blocks|key|3010535|text|首先，您正在使用关系数据库，所以不要使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3010536|+work_schedule+=+[3,4,7]|code-block|syntax|javascript|3010537|数据库规范化：1+Normal+Form|3010538|如果数据库满足以下条件，它将处于第一范式：|blockquote|3010539|只包含原子值|unordered-list-item|3010540|没有重复的组|3010541|原子值是一个不可分割的值。|3010542|不要在一列中存储多个值，除非您将来想问如何分割CSV数据，以及为什么它如此缓慢。|3010543|一种方法是每天使用多个旗子：|3010544|id
name
job
is_working_Monday
is_working_Tuesday
is_working_Wednesday
...|3010545|entityMap|0|LINK|mutability|MUTABLE|url|http://www.1keydata.com/database-normalization/first-normal-form-1nf.php^0|8|2|0|0|7|D|0|0|0|0|6|0|0|0|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|M|3|N|5|O|7|1I|8|@]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|S|3|T|5|R|7|1M|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1N|8|@$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1R|8|@]|D|@]|E|$]]|$1|10|3|11|5|H|7|1S|8|@]|D|@]|E|$I|J]]|$1|12|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|13|$14|$5|15|16|17|E|$18|19]]]]

First of all you are using Relational Databases, so don't use:

<pre><code> work_schedule = [3,4,7]
</code></pre>

Database normalization: <a href="http://www.1keydata.com/database-normalization/first-normal-form-1nf.php" rel="nofollow"><code>1 Normal Form</code></a>:

<blockquote>
 A database is in first normal form if it satisfies the following
 conditions:
 
 <ul>
 <li>Contains only atomic values</li>
 <li>There are no repeating groups</li>
 </ul>
 
 An atomic value is a value that cannot be divided.
</blockquote>

Do not store multiple values in one column, unless you want to ask in future how to split CSV data and why it is so slow. 

One way to go is multiple flags for every day:

<pre><code>id
name
job
is_working_Monday
is_working_Tuesday
is_working_Wednesday
...
</code></pre>

blocks|key|3010556|text|这其实是一个有趣的问题。有几种方法。我可以很容易地想到三个，其中任何一个可能适合在这种情况下。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3010557|有一个单独的表WorkSchedule为每一天的可能组合，当有人可以工作。|unordered-list-item|offset|length|style|CODE|3010558|拥有一个单独的WorkerDays表，该表为每个工作人员和他可以工作的每一天都有一个单独的行。|3010559|将信息存储在一行中。|3010560|中间一个是最“类似SQL”的，因为它是规范化的，并且应该灵活地满足大多数需求。|3010561|第三种选择似乎是你要走的路。一种典型的方法是每天存储一个单独的标志：MondayFlag、TuesdayFlag等。|3010562|另一种方法是将标志存储在单个列中，使用位掩码来标识您想要的信息。当然，这取决于您正在使用的数据库的比特篡改功能。|3010563|如何对数据建模的实际选择取决于如何使用数据。你需要考虑会被问到的工作日问题的类型。|3010564|entityMap^0|0|7|C|0|7|A|0|0|0|Y|A|19|B|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@$E|Y|F|Z|G|H]]|9|@]|A|$]]|$1|I|3|J|5|D|7|10|8|@$E|11|F|12|G|H]]|9|@]|A|$]]|$1|K|3|L|5|D|7|13|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|15|8|@$E|16|F|17|G|H]|$E|18|F|19|G|H]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1B|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|V|$]]

This is actually an interesting question. There are a handful of methods. I can readily think of three, any of which might be appropriate given the circumstances.

<ul>
<li>Have a separate table <code>WorkSchedule</code> for each possible combination of days when someone could work.</li>
<li>Have a separate table of <code>WorkerDays</code> that has a separate row for each worker and each day when s/he could work.</li>
<li>Store the information in one row.</li>
</ul>

The middle one is the most "SQL-like" in the sense that it is normalized, and should be flexible for most needs.

The third alternative seems to be the path you are going down. A typical method is to store a separate flag for each day: <code>MondayFlag</code>, <code>TuesdayFlag</code>, etc.

An alternative method is to store the flags within a single column, using bit-masks to identify the information you want. Of course, this depends on the bit-fiddling capabilities of the database you are working with.

The actual choice of how to model the data depends on how it will be used. You need to think about the types of questions that will be asked about work days.

blocks|key|2543038|text|使用按位运算。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2543039|使用此值常量|2543040|Monday++=+1
Tuesday+=+2
Wednesday+=+4
Thursday+=+8
Friday+=+16
Saturday+=+32
Sunday++=+64|code-block|syntax|javascript|2543041|那么，3，5，7的工作计划将是|2543042|SET+workschedule+=+4+%2B+16+%2B+64;|2543043|并选择在周三找到工作|2543044|SELECT+*
FROM+YourTable
WHERE+workschedule++&+4++>+0+|2543045|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Use bitwise operation.

use this values constant

<pre><code>Monday = 1
Tuesday = 2
Wednesday = 4
Thursday = 8
Friday = 16
Saturday = 32
Sunday = 64
</code></pre>

Then workschedule for 3, 5, 7 will be 

<pre><code>SET workschedule = 4 + 16 + 64;
</code></pre>

And select to get jobs on Wednesday will be

<pre><code>SELECT *
FROM YourTable
WHERE workschedule &amp; 4 &gt; 0 
</code></pre>

I Plan to create a database that consists of these attributes

<ol>
<li>id</li>
<li>name</li>
<li>job</li>
<li>work_schedule</li>
</ol>

For example, I want to input Mike the Janitor, and he works every Wednesday, Thursday and Sunday. How do I input this into SQL effectively ? I've tried to use array (For example work_schedule = [3,4,7]). But, is there anything method that is easier ?

Input multiple data into SQL column

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我计划创建一个由以下属性组成的数据库id名字工作work_schedule例如，我想输入看门人迈克，他每周三、周四和周日都工作。如何有效地将其输入SQL？我尝试使用数组(例如，work_schedule = 3,4,7)。但是，有什么方法更容易吗？

问将多个数据输入SQL列
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