多个字符串替换而不影响后续迭代中的替换文本

我之前已经发布过关于信件的内容，但这是另一个主题，我有一个包含 2 个对象的 json 响应，from and to , from是要改变的内容，并且to是它将被更改为的内容。

我的代码是：

// for example, the EnteredText is "ab b test a b" .
EnteredString = EnteredText.getText().toString();
for (int i = 0; i < m_jArry.length(); i++) {
    JSONObject jo_inside = m_jArry.getJSONObject(i);

    String Original = jo_inside.getString("from");
    String To = jo_inside.getString("to");

    if(isMethodConvertingIn){
        EnteredString = EnteredString.replace(" ","_");
        EnteredString = EnteredString.replace(Original,To + " ");
    } else {
        EnteredString = EnteredString.replace("_"," ");
        EnteredString = EnteredString.replace(To + " ", Original);
    }
}

LoadingProgress.setVisibility(View.GONE);
SetResultText(EnteredString);
ShowResultCardView();

例如，json 响应是：

{
    "Response":[
        {"from":"a","to":"bhduh"},{"from":"b","to":"eieja"},{"from":"tes","to":"neesj"}
    ]
}

String.replace()方法在这里不起作用，因为首先它将替换a to bhduh, then b to eieja，但是问题来了，它会转换b in bhduh to eieja，我不想这样做。

我想根据 Json 完美转换字符串中的字母和“单词”，但这就是我失败的地方。

新代码：

if(m_jArry.length() > 0){
    HashMap<String, String> m_li;

    EnteredString = EnteredText.getText().toString();

    Log.i("TestAf_","Before Converting: "  + EnteredString);

    HashMap<String,String> replacements = new HashMap<String,String>();
    for (int i = 0; i < m_jArry.length(); i++) {
        JSONObject jo_inside = m_jArry.getJSONObject(i);

        String Original = jo_inside.getString("from");
        String To = jo_inside.getString("to");

        if(isMethodConvertingIn){

            //EnteredString = EnteredString.replace(" ","_");

            replacements.put(Original,To);
            Log.i("TestAf_","From: " + Original + " - To: " + To + " - Loop: " + i);
            //EnteredString = EnteredString.replace(" ","_");
            //EnteredString = EnteredString.replace(Original,To + " ");

        } else {

            EnteredString = EnteredString.replace("_"," ");
            EnteredString = EnteredString.replace("'" + To + "'", Original);
        }

    }
    Log.i("TestAf_","After Converting: " + replaceTokens(EnteredString,replacements));

    // Replace Logic Here
    // When Finish, Do :
    LoadingProgress.setVisibility(View.GONE);
    SetResultText(replaceTokens(EnteredString,replacements));
    ShowResultCardView();

输出 ：

10-10 19:51:19.757 12113-12113/? I/TestAf_: Before Converting: ab a ba
10-10 19:51:19.757 12113-12113/? I/TestAf_: From: a - To: bhduh - Loop: 0
10-10 19:51:19.757 12113-12113/? I/TestAf_: From: b - To: eieja - Loop: 1
10-10 19:51:19.757 12113-12113/? I/TestAf_: From: o - To: neesj - Loop: 2
10-10 19:51:19.758 12113-12113/? I/TestAf_: After Converting: ab a ba

如果您给出该函数的预期输出，您的问题会更清楚。

假设是：ab b test a b >>>> bhduheieja eieja neesjt bhduh eieja

然后看下面的内容，Javadoc 中的关键点是“这不会重复”

http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/StringUtils.html#replaceEach(java.lang.String,%20java.lang.String[] ,%20java.lang.String[]) http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/StringUtils.html#replaceEach(java.lang.String,%20java.lang.String%5B%5D,%20java.lang.String%5B%5D)

替换另一个字符串中所有出现的字符串。 传递给此方法的空引用是无操作，或者如果有任何“搜索 string”或“要替换的字符串”为空，则替换将被忽略。这个就不再重复了。对于重复替换，请调用重载 方法。

实施例1

import org.apache.commons.lang3.StringUtils;

public class StringReplacer {

    public static void main(String[] args) {

        String input = "ab b test a b";
        String output = StringUtils.replaceEach(input, new String[] { "a", "b", "tes" },
                new String[] { "bhduh", "eieja", "neesj" });

        System.out.println(input + " >>>> " + output);
    }
}

实施例2

import org.apache.commons.lang3.StringUtils;

public class StringReplacer {

    public static void main(String[] args) {

        String input = "this is a test string with foo";
        String output = StringUtils.replaceEach(input, new String[] { "a", "foo" },
                new String[] { "foo", "bar"});

        System.out.println(input + " >>>> " + output);
    }
}