org.apache.commons.lang3.StringUtils.lastOrdinalIndexOf()方法的使用及代码示例

x33g5p2x  于2022-01-17 转载在 其他  
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本文整理了Java中org.apache.commons.lang3.StringUtils.lastOrdinalIndexOf()方法的一些代码示例,展示了StringUtils.lastOrdinalIndexOf()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。StringUtils.lastOrdinalIndexOf()方法的具体详情如下:
包路径:org.apache.commons.lang3.StringUtils
类名称:StringUtils
方法名:lastOrdinalIndexOf

StringUtils.lastOrdinalIndexOf介绍

[英]Finds the n-th last index within a String, handling null. This method uses String#lastIndexOf(String).

A null String will return -1.

StringUtils.lastOrdinalIndexOf(null, *, *)          = -1 
StringUtils.lastOrdinalIndexOf(*, null, *)          = -1 
StringUtils.lastOrdinalIndexOf("", "", *)           = 0 
StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 1)  = 7 
StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 2)  = 6 
StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 1)  = 5 
StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 2)  = 2 
StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 1) = 4 
StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 2) = 1 
StringUtils.lastOrdinalIndexOf("aabaabaa", "", 1)   = 8 
StringUtils.lastOrdinalIndexOf("aabaabaa", "", 2)   = 8

Note that 'tail(CharSequence str, int n)' may be implemented as:

str.substring(lastOrdinalIndexOf(str, "\n", n) + 1)

[中]查找字符串中最后第n个索引,处理null。此方法使用字符串#lastIndexOf(String)。
空字符串将返回-1。

StringUtils.lastOrdinalIndexOf(null, *, *)          = -1 
StringUtils.lastOrdinalIndexOf(*, null, *)          = -1 
StringUtils.lastOrdinalIndexOf("", "", *)           = 0 
StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 1)  = 7 
StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 2)  = 6 
StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 1)  = 5 
StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 2)  = 2 
StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 1) = 4 
StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 2) = 1 
StringUtils.lastOrdinalIndexOf("aabaabaa", "", 1)   = 8 
StringUtils.lastOrdinalIndexOf("aabaabaa", "", 2)   = 8

注意,“tail(CharSequence str,int n)”可以实现为:

str.substring(lastOrdinalIndexOf(str, "\n", n) + 1)

代码示例

代码示例来源:origin: JpressProjects/jpress

/**
 * 只匹配 h5 的模板 ,如果匹配不到 h5 ,返回 null
 * <p>
 * 例如:
 * 需要 aa_bb_cc_dd_h5.html
 * 寻找的顺序是:aa_bb_cc_h5.html  ->   aa_bb_h5.html  ->   aa_h5.html
 *
 * @param template
 * @return
 */
private String matchH5Template(String template) {
  while (StringUtils.countMatches(template, '_') > 1) {
    int sLastIndex = StringUtils.lastOrdinalIndexOf(template, "_", 2);
    template = template.substring(0, sLastIndex) + "_h5.html";
    if (htmls.contains(template)) {
      return template;
    }
  }
  return htmls.contains(template) ? template : null;
}

代码示例来源:origin: org.apache.commons/commons-lang3

@Test
public void testLastOrdinalIndexOf() {
  assertEquals(-1, StringUtils.lastOrdinalIndexOf(null, "*", 42) );
  assertEquals(-1, StringUtils.lastOrdinalIndexOf("*", null, 42) );
  assertEquals(0, StringUtils.lastOrdinalIndexOf("", "", 42) );
  assertEquals(7, StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 1) );
  assertEquals(6, StringUtils.lastOrdinalIndexOf("aabaabaa", "a", 2) );
  assertEquals(5, StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 1) );
  assertEquals(2, StringUtils.lastOrdinalIndexOf("aabaabaa", "b", 2) );
  assertEquals(4, StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 1) );
  assertEquals(1, StringUtils.lastOrdinalIndexOf("aabaabaa", "ab", 2) );
  assertEquals(8, StringUtils.lastOrdinalIndexOf("aabaabaa", "", 1) );
  assertEquals(8, StringUtils.lastOrdinalIndexOf("aabaabaa", "", 2) );
}

代码示例来源:origin: org.xworker/xworker_core

public static int lastOrdinalIndexOf(ActionContext actionContext){
  Thing self = actionContext.getObject("self");
  CharSequence cs1  = (CharSequence) self.doAction("getCs1", actionContext);
  CharSequence cs2  = (CharSequence) self.doAction("getCs2", actionContext);
  Integer startPos = (Integer) self.doAction("getStartPos", actionContext);
  return StringUtils.lastOrdinalIndexOf(cs1, cs2, startPos);
}

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