php - 根据第一个表的id查询多个表

Question

我需要用来自两个不同表的数据填充一个数据表。很简单，我想，只需加入或子查询。不幸的是，第二个表不是基于 ID，所以我不能过滤那个。即使可以，我也不知道如何将其放入数据表中。

我已经搜索了好几天了，还是没找到..

表 wp_mollie_forms_registrations 有:

#id  # description #
#----#-------------#
#100 # Race #
#101 # Pull #
####################

表 wp_mollie_forms_registration_fields 有:

#id  # field # value
#----#------#-------#
#100 # Naam # Theun #
#100 # E-mail # test@test.com #
#100 # Leeftijd # 28 #
#100 # Soort voertuig # Auto #
#100 # Betaalmethode # ideal #
#101 # Naam # Theun    #
#101 # E-mail# quest@write.nl #
#101 # Woonplaats # Groningen #
#101 # Merk en type # New Holland #
#101 # Gewichtsklasse # 2.8T #
#101 # Betaalmethode # ideal #
#####################

这是代码:

$query = "select * from A";
$items_result = mysqli_query($conn,$query) or die;

if ($items_result->num_rows > 0) {
echo "<table id='table_id' class='display'><thead><tr><th>ID</th> 
<th>description</th><th>Name</th><th>Age</th><th>Email</th></tr></thead> 
</tbody>";
while ($row = mysqli_fetch_assoc($items_result)){
echo "<tr><td>".$row["id"]."</td><td>".$row["description"]."</td> 
<td>".$Name."</td><td>".$row["Age"]."</td><td>".$row["Email"]."</td>
</tr>";
}

我将如何做以下事情？:从 table_A 中选择 * 并使用 id 来选择姓名、年龄和电子邮件，将此信息放入我的数据表中并转到下一行？

编辑:它有效，但没有显示 Naam(姓名)电子邮件和年龄(leeftijd)我现在有:

$query = "SELECT wp_mollie_forms_registrations.id, wp_mollie_forms_registrations.description, tn.value AS 'Naam', te.value AS 'E-mail', ta.value AS 'Leeftijd' ".
"FROM wp_mollie_forms_registrations".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'Naam') tn ON wp_mollie_forms_registrations.id = tn.registration_id".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'E-mail') te ON wp_mollie_forms_registrations.id = te.registration_id".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'Leeftijd') ta ON wp_mollie_forms_registrations.id = ta.registration_id";
if(!mysqli_query($conn, $query)){ echo "Error: ".mysqli_error($conn); }
$items_result = mysqli_query($conn,$query) or die;

if ($items_result->num_rows > 0) {
echo "<table id='table_id' class='display'><thead><tr><th>ID</th><th>description</th><th>Name</th><th>Age</th><th>Email</th></tr></thead></tbody>";
while ($row = mysqli_fetch_assoc($items_result)){
    echo "<tr><td>".$row["id"]."</td><td>".$row["description"]."</td><td>".$row["tn.value"]."</td><td>".$row["ta.value"]."</td><td>".$row["te.value"]."</td>
</tr>";
}
    echo "</tbody></table>";
} else {
    echo "0 results";
}

Answer 1

根据您的回答，这是应该有效的代码:

$query = "SELECT wp_mollie_forms_registrations.id as 'ID', wp_mollie_forms_registrations.description as 'Description', tn.value AS 'Naam', te.value AS 'Email', ta.value AS 'Leeftijd' ".
"FROM wp_mollie_forms_registrations".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'Naam') tn ON wp_mollie_forms_registrations.id = tn.registration_id".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'E-mail') te ON wp_mollie_forms_registrations.id = te.registration_id".
" INNER JOIN (SELECT registration_id, value FROM wp_mollie_forms_registration_fields WHERE field = 'Leeftijd') ta ON wp_mollie_forms_registrations.id = ta.registration_id";
if(!mysqli_query($conn, $query)){ echo "Error: ".mysqli_error($conn); }
$items_result = mysqli_query($conn,$query) or die;

if ($items_result->num_rows > 0) {
echo "<table id='table_id' class='display'><thead><tr><th>ID</th><th>description</th><th>Name</th><th>Age</th><th>Email</th></tr></thead></tbody>";
while ($row = mysqli_fetch_assoc($items_result)){
    echo "<tr><td>".$row["ID"]."</td><td>".$row["Description"]."</td><td>".$row["Naam"]."</td><td>".$row["Leeftijd"]."</td><td>".$row["Email"]."</td>
</tr>";
}
    echo "</tbody></table>";
} else {
    echo "0 results";
}

您没有让某些列显示的原因是因为您在 PHP 的“echo”语句中有错误的值。例如，tn.value as Naam 将 tn.value 定义为“Naam”。因此，更改 PHP 中的值以反射(reflect)应该让查询工作。

为了更详细地描述查询正在做什么，这里是一个超简化的版本。假设您有表 A 和 B，每个表都有一个字段“id”和“value”。看看这个查询:

SELECT A.value, B.value
FROM A
INNER JOIN B
ON A.id = B.id

这将从表 A 和 B 中选择值并返回它们，但前提是 A 中的 ID 与 B 中的 ID 之间存在匹配。您可以阅读有关联接的更多信息 Here .

子查询部分比较简单。它从表 B 中选择所有“registration_id”和“value”，其中字段是特定类型，例如“Naam”。