[英]Trim String based on Offset and Count
Is there an easy way to trim a String based on his Offset and Count values, that is, a function that returns OO
for a String FOOBAR
with an offset
of 1 and a count
of 2? 是否有一种简单的方法可以根据其偏移量和计数值来修剪字符串,也就是说,该函数为
offset
为1且count
为2的字符串FOOBAR
返回OO
?
Obvious, one could write a simple function for this task, but I wonder if their is not a predefined Java functionality for this? 显然,可以为此任务编写一个简单的函数,但是我想知道它们是否不是为此目的预定义的Java功能?
//edit: To clearify: I do mean the offset
and count
values defined within the given String
, not external Integer
values. // edit:澄清一下:我的意思是在给定
String
定义的offset
和count
数值,而不是外部Integer
值。
EDIT: Okay, now you've made your question clearer, it sounds like this is the scenario you're talking about, and the solution using the String(String)
constructor: 编辑:好的,现在您已经使问题更清楚了,听起来这就是您正在谈论的场景,以及使用
String(String)
构造函数的解决方案:
// offset = 0, count = 6, backing array = { 'F', 'O', 'O', 'B', 'A', 'R' }
String original = "FOOBAR";
// offset = 1, count = 2, backing array = { 'F', 'O', 'O', 'B', 'A', 'R' }
String substring = original.substring(2);
// offset = 0, count = 2, backing array = { 'O', 'O' }
String trimmed = new String(substring);
Yup, substring
: 是的,
substring
:
String substring = text.substring(offset, offset + count);
substring
takes two parameters - the "begin index" (inclusive) and the "end index" (exclusive) - hence the addition in the above code. substring
具有两个参数-“ begin index”(包含)和“ end index”(不含)-因此在上面的代码中添加了参数。
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