Php 在Joomla中构建查询的不同方法？

刚开始，这应该是一个简单的，但我还没有找到一个好的来源，特别是在1.5到1.6/1.7/2.5的转换

构建组件和其他问题会不断遇到语法问题

例如，我构建查询的一种方法是： [类型1]

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ')
->FROM (' #__users AS u')
->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ')
->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ')
->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');
$query->ORDER ('u.name ASC');

$query= "SELECT `u`.`id` as UserID
    , `u`.`name` AS Name
    , `uppi`.`profile_value` AS Hair
    , `uppi2`.`profile_value` AS Height
FROM `#__users` AS u
LEFT JOIN `#__user_profiles` AS uppi ON `u`.`id` = `uppi`.`user_id` AND `uppi`.`ordering` = 1 
LEFT JOIN `#__user_profiles` AS uppi2 ON `u`.`id` = `uppi2`.`user_id` AND `uppi2`.`ordering` = 2
    ";

另一种方式： [类型2]

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ')
->FROM (' #__users AS u')
->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ')
->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ')
->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');
$query->ORDER ('u.name ASC');

$query= "SELECT `u`.`id` as UserID
    , `u`.`name` AS Name
    , `uppi`.`profile_value` AS Hair
    , `uppi2`.`profile_value` AS Height
FROM `#__users` AS u
LEFT JOIN `#__user_profiles` AS uppi ON `u`.`id` = `uppi`.`user_id` AND `uppi`.`ordering` = 1 
LEFT JOIN `#__user_profiles` AS uppi2 ON `u`.`id` = `uppi2`.`user_id` AND `uppi2`.`ordering` = 2
    ";

令人困惑的是，像“LEFTJOIN”这样的mysql调用是一个词而不是两个词。因此，如果我想在重复密钥更新时使用

INSERT…

我完全迷路了

这是第三点： [类型3]

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ')
->FROM (' #__users AS u')
->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ')
->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ')
->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');
$query->ORDER ('u.name ASC');

$query= "SELECT `u`.`id` as UserID
    , `u`.`name` AS Name
    , `uppi`.`profile_value` AS Hair
    , `uppi2`.`profile_value` AS Height
FROM `#__users` AS u
LEFT JOIN `#__user_profiles` AS uppi ON `u`.`id` = `uppi`.`user_id` AND `uppi`.`ordering` = 1 
LEFT JOIN `#__user_profiles` AS uppi2 ON `u`.`id` = `uppi2`.`user_id` AND `uppi2`.`ordering` = 2
    ";

这是我看到的另一种生成查询但无法开始工作的方式[编辑的[/strong>-现在可以工作了，谢谢@cppl！] [Type 4]

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ')
->FROM (' #__users AS u')
->LEFTJOIN (' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ')
->LEFTJOIN (' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ')
->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');

$query->SELECT('u.id as UserID
    , u.name AS Name
    , uppi.profile_value AS Hair
    , uppi2.profile_value AS Height
    ');
$query->FROM (' #__users AS u');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi ON u.id = uppi.user_id AND uppi.ordering = 1 ');
$query->JOIN ('LEFT', ' #__user_profiles AS uppi2 ON u.id = uppi2.user_id AND uppi2.ordering = 2 ');
$query->GROUPBY (' u.id
    , u.name
    , uppi.profile_value
    ');
$query->ORDER ('u.name ASC');

$query= "SELECT `u`.`id` as UserID
    , `u`.`name` AS Name
    , `uppi`.`profile_value` AS Hair
    , `uppi2`.`profile_value` AS Height
FROM `#__users` AS u
LEFT JOIN `#__user_profiles` AS uppi ON `u`.`id` = `uppi`.`user_id` AND `uppi`.`ordering` = 1 
LEFT JOIN `#__user_profiles` AS uppi2 ON `u`.`id` = `uppi2`.`user_id` AND `uppi2`.`ordering` = 2
    ";

我已经阅读了Joomla MVC教程和Lynda…除了“这里有一些代码错误”之外，没有一个真正解决这个问题

问题：

1） 这两者之间有什么区别吗

2） 是否有理由使用一种方法而不是另一种方法

3） 是否有一个很好的来源来学习构建这些模型的不同方法？我搜索了几个小时，没有找到任何全面的

谢谢

类型1、2和3使用较新的JDatabaseQuery对象，而类型4使用较旧的样式将查询设置为SQL字符串

如果类型4不起作用，是否会出现SQL错误？（您是否已打开Joomla的调试模式，以便可以将SQL错误与应用程序配置文件一起转储？）

JDatabaseQuery用于提供支持多个SQL数据库而不仅仅是mySQL所需的抽象，这就是为什么

$query

对象具有类似

select（）

，

where（）

，

join（）

等函数的原因

请记住，

->leftJoin（）

是对名为

leftJoin（）

的JDatabaseQuery函数的PHP调用，因此必须是一个单词ie。它不能是两个

leftJoin（）

实际上只调用了

join（'LEFT'，$conditions）

，因此它是一个可读性实现，而不是一个功能性差异

因此，类型1、2和3实际上是相同的，它们只是具有不同程度的可读性（例如，IMHO类型2和3比类型1更难阅读和维护）

你似乎已经找到了Joomla的所有在线资源！文档，除了Joomla！谷歌集团（，和）

唯一的其他来源是Andrew Eddies“”。我自己也没用过安德鲁的东西，但我在上个月听到了很多关于它的好消息！我参加的那一天

编辑 基于这个错误，我会说这是因为您将标识符

u.id

封装为一个标识符。尝试以下方法：

SELECT `u`.`id` AS "UserID" FROM `#__users` AS `u`

---

我不知道Joomla的情况，但总的来说，我尽量不在重复密钥更新时使用

。我更喜欢尝试一个查询并生成一个异常/错误，然后用第二个查询处理该错误。我使用了很多不同的ORM和DBAL层，它们中没有一个从API中支持这一功能。通常，要实现这一功能，您必须还原到任何原始的SQL工具。啊，说得对！我现在所做的不会在其他任何地方使用，但在我学习的过程中，我肯定会保留这些东西。谢谢关于2）-调试器说“JDatabaseMySQLi:：query:1054-字段列表”中的未知列'u.id'。我可能没有正确的“或”-我看到了一些不同的变化，不确定什么是正确的。但如果这已经过时了，我甚至都不会担心。太棒了——真管用！我甚至都没想到那会起作用，那些小东西完全把我绊倒了。谢谢你两次！