blocks|key|3604457|text|您可以从YEAR、MONTH和DAY函数中获取返回的整数，并使用适当数量的零填充它们(在将它们转换为字符串之后)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3604458|PREFIX+xsd:+<http://www.w3.org/2001/XMLSchema#>
SELECT+*+WHERE+{
++++BIND(2022+AS+?yearInt)+#+this+would+come+from+your+YEAR(?dateTime)+call
++++BIND(2+AS+?monthInt)+++#+this+would+come+from+your+MONTH(?dateTime)+call
++++BIND(13+AS+?dayInt)++++#+this+would+come+from+your+DAY(?dateTime)+call

++++#+convert+to+strings
++++BIND(STR(?yearInt)+AS+?year)
++++BIND(STR(?monthInt)+AS+?month)
++++BIND(STR(?dayInt)+AS+?day)

++++#+pad+with+zeros
++++BIND(CONCAT("00",+?year)+AS+?paddedYear)
++++BIND(CONCAT("0000",+?month)+AS+?paddedMonth)
++++BIND(CONCAT("00",+?day)+AS+?paddedDay)

++++#+extract+the+right+number+of+digits+from+the+padded+strings
++++BIND(SUBSTR(?paddedYear,+STRLEN(?paddedYear)-3)+AS+?fourDigitYear)
++++BIND(SUBSTR(?paddedDay,+STRLEN(?paddedDay)-1)+AS+?twoDigitDay)
++++BIND(SUBSTR(?paddedMonth,+STRLEN(?paddedMonth)-1)+AS+?twoDigitMonth)
++++
++++#+put+it+all+back+together
++++BIND(CONCAT(?fourDigitYear,+"-",+?twoDigitMonth,+"-",+?twoDigitDay)+as+?date)
}|code-block|syntax|javascript|3604459|entityMap^0|4|4|9|5|F|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

You can take the integers you get back from the <code>YEAR</code>, <code>MONTH</code>, and <code>DAY</code> functions and pad them with the appropriate number of zeros (after turning them into strings):
<pre><code>PREFIX xsd: &lt;http://www.w3.org/2001/XMLSchema#&gt;
SELECT * WHERE {
 BIND(2022 AS ?yearInt) # this would come from your YEAR(?dateTime) call
 BIND(2 AS ?monthInt) # this would come from your MONTH(?dateTime) call
 BIND(13 AS ?dayInt) # this would come from your DAY(?dateTime) call

 # convert to strings
 BIND(STR(?yearInt) AS ?year)
 BIND(STR(?monthInt) AS ?month)
 BIND(STR(?dayInt) AS ?day)

 # pad with zeros
 BIND(CONCAT(&quot;00&quot;, ?year) AS ?paddedYear)
 BIND(CONCAT(&quot;0000&quot;, ?month) AS ?paddedMonth)
 BIND(CONCAT(&quot;00&quot;, ?day) AS ?paddedDay)

 # extract the right number of digits from the padded strings
 BIND(SUBSTR(?paddedYear, STRLEN(?paddedYear)-3) AS ?fourDigitYear)
 BIND(SUBSTR(?paddedDay, STRLEN(?paddedDay)-1) AS ?twoDigitDay)
 BIND(SUBSTR(?paddedMonth, STRLEN(?paddedMonth)-1) AS ?twoDigitMonth)
 
 # put it all back together
 BIND(CONCAT(?fourDigitYear, &quot;-&quot;, ?twoDigitMonth, &quot;-&quot;, ?twoDigitDay) as ?date)
}
</code></pre>

blocks|key|3697701|text|@gregory-williams给出了一个便携的答案。另一种方法是F&O+(XPath和XQuery函数和运算符3.1)中的函数“fn:format-.”|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3697702|我不确定各种triplestores中的内容-+Apache提供了fn:format-number，这是问题所需要的，但是fn-format-dateTime等等。|style|CODE|3697703|看见|3697704|https://www.w3.org/TR/xpath-functions-3/#formatting-the-number|3697705|https://www.w3.org/TR/xpath-functions-3/#formatting-dates-and-times|3697706|例如：|3697707|fn:format-number(1,"000")返回字符串"001“。|3697708|Apache还有一个本地扩展afn:sprintf，它使用sprintf的C或sprintf语法。|3697709|afn:sprintf("%2503d",+1)返回"001“。|3697710|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/3101395/gregory-williams|1|2^0|0|H|0|0|X|G|1P|I|0|0|0|1Q|1|0|0|1V|2|0|0|0|P|0|E|B|T|7|13|7|0|0|M|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|6|7|19|8|@$A|1A|B|1B|F|G]|$A|1C|B|1D|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|1F|8|@]|9|@$A|1G|B|1H|1|1I]]|C|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@$A|1K|B|1L|1|1M]]|C|$]]|$1|N|3|O|5|6|7|1N|8|@]|9|@]|C|$]]|$1|P|3|Q|5|6|7|1O|8|@$A|1P|B|1Q|F|G]]|9|@]|C|$]]|$1|R|3|S|5|6|7|1R|8|@$A|1S|B|1T|F|G]|$A|1U|B|1V|F|G]|$A|1W|B|1X|F|G]]|9|@]|C|$]]|$1|T|3|U|5|6|7|1Y|8|@$A|1Z|B|20|F|G]]|9|@]|C|$]]|$1|V|3|-4|5|6|7|21|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|K]]|14|$5|Y|Z|10|C|$11|M]]]]

<a href="https://stackoverflow.com/users/3101395/gregory-williams">@gregory-williams</a> gives a portable answer. An alternative is functions from F&amp;O (XPath and XQuery Functions and Operators 3.1) &quot;fn:format-....&quot;
I'm not sure of the coverage in various triplestores - Apache Jena provides <code>fn:format-number</code>, which is needed for the question, but not <code>fn-format-dateTime</code> etc
See 
<a href="https://www.w3.org/TR/xpath-functions-3/#formatting-the-number" rel="nofollow noreferrer">https://www.w3.org/TR/xpath-functions-3/#formatting-the-number</a> 
<a href="https://www.w3.org/TR/xpath-functions-3/#formatting-dates-and-times" rel="nofollow noreferrer">https://www.w3.org/TR/xpath-functions-3/#formatting-dates-and-times</a>
For example:
<code>fn:format-number(1,&quot;000&quot;)</code> returns the string &quot;001&quot;.
Apache Jena also has a local extension <code>afn:sprintf</code> using the C or Java syntax of <code>sprintf</code>:
<code>afn:sprintf(&quot;%03d&quot;, 1)</code> returns &quot;001&quot;.

I have a datetime in my SPARQL-query that I want to transform to a date.
Therefore I do:
<pre><code> BIND(CONCAT(YEAR(?dateTime), &quot;-&quot;,MONTH(?dateTime), &quot;-&quot;, DAY(?dateTime)) as ?date) 
</code></pre>
This part of code works but returns for example 2022-2-3, I want it to be 2022-02-03. If the dateTime is 2022-11-23, nothing should change.

SPARQL how to make DAY/MONTH always return 2 digits

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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我在SPARQL查询中有一个日期时间，我想将它转换为一个日期。因此，我要： BIND(CONCAT(YEAR(?dateTime), "-",MONTH(?dateTime), "-", DAY(?dateTime)) as ?date)  这部分代码工作，但返回例如2022-2-3，我希望它是2022-02-03。如...

问如何使日/月总是返回2位数
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问如何使日/月总是返回2位数EN