今天偶然发现Netbeans不能显示中文，本来以为是charset的问题，然后意识到错误的charset会导致乱码，而不是小方块。
经过测试，使用最常用的英文字体，例如Arial, New Times Roman 都可以正常显示中文。其余字体，比如编程常用的MonoSpace，Consolas，Monaco什么的会无法显示中文。如图：
经过一番搜索，找到原因，简单地说，原因在于这些字体里面不包含中文，fallback font也没有中文.
又搜了一大圈，发现了一个很好的第三方字体，使用Consolas+微软雅黑中文，非常漂亮，成功解决，如图：（如果细心就会发现字数不一样，所以只是我的一个例子么～）
下载链接：
http://ishare.iask.sina.com.cn/f/8965397.html
备用链接：
What You Want:
/path/to/www/root/
—————–/123/index.php ===> http://123.yourwebsite.com/
—————–/www/index.php ===> http://www.yourwebsite.com/
—————–/blog/index.php ===> http://blog.yourwebsite.com/
—————–/email/main/index.php ===> http://email.yourwebsite.com/main/
Set dns:
A (may be AAAA) * to your server’s IP address.
This is the rewrite code you should use:
RewriteEngine On
RewriteMap lowercase int:tolower
RewriteCond %{HTTP_HOST} ^(.*)\.yourwebsite\.com$
RewriteRule ^(.*)$ /path/to/www/root/${lowercase:%1}/$1 [L]
PS:
Options -Indexes
Will not working in this case. Since Apache can not find the subfolder if you hide them. All subdomains will return a 404 error.
/wp-include/class-wp-admin-bar.php
remove these two lines:
wp_enqueue_script( 'admin-bar' ); wp_enqueue_style( 'admin-bar' );
PHP has different functions which can be used to test the value of a variable. Three useful functions for this are isset(), empty() and is_null(). All these function return a boolean value. If these functions are not used in correct way they can cause unexpected results.
isset() and empty() are often viewed as functions that are opposite, however this is not always true. In this post I will explain the differences between these functions.
From PHP manual – isset():
isset — Determine if a variable is set and is not NULL
In other words, it returns true only when the variable is not null.
empty — Determine whether a variable is empty
In other words, it will return true if the variable is an empty string, false, array(), NULL, “0?, 0, and an unset variable.
is_null — Finds whether a variable is NULL
In other words, it returns true only when the variable is null. is_null() is opposite of isset(), except for one difference that isset() can be applied to unknown variables, but is_null() only to declared variables.
| Value of variable ($var) | isset($var) | empty($var) | is_null($var) |
|---|---|---|---|
| “” (an empty string) | bool(true) | bool(true) | |
| ” ” (space) | bool(true) | ||
| FALSE | bool(true) | bool(true) | |
| TRUE | bool(true) | ||
| array() (an empty array) | bool(true) | bool(true) | |
| NULL | bool(true) | bool(true) | |
| “0″ (0 as a string) | bool(true) | bool(true) | |
| 0 (0 as an integer) | bool(true) | bool(true) | |
| 0.0 (0 as a float) | bool(true) | bool(true) | |
| var $var; (a variable declared, but without a value) | bool(true) | bool(true) | |
| NULL byte (“\ 0″) | bool(true) |
Write a method named hasMidpoint that accepts three integers as parameters and returns true if one of the integers is the midpoint between the other two integers; that is, if one integer is exactly halfway between them. Your method should return false if no such midpoint relationship exists.
The integers could be passed in any order; the midpoint could be the 1st, 2nd, or 3rd. You must check all cases.
Calls such as the following should return true:
hasMidpoint(4, 6, 8)
hasMidpoint(2, 10, 6)
hasMidpoint(8, 8, 8)
hasMidpoint(25, 10, -5)
Calls such as the following should return false:
hasMidpoint(3, 1, 3)
hasMidpoint(1, 3, 1)
hasMidpoint(21, 9, 58)
hasMidpoint(2, 8, 16)
=========================================================
以下是我和晓天想出来的答案。我们最开始打算先进行排序，再求差，来判断，很是程序写起来会很长，而且我们现在没有讲各种排序法。
=========================================================
public static boolean hasMidpoint(int a,int b,int c){
int e=a+b+c;
if (e/3==a||e/3==b||e/3==c)
return true;
return false;
}
=============================================================
# name expected return your return result
1 hasMidpoint(1, 2, 3) true true pass
2 hasMidpoint(2, 10, 6) true true pass
3 hasMidpoint(0, -50, -25) true true pass
4 hasMidpoint(21, 9, 58) false false pass
5 hasMidpoint(-2, 9, 27) false false pass
=================================================
提交答案之后，我突然觉得这个程序存在问题，虽然在系统给的数值里没有测出问题，但是由于int类数据的特点，10/3是等于3的。所以输入以下测试代码。
==================================================
public class test{
public static void main(String[] args){
System.out.println(hasMidpoint(0,-51,-25));
}
public static boolean hasMidpoint(int a,int b,int c){
int e=a+b+c;
if (e/3==a||e/3==b||e/3==c)
return true;
return false;
}
}
//由于是复制过来的，就不能保持很好的代码格式了。
=============================================================
输出结果：true
=============================================================
很显然，-25不是-51和0的中点，刚刚的源码是有问题的。
=================================================
把if的条件语句作了如下调整：
if ((e/3==a||e/3==b||e/3==c)&&e%3==0)
再次调用
System.out.println(hasMidpoint(0,-501,-250));
输出是false
=================================================
现在基本已经不会存在问题了，通过了我的所有测试。
如果有人认为我的代码还存在问题，欢迎回复。