大约有 18,474 项符合查询结果（耗时：0.0142秒） [XML]
scopes with lambda and arguments in Rails 4 style?
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I think it should be:
scope :find_lazy, -> (id) { where(id: id) }
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Avoid duplicates in INSERT INTO SELECT query in SQL Server
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Using NOT EXISTS:
INSERT INTO TABLE_2
(id, name)
SELECT t1.id,
t1.name
FROM TABLE_1 t1
WHERE NOT EXISTS(SELECT id
FROM TABLE_2 t2
WHERE t2.id = t1.id)
Using NOT IN:
INSERT INTO TABLE_2
(id, name)
SELECT t1.id,
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Extension · App Inventor 2 中文网
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Table of Contents:
AI2Utils
AirPlaneState
AliSms
BaiduPan
BrowserPromptHelper
ClientSocketAI2Ext
Clipboard
CloudBaseAuth
CloudBaseDatabase
CloudBaseFunction
CloudBaseMySQL
CloudBaseStorage
CustomWebView
DeepSeek
DownloadHelper
DynamicRecycler
Dynamic...
How to create a date and time picker in Android? [closed]
Is there any android widget that enable to pick the date and the time at the same time ?
I already use the basic time picker and date picker .
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Rails Model find where not equal
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In Rails 4.x (See http://edgeguides.rubyonrails.org/active_record_querying.html#not-conditions)
GroupUser.where.not(user_id: me)
In Rails 3.x
GroupUser.where(GroupUser.arel_table[:user_id].not_eq(me))
To shorten the length, you could store GroupUser....
jquery selector for id starts with specific text [duplicate]
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Use jquery starts with attribute selector
$('[id^=editDialog]')
Alternative solution - 1 (highly recommended)
A cleaner solution is to add a common class to each of the divs & use
$('.commonClass').
But you can use the first one if html markup is not in your ...
What is the “N+1 selects problem” in ORM (Object-Relational Mapping)?
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SELECT * FROM Cars;
And then for each Car:
SELECT * FROM Wheel WHERE CarId = ?
In other words, you have one select for the Cars, and then N additional selects, where N is the total number of cars.
Alternatively, one could get all wheels and perform the lookups in memory:
SELECT * FROM Wheel
Thi...
SELECT * FROM X WHERE id IN (…) with Dapper ORM
...this directly. For example...
string sql = "SELECT * FROM SomeTable WHERE id IN @ids"
var results = conn.Query(sql, new { ids = new[] { 1, 2, 3, 4, 5 }});
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MySQL, update multiple tables with one query
... In case, we increase the number of books in a particular order with Order.ID = 1002 in Orders table then we also need to reduce that the total number of books available in our stock by the same number in Books table.
UPDATE Books, Orders
SET Orders.Quantity = Orders.Quantity + 2,
Books.InStock ...
Does the join order matter in SQL?
... (commutativity and associativity) properties:
a LEFT JOIN b
ON b.ab_id = a.ab_id
LEFT JOIN c
ON c.ac_id = a.ac_id
is equivalent to:
a LEFT JOIN c
ON c.ac_id = a.ac_id
LEFT JOIN b
ON b.ab_id = a.ab_id
but:
a LEFT JOIN b
ON b.ab_id = a.ab_id
LEFT JOIN c
ON c....