compareTo经常用于比较日期的大小，其实java里面想要比较两个Date用此方法很方便，直接进行调用即可，我们根据返回的数值来判断是大于等于还是小于，这里我们进行实际代码格式分析compareTo(Date time1,Date time2) 其中return 1表示time1 > time2 ,return 0 表示time1 = time2 ,return -1 表示time1 < time2.很简单吧！！下面我们进行源码分析，毕竟没有任何一个开发者仅仅想知道知道这个结论！
/**
* Compares two Dates for ordering.
*
* @param anotherDate the <code>Date</code> to be compared.
* @return the value <code>0</code> if the argument Date is equal to
* this Date; a value less than <code>0</code> if this Date
* is before the Date argument; and a value greater than
* <code>0</code> if this Date is after the Date argument.
* @since 1.2
* @exception NullPointerException if <code>anotherDate</code> is null.
*/
public int compareTo(Date anotherDate) {
long thisTime = getMillisOf(this); //this表示当前当前时间转化成毫秒数
long anotherTime = getMillisOf(anotherDate); //anotherDate此时间转化成毫秒数
return (thisTime<anotherTime ? -1 : (thisTime==anotherTime ? 0 : 1)); //比较运算返回简化形式
}这是jdk里面的源码，很简单，这里使用的是将Date转换成毫秒进行直接比较，其实这就相当于两个Long型的数值进行比较大小，然后进行返回-1 0 1 这就得到我们的返回结果，应征了上面的return
那如果我们传入的日期本身就是Date类型的呢？我们返回的结果又是怎样的呢？
我们首先把运行结果粘贴出来
百思不得其解！！！
自己写的测试用例，开始怀疑自己，本来数据库查询出来进行判断使用的1 0 -1 结果现在没有一个结果符合，开始查错， 最后总结对于String类型的日期比较，如果大于的话返回的是正整数，等于是0，小于的话就是负整数，而不仅仅局限于1,0和-1，下面讲解为什么。
我们看一下JDK源码，很多时间我们对运行结果不理解往往是因为对底层不熟悉，下面直接对源码进行分析
/**
* Compares two strings lexicographically.
* The comparison is based on the Unicode value of each character in
* the strings. The character sequence represented by this
* <code>String</code> object is compared lexicographically to the
* character sequence represented by the argument string. The result is
* a negative integer if this <code>String</code> object
* lexicographically precedes the argument string. The result is a
* positive integer if this <code>String</code> object lexicographically
* follows the argument string. The result is zero if the strings
* are equal; <code>compareTo</code> returns <code>0</code> exactly when
* the {@link #equals(Object)} method would return <code>true</code>.
* <p>
* This is the definition of lexicographic ordering. If two strings are
* different, then either they have different characters at some index
* that is a valid index for both strings, or their lengths are different,
* or both. If they have different characters at one or more index
* positions, let <i>k</i> be the smallest such index; then the string
* whose character at position <i>k</i> has the smaller value, as
* determined by using the < operator, lexicographically precedes the
* other string. In this case, <code>compareTo</code> returns the
* difference of the two character values at position <code>k</code> in
* the two string -- that is, the value:
* <blockquote><pre>
* this.charAt(k)-anotherString.charAt(k)
* </pre></blockquote>
* If there is no index position at which they differ, then the shorter
* string lexicographically precedes the longer string. In this case,
* <code>compareTo</code> returns the difference of the lengths of the
* strings -- that is, the value:
* <blockquote><pre>
* this.length()-anotherString.length()
* </pre></blockquote>
*
* @param anotherString the <code>String</code> to be compared.
* @return the value <code>0</code> if the argument string is equal to
* this string; a value less than <code>0</code> if this string
* is lexicographically less than the string argument; and a
* value greater than <code>0</code> if this string is
* lexicographically greater than the string argument.
*/
public int compareTo(String anotherString) {
int len1 = value.length;
int len2 = anotherString.value.length;
int lim = Math.min(len1, len2); //调用方法得到最小的字符串长度 lim
char v1[] = value; //转化成字符串数组进行存储
char v2[] = anotherString.value;
int k = 0;
while (k < lim) { //进行遍历比较
char c1 = v1[k];
char c2 = v2[k];
if (c1 != c2) {
return c1 - c2; //进行求差运算
}
k++;
}
return len1 - len2; //最后返回长度差
}由此可见，同一个方法，对于不同的数据类型在JDK源码中使用完全不同的解决思路，可见算法在我们开发过程中的重要性，所以衡量一个合格程序员的首要因素就应该是算法的能力！！！
以后争取每天都能出一些用来研究底层的代码和算法的实现，毕竟他们的好坏直接或者间接决定了你的高度！是码农还是程序设计师。这也就很好的解决了上面的运行结果，可见追根溯源还是底层实现的不同。
因为我这边是从数据库查询对数据进行过滤，所以我这里就不直接进行粘贴代码了，毕竟公司电脑码的代码所有知识产权都是公司，只有能力属于自己！！！