我对此的解决方案在底部
我的问题是：我正在尝试显示外键数据，但由于有多个外键，我得到每个外键的“重复”查询。
http://i.imgur.com/Gfqx497.png
如您所见，我可以查询正确的数据，但我不知道如何将其他外键数据附加到相同的“单行输出”。
我已经潜伏了一段时间的 stackoverflow 来找到我的问题的答案，但我束手无策。我发现了很多线程，例如下面的两个链接，我相信人们在问同样的事情，但是我似乎无法全神贯注地让解决方案在我的情况下工作。据我了解，我需要为表使用别名，但是我尝试了多种不同的解决方案解释，但无法重新创建解决方案。
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我有两张表（“奴才和能力”），其中一张有四个外键链接到另一张。
http://i.imgur.com/ctpFHur.png
这是我用于查询的 php 代码，主要取自 PHP 和 MySQL Web Development 第 4 版（Welling，Thomson），我购买它是为了让我开始使用 php 和 mysql。
$query = "SELECT minions.name, minions.summon, minions.attack,
minions.health, minions.race, minions.rarity,
minions.ability1, minions.ability2, minions.ability3,
minions.ability4, minions.imagebig,
ability.ability
AS ability FROM minions
INNER JOIN ability on
minions.ability1 = ability.abilityid
OR minions.ability2=ability.abilityid";
//Only trying for 2 foreign keys to try get it to work
$result = $db->query($query);
$num_results = $result->num_rows;
echo "<p>Number of items found: ".$num_results."</p>";
for ($i=0; $i <$num_results; $i++){
$row = $result->fetch_assoc();
//echo "<p><strong>".($i+1).". Name: ";
echo "<p><strong>";
echo htmlspecialchars(stripslashes($row['name']));
echo "</strong><br />Summoning cost: ";
echo stripslashes($row['summon']);
echo "<br />Attack: ";
echo stripslashes($row['attack']);
echo "<br />Health: ";
echo stripslashes($row['health']);
echo "<br />Race: ";
echo stripslashes($row['race']);
echo "<br />Rarity: ";
echo stripslashes($row['rarity']);
//if (stripslashes($row['ability'] != NULL)){
echo "<br />Abilty: ";
echo stripslashes($row['ability']);
//}
echo "<br />";
$imageMinion = stripslashes($row['imagebig']);
// $iwidth = 25;
// $iheight = 100;
// echo '<img src="img/'.$imageMinion.'.png" style="width:'.$iwidth.'px;height:'.$iheight.'px;">';
//echo "<br />";
echo '<img src="img/'.$imageMinion.'.png">';
echo "</p>";
有人可以指导我让它正确显示吗？我试图遵循其他解决方案，但似乎无法正确命名别名，如果我认为这是解决方案是正确的。
========编辑关于verbumSapienti的答案===========
我很尴尬地无法让您的答案正常工作。这就是代码的外观。
$query = "SELECT minions.name, minions.summon, minions.attack, minions.health,
minions.race, minions.rarity, minions.ability1, minions.ability2,
minions.ability3, minions.ability4, minions.imagebig,
ability.ability
AS ability
FROM minions
INNER JOIN ability
ON minions.ability1 = ability.abilityid
OR minions.ability2 = ability.abilityid
OR minions.ability3 = ability.abilityid
OR minions.ability4 = ability.abilityid";
$result = $db->query($query);
$num_results = $result->num_rows;
echo "<p>Number of items found: ".$num_results."</p>";
for ($i=0; $i <$num_results; $i++){
$row = $result->fetch_assoc();
$abilities = array('ability1', 'ability2', 'ability3', 'ability4');
foreach($abilities as $ability)
{
$q = "SELECT $ability FROM minions WHERE name={$row['name']}";
$result = $db->query($q);
$row2 = $result->fetch_assoc();
$abilitiesArr[] = $row2[$ability];
}
echo "<p><strong>";
echo htmlspecialchars(stripslashes($row['name']));
echo "</strong><br />Summoning cost: ";
echo stripslashes($row['summon']);
echo "<br />Attack: ";
echo stripslashes($row['attack']);
echo "<br />Health: ";
echo stripslashes($row['health']);
echo "<br />Race: ";
echo stripslashes($row['race']);
echo "<br />Rarity: ";
echo stripslashes($row['rarity']);
foreach($abilitiesArr as $ability)
{
$q = "SELECT $ability FROM ability";
$result = $db->query($q);
$row = $result->fetch_assoc();
echo "<br />Ability: $row";
}
/*if (stripslashes($row['ability'] != NULL)){
echo "<br />Abilty: ";
echo stripslashes($row['ability']);
}*/
echo "<br />";
$imageMinion = stripslashes($row['imagebig']);
echo '<img src="img/'.$imageMinion.'.png">';
echo "</p>";
}
我尝试过改变一些事情，但没有取得任何成功。照原样，我收到以下错误：
Fatal error: Call to a member function fetch_assoc() on a non-object in D:\Xampp\htdocs\ocduels\results.php on line 87
这是：
$row2 = $result->fetch_assoc();
在：
$abilities = array('ability1', 'ability2', 'ability3', 'ability4');
foreach($abilities as $ability)
{
$q = "SELECT $ability FROM minions WHERE name={$row['name']}";
$result = $db->query($q);
$row2 = $result->fetch_assoc();
$abilitiesArr[] = $row2[$ability];
}
::我对此的解决方案:: 这似乎有效。我认为它效率不高，但足以让我继续学习。谢谢大家的回复。这允许我找到一个“Minion”，并且当有超过 1 个带数据的外键时，只有一个“Minion”实例。
$query = "SELECT
m.name as m_name,
m.summon as m_summon,
m.attack as m_attack,
m.health as m_health,
m.race as m_race,
m.rarity as m_rarity,
m.ability1 as m_ability1,
m.ability2 as m_ability2,
aa.ability as a_ability,
ab.ability as b_ability,
m.imagebig as m_imagebig
FROM minions m
LEFT JOIN ability aa
ON m.ability1 = aa.abilityid
LEFT JOIN ability ab
ON m.ability2 = ab.abilityid";
$result = $db->query($query);
$num_results = $result->num_rows;
echo "<p>Number of items found: ".$num_results."</p>";
for ($i=0; $i <$num_results; $i++){
$row = $result->fetch_assoc();
echo "<p><strong>";
echo htmlspecialchars(stripslashes($row['m_name']));
echo "</strong><br />Summoning cost: ";
echo stripslashes($row['m_summon']);
echo "<br />Attack: ";
echo stripslashes($row['m_attack']);
echo "<br />Health: ";
echo stripslashes($row['m_health']);
echo "<br />Race: ";
echo stripslashes($row['m_race']);
echo "<br />Rarity: ";
echo stripslashes($row['m_rarity']);
if (stripslashes($row['a_ability'] != NULL)){
echo "<br />Ability 1: ";
echo stripslashes($row['a_ability']);
}
if (stripslashes($row['b_ability'] != NULL)){
echo "<br />Ability 2: ";
echo stripslashes($row['b_ability']);
}
echo "<br />";
$imageMinion = stripslashes($row['m_imagebig']);
echo '<img src="img/'.$imageMinion.'.png">';
echo "</p>";
}