原文引自：http://www.3doing.net/forums/dispbbs.asp?boardID=57&ID=1351&page=1
StringUtils類使用
檢查字符串是否為空或null或僅僅包含空格
??String test = "";
??String test1=" ";
??String test2 = "\n\n\t";
??String test3 = null;
??System.out.println( "test blank? " + StringUtils.isBlank( test ) );
??System.out.println( "test1 blank? " + StringUtils.isBlank( test1 ) );
??System.out.println( "test2 blank? " + StringUtils.isBlank( test2 ) );
??System.out.println( "test3 blank? " + StringUtils.isBlank( test3 ) );
??運行結果：
??test blank? true
??test1 blank? true
??test2 blank? true
??test3 blank? true
??相對應的還有一個StringUtils.isNotBlank(String str)
??StringUtils.isEmpty(String str)則檢查字符串是否為空或null（不檢查是否僅僅包含空格）
??
??分解字符串
??StringUtils.split(null, *, *)????????????= null
??StringUtils.split("", *, *)??????????????= []
??StringUtils.split("ab de fg", null, 0)?? = ["ab", "cd", "ef"]
??StringUtils.split("ab?? de fg", null, 0) = ["ab", "cd", "ef"]
??StringUtils.split("ab:cd:ef", ":", 0)????= ["ab", "cd", "ef"]
??StringUtils.split("ab:cd:ef", ":", 1)????= ["ab:cd:ef"]
??StringUtils.split("ab:cd:ef", ":", 2)????= ["ab", "cd:ef"]
??StringUtils.split(String str,String separatorChars,int max) str為null時返回null
??separatorChars為null時默認為按空格分解，max為0或負數時分解沒有限制，max為1時返回整個字符串，max為分解成的個數（大于實際則無效）
??
??去除字符串前后指定的字符
??StringUtils.strip(null, *)??????????= null
??StringUtils.strip("", *)????????????= ""
??StringUtils.strip("abc", null)??????= "abc"
??StringUtils.strip(" abc ", null)????= "abc"
??StringUtils.strip("??abcyx", "xyz") = "??abc"
??StringUtils.strip(String str,String stripChars) str為null時返回null,stripChars為null時默認為空格
??創建醒目的Header（調試時用）
??public String createHeader( String title ) {
????int width = 30;
????String stars = StringUtils.repeat( "*", width);
????String centered = StringUtils.center( title, width, "*" );
????String heading = StringUtils.join(new Object[]{stars, centered, stars}, "\n");
????return heading;
??}
??調用createHeader("TEST")的輸出結果:
??******************************
??************ TEST ************
??******************************
??字符的全部反轉及以單個詞為單位的反轉
??String original = "In time, I grew tired of his babbling nonsense.";
??StringUtils.reverse( original )?? = ".esnesnon gnilbbab sih fo derit werg I ,emit nI"
??以單個詞為單位的反轉
??public Sentence reverseSentence(String sentence) {
????String reversed = StringUtils.chomp( sentence, "." );
????reversed = StringUtils.reverseDelimited( reversed, ' ' );
????reversed = reversed + ".";
????return reversed;
??}
??String sentence = "I am Susan."
??reverseSentence( sentence ) )?? = "Susan am I."
??檢查字符串是否僅僅包含數字、字母或數字和字母的混合
??String test1 = "ORANGE";
??String test2 = "ICE9";
??String test3 = "ICE CREAM";
??String test4 = "820B Judson Avenue";
??String test5 = "1976";
??結果：
??boolean t1val = StringUtils.isAlpha( test1 ); // returns true
??boolean t2val = StringUtils.isAlphanumeric( test2 ); // returns true
??boolean t3val = StringUtils.isAlphaSpace( test3 ); // returns true
??boolean t4val = StringUtils.isAlphanumericSpace( test4 ); // returns true
??boolean t5val = StringUtils.isNumeric( test5 ); // returns true
ArrayUtils類使用
primitive 數組克隆及反轉排序
??long[] array = { 1, 3, 2, 3, 5, 6 };
??long[] reversed = ArrayUtils.clone( array );
??ArrayUtils.reverse( reversed );
??System.out.println( "Original: " + ArrayUtils.toString( array ) );?? //打印
??System.out.println( "Reversed: " + ArrayUtils.toString( reversed ) );
??
??對象數組克隆及反轉排序
??Long[] array = new Long[] { new Long(3), new Long(56), new Long(233) };
??Long[] reversed = ArrayUtils.clone( array );
??ArrayUtils.reverse( reversed );
??
??primitive 數組與對象數組之間的轉換
??long[] primitiveArray = new long[] { 12, 100, 2929, 3323 };
??Long[] objectArray = ArrayUtils.toObject( primitiveArray );
??Double[] doubleObjects = new Double[] { new Double( 3.22, 5.222, 3.221 ) };
??double[] doublePrimitives = ArrayUtils.toPrimitive( doubleObject );
??注意：對象數組可以含有null元素，primitive 數組則不容許含有null元素，所以對象數組轉換為primitive 數組時，可以添入第二個參數，當碰到為null的元素時用其代替（如下，Double.NaN）。如果不添入第二個參數，當碰到為null的元素時，則會拋出NullPointerException 。
??double[] result = ArrayUtils.toPrimitive( resultObjArray, Double.NaN??);
??
??查找一個數組中是否含有特定的元素（查找對象數組時，比較的是對象的equals()方法），及特定元素的第一次出現位置和最后一次出現位置
??String[] stringArray = { "Red", "Orange", "Blue", "Brown", "Red" };
??boolean containsBlue = ArrayUtils.contains( stringArray, "Blue" );
??int indexOfRed = ArrayUtils.indexOf( stringArray, "Red");
??int lastIndexOfRed = ArrayUtils.lastIndexOf( string, "Red" );??
??
??由二維對象數組創建一個 Map
??Object[] weightArray =
????new Object[][] { {"H" , new Double( 1.007)},
???????????????????? {"He", new Double( 4.002)},
???????????????????? {"Li", new Double( 6.941)},
???????????????????? {"Be", new Double( 9.012)},
???????????????????? {"B",??new Double(10.811)},
???????????????????? {"C",??new Double(12.010)},
???????????????????? {"N",??new Double(14.007)},
???????????????????? {"O",??new Double(15.999)},
???????????????????? {"F",??new Double(18.998)},
???????????????????? {"Ne", new Double(20.180)} };
??Map weights = ArrayUtils.toMap( weightArray );
??Double hydrogenWeight = （Double)weights.get( "H" );
??注意：當二維對象數組"key"值重復時，創建的Map，后面的鍵-值對會把前面的覆蓋掉