具有多个查询的 php/mysql

Question

<?php

$query1 = "CREATE VIEW current_rankings AS SELECT * FROM main_table WHERE date = X";

$query2 = "CREATE VIEW previous_rankings AS SELECT rank FROM main_table WHERE date = date_sub('X', INTERVAL 1 MONTH)";

$query3 = "CREATE VIEW final_output AS SELECT current_rankings.player, current_rankings.rank as current_rank LEFT JOIN previous_rankings.rank as prev_rank
             ON (current_rankings.player = previous_rankings.player)";

$query4 = "SELECT *, @rank_change = prev_rank - current_rank as rank_change from final_output";

$result = mysql_query($query4) or die(mysql_error()); 

while($row = mysql_fetch_array($result)) {
echo $row['player']. $row['current_rank']. $row['prev_rank']. $row['rank_change'];
}

?>

所有的查询都是独立工作的，但我真的很难将所有的部分放在一个单一的结果中，所以我可以将它与 mysql_fetch_array 一起使用。

我试图创建视图以及临时表，但每次它要么说表不存在，要么返回一个空的获取数组循环......逻辑在那里，但语法混乱，我认为这是我第一次不得不这样做处理多个查询我需要合并在一起。 期待一些支持。 非常感谢。

Answer 1

感谢 php.net，我想出了一个解决方案：您必须使用(mysqli_multi_query($link, $query))来运行多个串联查询。

 /* create sql connection*/
$link = mysqli_connect("server", "user", "password", "database");

$query = "SQL STATEMENTS;"; /*  first query : Notice the 2 semicolons at the end ! */
$query .= "SQL STATEMENTS;"; /* Notice the dot before = and the 2 semicolons at the end ! */
$query .= "SQL STATEMENTS;"; /* Notice the dot before = and the 2 semicolons at the end ! */
$query .= "SQL STATEMENTS"; /* last query : Notice the dot before = at the end ! */

/* Execute queries */

if (mysqli_multi_query($link, $query)) {
do {
    /* store first result set */
    if ($result = mysqli_store_result($link)) {
        while ($row = mysqli_fetch_array($result)) 

/* print your results */    
{
echo $row['column1'];
echo $row['column2'];
}
mysqli_free_result($result);
}   
} while (mysqli_next_result($link));
}

编辑 - 如果您真的想做一个大查询，上面的解决方案有效，但也可以根据需要执行任意数量的查询并分别执行它们。

$query1 = "Create temporary table A select c1 from t1"; 
$result1 = mysqli_query($link, $query1) or die(mysqli_error());

$query2 = "select c1 from A"; 
$result2 = mysqli_query($link, $query2) or die(mysqli_error());

while($row = mysqli_fetch_array($result2)) {

echo $row['c1'];
    }  

Answer 2

看来您没有执行 $query1 - $query3。 您刚刚跳到 $query4 ，如果其他人没有先执行，它将不起作用。

还

$query4 = "SELECT *, @rank_change = prev_rank - current_rank as rank_change from final_output";

应该是

$query4 = "SELECT *, @rank_change := prev_rank - current_rank as rank_change from final_output";

否则 rank_change 的值将只是一个布尔值，如果@rank_change 等于 (prev_rank - current_rank)，则为 true，否则为 false。 但是你真的需要@rank_change 吗？ 您会在后续查询中使用它吗？ 也许你可以完全删除它。

更好的是，您可以将所有查询合并为一个，如下所示：

SELECT 
    curr.player,
    curr.rank AS current_rank,
    @rank_change := prev.rank - curr.rank AS rank_change
FROM
    main_table AS curr
    LEFT JOIN main_table AS prev
        ON curr.player = prev.player    
WHERE 
    curr.date = X
    AND prev.date = date_sub('X', INTERVAL 1 MONTH)

Answer 3

你应该连接它们：

<?php

$query = "CREATE VIEW current_rankings AS SELECT * FROM main_table WHERE date = X";

$query .= " CREATE VIEW previous_rankings AS SELECT rank FROM main_table WHERE date =     date_sub('X', INTERVAL 1 MONTH)";

$query .= " CREATE VIEW final_output AS SELECT current_rankings.player,     current_rankings.rank as current_rank LEFT JOIN previous_rankings.rank as prev_rank
         ON (current_rankings.player = previous_rankings.player)";

$query .= " SELECT *, @rank_change = prev_rank - current_rank as rank_change from final_output";

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

while($row = mysql_fetch_array($result)) {
echo $row['player']. $row['current_rank']. $row['prev_rank']. $row['rank_change'];
}

?>