blocks|key|1355174|text|只需储存同等数量的青铜。然后使用以下算法尽可能多地转换成黄金，从剩余的尽可能多地转换成银，其余的则保持在青铜中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1355175|1355176|将黄金的数量设置为零，将银设为零，将青铜的数目设置为任何数据库所说的。|ordered-list-item|1355177|将银的数量设置为等于青铜的数量除以10+(使用整数除法)。设置青铜的数量等于青铜的数量10。|1355178|设置黄金的数量等于银的数量除以10(使用整数除法)。将白银的数量设置为等于银mod+10的数目.|1355179|1355180|所以如果你有9327个青铜：|1355181|bronze=9,327|1355182|gold=0，|unordered-list-item|1355183|
|1355184|gold=0，silver=0，+silver=932，bronze=7|1355185|gold=93，silver=2，bronze=7|1355186|1355187|1355188|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|13|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|14|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|15|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|17|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|19|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1A|8|@]|9|@]|A|$]]|$1|T|3|U|5|Q|7|1B|8|@]|9|@]|A|$]]|$1|V|3|W|5|E|7|1C|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|10|$]]

Just store the equivalent number of bronze. Then use the following algorithm to get as much as possible converted to gold, from the remainder as much as possible to silver and the rest stays in bronze:

<ol>
<li>Set the number of gold to zero, silver to zero and bronze to whatever the database says.</li>
<li>Set the number of silver equal to the number of bronze divided by 10 (use integer division). Set the number of bronze equal to the number of bronze mod 10.</li>
<li>Set the number of gold equal to the number of silver divided by 10 (use integer division). Set the number of silver equal to the number of silver mod 10.</li>
</ol>

So if you have 9,327 bronze:

<ol>
<li>gold=0, silver=0, bronze=9,327</li>
<li>gold=0, silver=932, bronze=7</li>
<li>gold=93, silver=2, bronze=7</li>
</ol>

blocks|key|3737343|text|我以前也做过类似的事情，我同意你应该有“货币”一栏，并根据需要计算其他货币。如果有单独的列，则必须依赖触发器或应用程序逻辑来保持它们的同步。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3737344|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I've done something similar before, and I agree that you should just have the one column for "money", and compute other currencies as-needed. If you had separate columns, you'd have to rely on a trigger or application logic to keep them in-sync.

blocks|key|1355150|text|我会把所有的东西都变成青铜，然后把它储存起来。任何类型的交易也将首先被转换为铜牌。你唯一需要转换成白银或黄金的时间就是展示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1355151|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I would convert everything to bronze and just store that value. Any sort of transactions would also be converted to bronze first. The only time you need to convert to silver or gold would be for display.

blocks|key|1353597|text|一种方法是只将所有货币存储在青铜中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1353598|那样的话，如果你有123个青铜，你就可以转换它们。|1353599|123+/+100+++//Gold+with+integer+division+which+should+be+1|code-block|syntax|javascript|1353600|剩下的是123+-+NumGold+*+100+ie+23|offset|length|style|CODE|BOLD|1353601|所以你有|1353602|23+/+10++//+Silver+with+integer+division+should+be+2|1353603|剩下的是23+-+NumSilver+*+10+ie+3铜币。|1353604|这是SQL只知道的铜币，并将它们显示为黄金，银币都在应用层。|1353605|entityMap^0|0|0|0|4|J|R|2|0|0|0|4|J|R|1|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|12|8|@$K|13|L|14|M|N]|$K|15|L|16|M|O]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|17|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|T|3|U|5|6|7|19|8|@$K|1A|L|1B|M|N]|$K|1C|L|1D|M|O]]|9|@]|A|$]]|$1|V|3|W|5|6|7|1E|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|Y|$]]

One way is to only store all currency in Bronze.

That way if you have 123 Bronze you can convert them 

<pre><code>123 / 100 //Gold with integer division which should be 1
</code></pre>

Remaining are <code>123 - NumGold * 100</code> ie 23

so you have 

<pre><code>23 / 10 // Silver with integer division should be 2
</code></pre>

Remaining are <code>23 - NumSilver * 10</code> ie 3 bronze coins.

This was the SQL knows only about Bronze coins and showing them as Gold , Silver is all in the application layer

blocks|key|3720356|text|被接受的答案没有考虑到9999以上的数值，因为它将增加10个以上的银，并给出一个错误的输出。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3720357|最好的解决办法是：|3720358|var+bronze,+silver,+gold;
var+myMoney+=+32141;+//+number+from+database

bronze+=+myMoney+%25+10;
myMoney+=+(myMoney+-+bronze)+/+10;
silver+=+myMoney+%25+10;
gold+=+(myMoney+-+silver)+/+10;|code-block|syntax|javascript|3720359|输出：|3720360|gold+=+321
silver+=+4
bronze+=+1|3720361|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The accepted answer doesn’t account for values over 9999 since it will add 10 more to silver and give a false output. 

The best solution is this:

<pre><code>var bronze, silver, gold;
var myMoney = 32141; // number from database

bronze = myMoney % 10;
myMoney = (myMoney - bronze) / 10;
silver = myMoney % 10;
gold = (myMoney - silver) / 10;
</code></pre>

Output:

<pre><code>gold = 321
silver = 4
bronze = 1
</code></pre>

I have a table called <code>users</code> which currently contains column <code>money</code> of type integer which contains the amount of money the user currently has.

However, I would like to expand and have a couple of more currencies that are based on each other. 1 Gold = 10 Silver = 100 Bronze (or something like that). When you get 100 Bronze, they will convert into 1 Gold. The same with 10 Bronze = 1 Silver. All these currencies have their own column in my table. I have a hook function that is called on every request and updates the money, if possible.

I was told though I could use just one column <code>money</code> and then I can play with variables and take out the different currencies.

But my question is: What would be the best method?

In case the latter method is best: How do I put variables to an already existing object (an object which contains user data; I think it’s called <code>stdClass</code> or something like that)? <code>$user-&gt;username</code>, etc.?

How to store more than one virtual currency in SQL

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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我有一个名为users的表，当前包含列、整数类型的，其中包含用户当前的钱数。然而，我想扩大并有更多的基于彼此的货币。1金= 10银= 100铜(或类似的东西)。当你得到100铜，他们将转换为1金。同样的10铜=1银。所有这些货币在我的表中都有它们自己的列。我有一个钩子函数，它在每个请求中调用并在可能的情况下更新货币。我被告知，虽然我可以使用，只需一列，然后我就可以使用变量并取出不同的货币。但我的问

问如何在SQL中存储多个虚拟货币
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