将EDITTEXT限制为一定数量的单词计数

Question

我们如何将 EditText的长度限制为某些单词计数（100个单词）。

即使从某些地方复制一个大文本，在粘贴的同时，它应该只接受100个单词。我知道我们可以按角色计数（maxLength）

限制它

我尝试了以下代码，但它不会完全限制100个单词，并且允许复制大型文本。

editText.addTextChangedListener(new TextWatcher() {
            @Override
            public void beforeTextChanged(CharSequence charSequence, int start, int count, int after) {
                int wordsLength = countWords(charSequence.toString());// words.length;
                // count == 0 means a new word is going to start
                if (count == 0 && wordsLength >= MAX_WORD_LIMIT) {
                    int charLength = mDescription.getText().length();
                    setCharLimit(mDescription, charLength > 0 ? charLength - 2 : charLength);
                } else {
                    removeFilter(mDescription);
                }

和

private InputFilter filter;
    private void setCharLimit(EditText et, int max) {
        filter = new InputFilter.LengthFilter(max);
        et.setFilters(new InputFilter[]{filter});
    }
    private void removeFilter(EditText et) {
        if (filter != null) {
            et.setFilters(new InputFilter[0]);
            filter = null;
        }
    }

Answer 1

    edittext.addTextChangedListener(new TextWatcher() {
            @Override
            public void beforeTextChanged(CharSequence s, int start, int count, int after) {
            }
   @Override
   public void onTextChanged(CharSequence charSequence, int start, int before, int count) {
            //To fix word count
            String yourText= editText.getText().toString().replace(String.valueOf((char) 160), " ");
            if (yourText.split("\s+").length > MAX_WORD_LIMIT ) {
                int space = 0;
                int length = 0;
                for (int i = 0; i < yourText.length(); i++) {
                    if (yourText.charAt(i) == ' ') {
                        space++;
                        if (space >= MAX_WORD_LIMIT) {
                            length = i;
                            break;
                        }
                    }
                }
                if (length > 1) {
                    editText.getText().delete(length, yourSelf.length()); // deleting last space
                    setCharLimit(editText, length - 1); //or limit edit text
                }
            } else {
                removeFilter(editText);
            }

        }
            @Override
            public void afterTextChanged(Editable s) {
        });
          private InputFilter filter;
          private void setCharLimit(EditText et, int max) {
            filter = new InputFilter.LengthFilter(max);
            et.setFilters(new InputFilter[] { filter });
            }
          private void removeFilter(EditText et) {
            if (filter != null) {
              et.setFilters(new InputFilter[0]);
             filter = null;
             }
           }

这里160是非断裂空间

Answer 2

您可以使用：

android:maxLength="50"

在您的Edittext中。

Answer 3

希望它为您工作

edittext.addTextChangedListener(new TextWatcher() {
        @Override
        public void beforeTextChanged(CharSequence s, int start, int count, int after) {
        }
        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
        }
        @Override
        public void afterTextChanged(Editable s) {
           // Check your edittext length here
           if (edittext.getText().toString().length() > 100)
              edittext.setText(edittext.getText().toString().substring(0, 100);
        }
    });

Answer 4

kotlin语言 - 要检查最大单词需要检查总空间

->步骤1

var filter: InputFilter?=null

-> step-2

edt.addTextChangedListener(object : TextWatcher {
            override fun afterTextChanged(s: Editable?) {
                var count : Int = 0
                if(s.toString().length > 0)
                    for(i in 0..s.toString().length-1){
                        if(s.toString()[i].toString() == " "){
                            count++
                        }
                        if(i > 1 && s.toString()[i-1].toString() == " " && s.toString()[i].toString() == " "){
                            count--
                        }
                    }
                if (count >= MAX_COUNTS_WORDS){
                    filter = InputFilter.LengthFilter(edt.text.toString().length)
                    edt.filters = arrayOf<InputFilter>(filter ?: return)
                }
                else if (filter != null) {
                    edt.filters = arrayOfNulls(0)
                    filter = null
                }

            }
            override fun beforeTextChanged(s: CharSequence?, p1: Int, p2: Int, p3: Int) {
            }
            override fun onTextChanged(p0: CharSequence?, p1: Int, p2: Int, p3: Int) {
            }
        })

Answer 5

如果要找到单词计数，则可以使用Java逻辑。

String text = "More than 100 word sentence";
String[] txtArr = text.split(" ");

现在仅选择第一个100个索引单词。

这只是一个黑客。可能还有更多最佳方法。

Answer 6

您可以使用功能索引的拆分。对不起，这么长的字符串，希望您能得到这个主意。只需提取一些变量即可。

edittext.addTextChangedListener(new TextWatcher() {
            @Override
            public void beforeTextChanged(CharSequence s, int start, int count, int after) {
            }
            @Override
            public void onTextChanged(CharSequence s, int start, int before, int count) {
            }
            @Override
            public void afterTextChanged(Editable s) {
               if (edittext.getText().toString().split(" ").size() > 100)
                  edittext.setText(edittext.getText().toString().substring(0, edittext.getText().toString().indexOf(edittext.getText().toString().split(" ").get(100))));
            }
        });

Answer 7

使用 android：maxlength deDittext的属性解决您的问题。