加入問題MYSQL / PHP

Question

我有一個問題，我以前從未使用過JOIN，這是第一次，我遇到了一些問題：

<?php
//$count to keep the counter go from 0 to new value
    $count=0;

//I need to select the level for the users building first, meanwhile i also 
//need to get the money_gain from the table buildings, which is a table that 
//is common for each member, which means it doesnt have a userid as the other table!
//And then for each of the buildings there will be a $counting 

    $qu1 = mysql_query("SELECT building_user.level,buildings.money_gain FROM building_user,buildings WHERE building_user.userid=$user");

    while($row = mysql_fetch_assoc($qu1))
    {
     $count=$count+$row['level'];
     echo $row['level'];
        }
?>

因此，基本上我已經聽說過你應該將它們與一個公共專欄聯系在一起，但是在這種情況下，他們還沒有..我現在就迷路了嗎？

編輯哦，對了，我也需要使用正確的money_gain來獲取正確的級別，在building_user中使用它的“ buildingid”，在建築物中使用它的“ id”！ 不知道如何發表共同聲明！

Answer 1

我認為您的問題是，MySQL不知道如何聯接兩個表。 因此，您必須告訴MySQL如何做到這一點。

例

SELECT * FROM TABLE1 INNER JOIN Table1.col1 ON TABLE2.col2;

其中col1和col2是要連接的列（唯一標識符）

Answer 2

<?php

$count=0;

$qu1 = mysql_query("SELECT bu.level, bs.money_gain FROM building_user bu, buildings bs WHERE bu.userid=$user");

while($row = mysql_fetch_assoc($qu1))
{
 $count=$count+$row['level'];
 echo $row['level'];
}

?>

Answer 3

根據您的修改，

SELECT building_user.level,buildings.money_gain FROM building_user,buildings WHERE building_user.userid=$user AND building_user.buildingid = building.id;

從本質上來說，您可以獲得在建築物ID中加入它們的用戶的記錄

從性能角度來看，聯接是一個更好的選擇，但是對於諸如此類的輕量級查詢，上面的查詢應該可以正常工作。

您還可以為每列指定更整潔的名稱

SELECT building_user.level as level,buildings.money_gain as money gain FROM building_user,buildings WHERE building_user.userid=$user AND building_user.buildingid = building.id;

並以

$level = $row['level'];
$gain  = $row['gain'];

Answer 4

嘗試這個

$qu1 = mysql_query("SELECT building_user.level,buildings.money_gain 
                    FROM building_user 
                    JOIN buildings  ON building_user.building_id = buildings.id 
                    WHERE building_user.userid=$user");

building_user.building_id是表building_user的前鍵，並且

buildings.id是餐桌buildings的主鍵