我将使用包含方法和toUpper方法的组合，它们是String类的一部分。下面是一个例子:

String string1 = "AAABBBCCC"; 
String string2 = "DDDEEEFFF";
String searchForThis = "AABB";

System.out.println("Search1="+string1.toUpperCase().contains(searchForThis.toUpperCase()));

System.out.println("Search2="+string2.toUpperCase().contains(searchForThis.toUpperCase()));

这将返回:

搜索 1=真 搜索 2=假

我将使用包含方法和toUpper方法的组合，它们是String类的一部分。下面是一个例子:

String string1 = "AAABBBCCC"; 
String string2 = "DDDEEEFFF";
String searchForThis = "AABB";

System.out.println("Search1="+string1.toUpperCase().contains(searchForThis.toUpperCase()));

System.out.println("Search2="+string2.toUpperCase().contains(searchForThis.toUpperCase()));

这将返回:

搜索 1=真 搜索 2=假

你可以使用toLowerCase()方法:

public boolean contains( String haystack, String needle ) {
  haystack = haystack == null ? "" : haystack;
  needle = needle == null ? "" : needle;

  // Works, but is not the best.
  //return haystack.toLowerCase().indexOf( needle.toLowerCase() ) > -1

  return haystack.toLowerCase().contains( needle.toLowerCase() )
}

然后调用它使用:

if( contains( str1, str2 ) ) {
  System.out.println( "Found " + str2 + " within " + str1 + "." );
}

注意，通过创建自己的方法，可以重用它。然后，当有人指出您应该使用contains而不是indexOf时，您只需要修改一行代码。

我也支持RegEx解决方案。代码将更加清晰。在我知道字符串会很大的情况下，我会犹豫使用toLowerCase()，因为字符串是不可变的，必须被复制。此外，matches()解决方案可能令人困惑，因为它将正则表达式作为参数(搜索“Need$le”可能会有问题)。

以上面的一些例子为基础:

public boolean containsIgnoreCase( String haystack, String needle ) {
  if(needle.equals(""))
    return true;
  if(haystack == null || needle == null || haystack .equals(""))
    return false; 

  Pattern p = Pattern.compile(needle,Pattern.CASE_INSENSITIVE+Pattern.LITERAL);
  Matcher m = p.matcher(haystack);
  return m.find();
}

example call: 

String needle = "Need$le";
String haystack = "This is a haystack that might have a need$le in it.";
if( containsIgnoreCase( haystack, needle) ) {
  System.out.println( "Found " + needle + " within " + haystack + "." );
}

(注意:你可能想要根据你的需要不同地处理NULL和空字符串。我认为他们的方式，我有它更接近Java规范的字符串。)

Speed critical solutions could include iterating through the haystack character by character looking for the first character of the needle. When the first character is matched (case insenstively), begin iterating through the needle character by character, looking for the corresponding character in the haystack and returning "true" if all characters get matched. If a non-matched character is encountered, resume iteration through the haystack at the next character, returning "false" if a position > haystack.length() - needle.length() is reached.

那么matches()呢?

String string = "Madam, I am Adam";

// Starts with
boolean  b = string.startsWith("Mad");  // true

// Ends with
b = string.endsWith("dam");             // true

// Anywhere
b = string.indexOf("I am") >= 0;        // true

// To ignore case, regular expressions must be used

// Starts with
b = string.matches("(?i)mad.*");

// Ends with
b = string.matches("(?i).*adam");

// Anywhere
b = string.matches("(?i).*i am.*");

如果你能够使用org.apache.commons.lang.StringUtils，我建议使用以下方法:

String container = "aBcDeFg";
String content = "dE";
boolean containerContainsContent = StringUtils.containsIgnoreCase(container, content);