string 如何将 Int 转换为给定长度的字符串，并带有前导零以对齐？
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How to convert an Int to a String of a given length with leading zeros to align?
提问by Ivan
How can I convert an Intto a 7-character long String, so that 123is turned into "0000123"?
如何将 an 转换Int为 7 个字符的 long String，从而123变成"0000123"？
回答by huynhjl
The Java library has pretty good (as in excellent) number formatting supportwhich is accessible from StringOpsenriched String class:
Java 库具有非常好的（如出色的）数字格式支持，可从StringOps丰富的 String 类访问：
scala> "%07d".format(123)
res5: String = 0000123
scala> "%07d".formatLocal(java.util.Locale.US, 123)
res6: String = 0000123
Edit post Scala 2.10: as suggested by fommil, from 2.10 on, there is also a formatting string interpolator (does not support localisation):
编辑 Scala 2.10 帖子：按照 fommil 的建议，从 2.10 开始，还有一个格式化字符串插值器（不支持本地化）：
val expr = 123
f"$expr%07d"
f"${expr}%07d"
Edit Apr 2019:
2019 年 4 月编辑：
- If you want leading spaces, and not zero, just leave out the
0from the format specifier. In the above case, it'd bef"$expr%7d".Tested in 2.12.8 REPL. No need to do the string replacement as suggested in a comment, or even put an explicit space in front of7as suggested in another comment. - If the length is variable,
s"%${len}d".format("123")
- 如果您想要前导空格而不是零，只需
0从格式说明符中删除。在上述情况下，它将f"$expr%7d"在 2.12.8 REPL 中进行 .Tested。无需按照注释中的建议进行字符串替换，甚至无需7按照另一条注释中的建议在前面放置显式空格。 - 如果长度是可变的，
s"%${len}d".format("123")
回答by Pablo Fernandez
Short answer:
简短的回答：
"1234".reverse.padTo(7, '0').reverse
Long answer:
长答案：
Scala StringOps(which contains a nice set of methods that Scala string objects have because of implicit conversions) has a padTomethod, which appends a certain amount of characters to your string. For example:
Scala StringOps（它包含一组很好的方法，Scala 字符串对象由于隐式转换而具有这些padTo方法）有一个方法，它将一定数量的字符附加到您的字符串中。例如：
"aloha".padTo(10,'a')
Will return "alohaaaaaa". Note the element type of a String is a Char, hence the single quotes around the 'a'.
将返回“alohaaaaaa”。请注意，String 的元素类型是 Char，因此'a'.
Your problem is a bit different since you need to prependcharacters instead of appendingthem. That's why you need to reverse the string, append the fill-up characters (you would be prepending them now since the string is reversed), and then reverse the whole thing again to get the final result.
您的问题有点不同，因为您需要预先添加字符而不是附加它们。这就是为什么您需要反转字符串，附加填充字符（由于字符串已反转，您现在将在前面添加它们），然后再次反转整个内容以获得最终结果。
Hope this helps!
希望这可以帮助！
回答by maxmithun
The paddingis denoted by %02dfor 0to be prefixed to make the length 2:
的padding是由表示%02d为0被前缀以使长度2：
scala> val i = 9
i: Int = 9
scala> val paddedVal = f"${num}%02d"
paddedVal: String = 09
scala> println(paddedVal)
09
回答by Luigi Plinge
huynhjl beat me to the right answer, so here's an alternative:
huynhjl 击败了我的正确答案，所以这里有一个替代方案：
"0000000" + 123 takeRight 7
回答by zeromem
def leftPad(s: String, len: Int, elem: Char): String = {
elem.toString * (len - s.length()) + s
}
回答by som-snytt
In case this Q&A becomes the canonical compendium,
如果此问答成为规范纲要，
scala> import java.text._
import java.text._
scala> NumberFormat.getIntegerInstance.asInstanceOf[DecimalFormat]
res0: java.text.DecimalFormat = java.text.DecimalFormat@674dc
scala> .applyPattern("0000000")
scala> res0.format(123)
res2: String = 0000123
回答by 0__
Do you need to deal with negative numbers? If not, I would just do
你需要处理负数吗？如果没有，我只会做
def str(i: Int) = (i % 10000000 + 10000000).toString.substring(1)
or
或者
def str(i: Int) = { val f = "000000" + i; f.substring(f.length() - 7) }
Otherwise, you can use NumberFormat:
否则，您可以使用NumberFormat：
val nf = java.text.NumberFormat.getIntegerInstance(java.util.Locale.US)
nf.setMinimumIntegerDigits(7)
nf.setGroupingUsed(false)
nf.format(-123)
