分割段落的Java：我想在一个变量前50个字从字符串

Question

我

String explanation = "The image-search feature will start rolling out in the next few days, said Johanna Wright, a Google search director. "Every picture has a story, and we want to help you discover that story she said."; 

有字的总数为300

在Java中，我如何才能在第一50个字从字符串？

Answer 1

在这里你去，完美的解释：对于50

搜索：http://www.aliaspooryorik.com/blog/index.cfm/e/posts.details/post/show-the-first-n-and-last-n-words-232

Answer 2

根据您的一个字的定义，这可能会为你做个空格字符，然后从0提取子到那个指数。

这是给你一些示例代码：

public static int nthOccurrence(String str, char c, int n) { 
    int pos = str.indexOf(c, 0); 
    while (n-- > 0 && pos != -1) 
     pos = str.indexOf(c, pos+1); 
    return pos; 
} 


public static void main(String[] args) { 
    String text = "Lorem ipsum dolor sit amet."; 

    int numWords = 4; 
    int i = nthOccurrence(text, ' ', numWords - 1); 
    String intro = i == -1 ? text : text.substring(0, i); 

    System.out.println(intro); // prints "Lorem ipsum dolor sit" 
} 

相关问题：

• How to find nth occurrence of character in a string?

Answer 3

斯普利特用正则表达式的输入数据，边界检查，然后重建前50个字。

String[] words = data.split(" "); 
String firstFifty = ""; 
int max = words.length; 
if (max > 50) 
    max = 50; 
for (int i = 0; i < max; ++i) 
    firstFifty += words[i] + " "; 

Answer 4

你可以尝试这样的事情（如果你想第一个50个字）：

String explanation="The image-search feature will start rolling out in the next few days, said Johanna Wright, a Google search director. "Every picture has a story, and we want to help you discover that story she said." 

String[] words = explanation.split(" "); 
StringBuilder sb = new StringBuilder(); 
for (int i = 0; i < Math.min(50, words.length); i++) 
{ 
sb.append(words[i] + " "); 
} 
System.out.println("The first 50 words are: " + sb.toString()); 

或像这样的东西，如果你想要的前50个字符：

String explanation="The image-search feature will start rolling out in the next few days, said Johanna Wright, a Google search director. "Every picture has a story, and we want to help you discover that story she said." 

String truncated = explanation.subString(0, Math.min(49, explanation.length()));