多个字符串替换,不会影响后续迭代中的替换文本

Question

我之前发过关于字母的信,但这是另一个主题,我有一个包含2个对象的json响应,from并且to,from它是要改变的,并且to它将被改变为什么.

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()方法不会在这里工作,因为首先它会替换a到bhduh,然后b到eieja,但这里的问题,它会转换b在bhduh给eieja,我不想.

我希望根据Json完美地转换字符串中的字母和"单词",但那是我失败的原因.

新守则:

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

Answer 1

这是一个严格来说只是 Java 的方法。我尝试在这里不使用任何 Java 8 方法。

public static String translate(final String str, List<String> from, List<String> to, int index) {
    StringBuilder components = new StringBuilder();
    String token, replace;
    int p;

    if (index < from.size()) {
        token = from.get(index);
        replace = to.get(index);
        p = 0;

        for (int i = str.indexOf(token, p); i != -1; i = str.indexOf(token, p)) {
            if (i != p) {
                components.append(translate(str.substring(p, i), from, to, index + 1));
            }
            components.append(replace);
            p = i + token.length();
        }
        return components.append(translate(str.substring(p), from, to, index + 1)).toString();
    }

    return str;
}

public static String translate(final String str, List<String> from, List<String> to) {
    if (null == str) {
        return null;
    }
    return translate(str, from, to, 0);
}

测试程序示例

public static void main(String []args) {
    String EnteredString = "aa  hjkyu  batesh  a";
    List<String> from = new ArrayList<>(Arrays.asList("a", "b", "tes"));
    List<String> to = new ArrayList<>(Arrays.asList("bhduh", "eieja", "neesj"));

    System.out.println(translate(EnteredString, from, to));
}

输出：

bhduhbhduh  hjkyu  eiejabhduhneesjh  bhduh

说明

该算法是递归的，它简单地执行以下操作

• 如果字符串中找到的模式与列表中的模式from匹配
  • 如果该模式之前有任何字符串，则将该算法应用于该字符串
  • 将找到的模式替换为列表中相应的to模式
  • 将替换附加到新字符串
  • 丢弃列表中的模式from并对字符串的其余部分重复该算法
• 否则将字符串的其余部分附加到新字符串