @rownum = 0;”时出错用PHP

2 个答案: