如何删除Java中的代理字符？

问题描述

我面临的情况是我在文本中获取Surrogate字符，我将保存到MySql 5.1。由于这不支持UTF-16，我想在将它保存到数据库之前通过java方法手动删除这些代理对。

I am facing a situation where i get Surrogate characters in text that i am saving to MySql 5.1. As the UTF-16 is not supported in this, I want to remove these surrogate pairs manually by a java method before saving it to the database.

我写了以下内容现在的方法，我很想知道是否有一种直接和最佳的方法来处理这个问题。

I have written the following method for now and I am curious to know if there is a direct and optimal way to handle this.

预先感谢您的帮助。

public static String removeSurrogates(String query) {
    StringBuffer sb = new StringBuffer();
    for (int i = 0; i < query.length() - 1; i++) {
        char firstChar = query.charAt(i);
        char nextChar = query.charAt(i+1);
        if (Character.isSurrogatePair(firstChar, nextChar) == false) {
            sb.append(firstChar);
        } else {
            i++;
        }
    }
    if (Character.isHighSurrogate(query.charAt(query.length() - 1)) == false
            && Character.isLowSurrogate(query.charAt(query.length() - 1)) == false) {
        sb.append(query.charAt(query.length() - 1));
    }

    return sb.toString();
}

推荐答案

以下是一些事情：

• ：

当且仅当它是低代理代码单元或高代理代码单元时，char值是代理代码单元。

检查配对似乎毫无意义，为什么不删除所有代理？

Checking for pairs seems pointless, why not just remove all surrogates?

x == false 相当于！x

更好在你不需要同步的情况下（比如一个永远不会离开本地范围的变量）。

StringBuilder is better in cases where you don't need synchronization (like a variable that never leaves local scope).

我建议这个：

public static String removeSurrogates(String query) {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < query.length(); i++) {
        char c = query.charAt(i);
        // !isSurrogate(c) in Java 7
        if (!(Character.isHighSurrogate(c) || Character.isLowSurrogate(c))) {
            sb.append(firstChar);
        }
    }
    return sb.toString();
}

如果 >声明

您询问此声明：

Breaking down the if statement

You asked about this statement:

if (!(Character.isHighSurrogate(c) || Character.isLowSurrogate(c))) {
    sb.append(firstChar);
}

理解它的一种方法是将每个操作分解为自己的函数，所以你可以看到这个组合符合你的期望：

One way to understand it is to break each operation into its own function, so you can see that the combination does what you'd expect:

static boolean isSurrogate(char c) {
    return Character.isHighSurrogate(c) || Character.isLowSurrogate(c);
}

static boolean isNotSurrogate(char c) {
    return !isSurrogate(c);
}

...

if (isNotSurrogate(c)) {
    sb.append(firstChar);
}

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