Java 如何将 StringBuilder 最好地转换为 String[]？
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原文地址: http://stackoverflow.com/questions/3472234/
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How can a StringBuilder best be converted to a String[]?
提问by Hymannad
The following code sort of works, but fixes the number of elements in String[]. Is there a way to make a String[] add the number of elements needed dynamically?
以下代码可以正常工作，但修复了 String[] 中元素的数量。有没有办法让 String[] 添加动态所需的元素数量？
private static StringBuilder names = new StringBuilder();
...
public String[] getNames() {
int start = 0;
int end = 0;
int i = 0;
String[] nameArray = {"","","",""};
while (-1 != end) {
end = names.indexOf(TAB, start);
nameArray[i++] = names.substring(start, end);
start = ++end; // The next name is after the TAB
}
return nameArray;
}
采纳答案by Jon Skeet
So you're just trying to split on tab? How about:
所以你只是想在选项卡上拆分？怎么样：
return names.toString().split(TAB);
Note that splittakes a regular expression pattern - so don't expect split(".")to split just on dots, for example :)
请注意，split它采用正则表达式模式 - 因此不要期望split(".")仅在点上拆分，例如 :)
回答by Pablo Santa Cruz
回答by Alexander Pogrebnyak
To dynamically grow array, use ArrayList<String>, you can even convert the result to String[]if that's what your API requires.
要动态增长数组，请使用ArrayList<String>，您甚至可以将结果转换为String[]您的 API 要求的结果。
ArrayList<String> namesList = new ArrayList<String>( );
while (-1 != end) {
end = names.indexOf(TAB, start);
namesList.add( names.substring(start, end) );
start = ++end; // The next name is after the TAB
}
return namesList.toArray( new String[ namesList.size( ) ] );
That said, for your purposes use splitas suggested by others
也就是说，出于您的目的，请split按照其他人的建议使用
回答by x4u
You can use a recursive implementation to use the program stack as a temporary array.
您可以使用递归实现将程序堆栈用作临时数组。
public String[] getNames()
{
return getNamesRecursively( names, 0, TAB, 0 );
}
private static String[] getNamesRecursively( StringBuilder str, int pos, String delimiter, int cnt )
{
int end = str.indexOf( delimiter, pos );
String[] res;
if( end >= 0 )
res = getNamesRecursively( str, end + delimiter.length(), delimiter, cnt + 1 );
else
{
res = new String[ cnt + 1 ];
end = str.length();
}
res[ cnt ] = str.substring( pos, end );
return res;
}
回答by Tuhin
StringBuilder t= new StringBuilder();
String s= t.toString();
回答by Ashwin H
String myLocation = builder.toString();
String myLocation = builder.toString();
