Kon*_*itz 119 java android android-edittext
我正在尝试编写一款可以帮助您管理财务状况的应用.我正在使用一个EditText用户可以指定金额的字段.
我设置inputType到numberDecimal其工作正常,但,这让人们输入数字,如123.122这是不完美的钱.
有没有办法将小数点后的字符数限制为两个？
Asa*_*ssi 108
更优雅的方式是使用正则表达式(正则表达式),如下所示:
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter(int digitsBeforeZero,int digitsAfterZero) {
mPattern=Pattern.compile("[0-9]{0," + (digitsBeforeZero-1) + "}+((\\.[0-9]{0," + (digitsAfterZero-1) + "})?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
Matcher matcher=mPattern.matcher(dest);
if(!matcher.matches())
return "";
return null;
}
}
使用它做:
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});
pec*_*eps 63
更简单的解决方案,不使用正则表达式:
import android.text.InputFilter;
import android.text.Spanned;
/**
* Input filter that limits the number of decimal digits that are allowed to be
* entered.
*/
public class DecimalDigitsInputFilter implements InputFilter {
private final int decimalDigits;
/**
* Constructor.
*
* @param decimalDigits maximum decimal digits
*/
public DecimalDigitsInputFilter(int decimalDigits) {
this.decimalDigits = decimalDigits;
}
@Override
public CharSequence filter(CharSequence source,
int start,
int end,
Spanned dest,
int dstart,
int dend) {
int dotPos = -1;
int len = dest.length();
for (int i = 0; i < len; i++) {
char c = dest.charAt(i);
if (c == '.' || c == ',') {
dotPos = i;
break;
}
}
if (dotPos >= 0) {
// protects against many dots
if (source.equals(".") || source.equals(","))
{
return "";
}
// if the text is entered before the dot
if (dend <= dotPos) {
return null;
}
if (len - dotPos > decimalDigits) {
return "";
}
}
return null;
}
}
使用:
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(2)});
Kon*_*itz 33
这种实现InputFilter解决了这个问题.
import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.DigitsKeyListener;
public class MoneyValueFilter extends DigitsKeyListener {
public MoneyValueFilter() {
super(false, true);
}
private int digits = 2;
public void setDigits(int d) {
digits = d;
}
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
CharSequence out = super.filter(source, start, end, dest, dstart, dend);
// if changed, replace the source
if (out != null) {
source = out;
start = 0;
end = out.length();
}
int len = end - start;
// if deleting, source is empty
// and deleting can't break anything
if (len == 0) {
return source;
}
int dlen = dest.length();
// Find the position of the decimal .
for (int i = 0; i < dstart; i++) {
if (dest.charAt(i) == '.') {
// being here means, that a number has
// been inserted after the dot
// check if the amount of digits is right
return (dlen-(i+1) + len > digits) ?
"" :
new SpannableStringBuilder(source, start, end);
}
}
for (int i = start; i < end; ++i) {
if (source.charAt(i) == '.') {
// being here means, dot has been inserted
// check if the amount of digits is right
if ((dlen-dend) + (end-(i + 1)) > digits)
return "";
else
break; // return new SpannableStringBuilder(source, start, end);
}
}
// if the dot is after the inserted part,
// nothing can break
return new SpannableStringBuilder(source, start, end);
}
}
Fav*_* Kv 33
下面是一个示例InputFilter,它只允许小数点前最多4位数,之后最多1位数.
价值观是允许的EditText:555.2,555,0.2
这EDITTEXT块值:55555.2,055.2,555.42
InputFilter filter = new InputFilter() {
final int maxDigitsBeforeDecimalPoint=4;
final int maxDigitsAfterDecimalPoint=1;
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
StringBuilder builder = new StringBuilder(dest);
builder.replace(dstart, dend, source
.subSequence(start, end).toString());
if (!builder.toString().matches(
"(([1-9]{1})([0-9]{0,"+(maxDigitsBeforeDecimalPoint-1)+"})?)?(\\.[0-9]{0,"+maxDigitsAfterDecimalPoint+"})?"
)) {
if(source.length()==0)
return dest.subSequence(dstart, dend);
return "";
}
return null;
}
};
mEdittext.setFilters(new InputFilter[] { filter });
and*_*dev 19
我为@Pinhassi解决方案做了一些修复.它处理一些情况:
1.你可以将光标移动到任何地方
2.minus标志处理
3.digitsbefore = 2和digitsafter = 4,然后输入12.4545.然后,如果你想删除".",它将不允许.
public class DecimalDigitsInputFilter implements InputFilter {
private int mDigitsBeforeZero;
private int mDigitsAfterZero;
private Pattern mPattern;
private static final int DIGITS_BEFORE_ZERO_DEFAULT = 100;
private static final int DIGITS_AFTER_ZERO_DEFAULT = 100;
public DecimalDigitsInputFilter(Integer digitsBeforeZero, Integer digitsAfterZero) {
this.mDigitsBeforeZero = (digitsBeforeZero != null ? digitsBeforeZero : DIGITS_BEFORE_ZERO_DEFAULT);
this.mDigitsAfterZero = (digitsAfterZero != null ? digitsAfterZero : DIGITS_AFTER_ZERO_DEFAULT);
mPattern = Pattern.compile("-?[0-9]{0," + (mDigitsBeforeZero) + "}+((\\.[0-9]{0," + (mDigitsAfterZero)
+ "})?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String replacement = source.subSequence(start, end).toString();
String newVal = dest.subSequence(0, dstart).toString() + replacement
+ dest.subSequence(dend, dest.length()).toString();
Matcher matcher = mPattern.matcher(newVal);
if (matcher.matches())
return null;
if (TextUtils.isEmpty(source))
return dest.subSequence(dstart, dend);
else
return "";
}
}
Gun*_*lan 17
我通过TextWatcher以下方式帮助实现了这一目标
final EditText et = (EditText) findViewById(R.id.EditText1);
et.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2,int arg3) {
}
public void beforeTextChanged(CharSequence arg0, int arg1,int arg2, int arg3) {
}
public void afterTextChanged(Editable arg0) {
if (arg0.length() > 0) {
String str = et.getText().toString();
et.setOnKeyListener(new OnKeyListener() {
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_DEL) {
count--;
InputFilter[] fArray = new InputFilter[1];
fArray[0] = new InputFilter.LengthFilter(100);
et.setFilters(fArray);
//change the edittext's maximum length to 100.
//If we didn't change this the edittext's maximum length will
//be number of digits we previously entered.
}
return false;
}
});
char t = str.charAt(arg0.length() - 1);
if (t == '.') {
count = 0;
}
if (count >= 0) {
if (count == 2) {
InputFilter[] fArray = new InputFilter[1];
fArray[0] = new InputFilter.LengthFilter(arg0.length());
et.setFilters(fArray);
//prevent the edittext from accessing digits
//by setting maximum length as total number of digits we typed till now.
}
count++;
}
}
}
});
此解决方案不允许用户在小数点后输入两位以上的数字.您也可以在小数点前输入任意位数.请参阅此博客http://v4all123.blogspot.com/2013/05/set-limit-for-fraction-in-decimal.html为多个EditText设置过滤器.我希望这将有所帮助.谢谢.
Apo*_*leo 13
我不喜欢其他解决方案而且我创建了自己的解决方案.使用此解决方案,您不能在该点之前输入超过MAX_BEFORE_POINT数字,并且小数不能超过MAX_DECIMAL.
你只是不能输入过多的数字,没有其他影响!另外如果你写"." 它输入"0"
将布局中的EditText设置为:
安卓的inputType = "numberDecimal"
在onCreate中添加监听器.如果你想修改点之前和之后的数字编辑调用PerfectDecimal(str,NUMBER_BEFORE_POINT,NUMBER_DECIMALS),这里设置为3和2
EditText targetEditText = (EditText)findViewById(R.id.targetEditTextLayoutId);
targetEditText.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {}
public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {}
public void afterTextChanged(Editable arg0) {
String str = targetEditText.getText().toString();
if (str.isEmpty()) return;
String str2 = PerfectDecimal(str, 3, 2);
if (!str2.equals(str)) {
targetEditText.setText(str2);
int pos = targetEditText.getText().length();
targetEditText.setSelection(pos);
}
}
});
包括这个Funcion:
public String PerfectDecimal(String str, int MAX_BEFORE_POINT, int MAX_DECIMAL){
if(str.charAt(0) == '.') str = "0"+str;
int max = str.length();
String rFinal = "";
boolean after = false;
int i = 0, up = 0, decimal = 0; char t;
while(i < max){
t = str.charAt(i);
if(t != '.' && after == false){
up++;
if(up > MAX_BEFORE_POINT) return rFinal;
}else if(t == '.'){
after = true;
}else{
decimal++;
if(decimal > MAX_DECIMAL)
return rFinal;
}
rFinal = rFinal + t;
i++;
}return rFinal;
}
它已经完成了!
Dom*_*icz 11
我的解决方案很简单,工作完美!
public class DecimalInputTextWatcher implements TextWatcher {
private String mPreviousValue;
private int mCursorPosition;
private boolean mRestoringPreviousValueFlag;
private int mDigitsAfterZero;
private EditText mEditText;
public DecimalInputTextWatcher(EditText editText, int digitsAfterZero) {
mDigitsAfterZero = digitsAfterZero;
mEditText = editText;
mPreviousValue = "";
mRestoringPreviousValueFlag = false;
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
if (!mRestoringPreviousValueFlag) {
mPreviousValue = s.toString();
mCursorPosition = mEditText.getSelectionStart();
}
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public void afterTextChanged(Editable s) {
if (!mRestoringPreviousValueFlag) {
if (!isValid(s.toString())) {
mRestoringPreviousValueFlag = true;
restorePreviousValue();
}
} else {
mRestoringPreviousValueFlag = false;
}
}
private void restorePreviousValue() {
mEditText.setText(mPreviousValue);
mEditText.setSelection(mCursorPosition);
}
private boolean isValid(String s) {
Pattern patternWithDot = Pattern.compile("[0-9]*((\\.[0-9]{0," + mDigitsAfterZero + "})?)||(\\.)?");
Pattern patternWithComma = Pattern.compile("[0-9]*((,[0-9]{0," + mDigitsAfterZero + "})?)||(,)?");
Matcher matcherDot = patternWithDot.matcher(s);
Matcher matcherComa = patternWithComma.matcher(s);
return matcherDot.matches() || matcherComa.matches();
}
}
用法:
myTextEdit.addTextChangedListener(new DecimalInputTextWatcher(myTextEdit, 2));
Man*_*esh 11
要求是十进制后的2位数.小数点前的数字应该没有限制.所以,解决方案应该是,
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter() {
mPattern = Pattern.compile("[0-9]*+((\\.[0-9]?)?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
Matcher matcher = mPattern.matcher(dest);
if (!matcher.matches())
return "";
return null;
}
}
用它作为,
mEditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter()});
感谢@Pinhassi的灵感.
Luk*_*uke 11
我想出的InputFilter允许您配置小数位前后的位数.此外,它不允许前导零.
public class DecimalDigitsInputFilter implements InputFilter
{
Pattern pattern;
public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal)
{
pattern = Pattern.compile("(([1-9]{1}[0-9]{0," + (digitsBeforeDecimal - 1) + "})?||[0]{1})((\\.[0-9]{0," + digitsAfterDecimal + "})?)||(\\.)?");
}
@Override public CharSequence filter(CharSequence source, int sourceStart, int sourceEnd, Spanned destination, int destinationStart, int destinationEnd)
{
// Remove the string out of destination that is to be replaced.
String newString = destination.toString().substring(0, destinationStart) + destination.toString().substring(destinationEnd, destination.toString().length());
// Add the new string in.
newString = newString.substring(0, destinationStart) + source.toString() + newString.substring(destinationStart, newString.length());
// Now check if the new string is valid.
Matcher matcher = pattern.matcher(newString);
if(matcher.matches())
{
// Returning null indicates that the input is valid.
return null;
}
// Returning the empty string indicates the input is invalid.
return "";
}
}
// To use this InputFilter, attach it to your EditText like so:
final EditText editText = (EditText) findViewById(R.id.editText);
EditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter(4, 4)});
小智 7
实现这一目标的最简单方法是：
et.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
String text = arg0.toString();
if (text.contains(".") && text.substring(text.indexOf(".") + 1).length() > 2) {
et.setText(text.substring(0, text.length() - 1));
et.setSelection(et.getText().length());
}
}
public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
}
public void afterTextChanged(Editable arg0) {
}
});
在将字符串放入TextView之前,请尝试使用NumberFormat.getCurrencyInstance()格式化字符串.
就像是:
NumberFormat currency = NumberFormat.getCurrencyInstance();
myTextView.setText(currency.format(dollars));
编辑 - 我没有在文档中找到的货币输入类型.我想这是因为有些货币不遵循相同的小数位规则,例如日元.
作为LeffelMania提到的,您可以使用上面的代码与正确的用户输入TextWatcher是在你的设置EditText.
稍微改进了@Pinhassi解决方案.
效果很好.它验证连接的字符串.
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter() {
mPattern = Pattern.compile("([1-9]{1}[0-9]{0,2}([0-9]{3})*(\\.[0-9]{0,2})?|[1-9]{1}[0-9]{0,}(\\.[0-9]{0,2})?|0(\\.[0-9]{0,2})?|(\\.[0-9]{1,2})?)");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String formatedSource = source.subSequence(start, end).toString();
String destPrefix = dest.subSequence(0, dstart).toString();
String destSuffix = dest.subSequence(dend, dest.length()).toString();
String result = destPrefix + formatedSource + destSuffix;
result = result.replace(",", ".");
Matcher matcher = mPattern.matcher(result);
if (matcher.matches()) {
return null;
}
return "";
}
}
小智 6
我已修改上述解决方案并创建了以下解决方案.您可以设置小数点前后的位数.
public class DecimalDigitsInputFilter implements InputFilter {
private final Pattern mPattern;
public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
mPattern = Pattern.compile(String.format("[0-9]{0,%d}(\\.[0-9]{0,%d})?", digitsBeforeZero, digitsAfterZero));
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
Matcher matcher = mPattern.matcher(createResultString(source, start, end, dest, dstart, dend));
if (!matcher.matches())
return "";
return null;
}
private String createResultString(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String sourceString = source.toString();
String destString = dest.toString();
return destString.substring(0, dstart) + sourceString.substring(start, end) + destString.substring(dend);
}
}
小智 6
在 Android kotlin 中创建一个名为 DecimalDigitsInputFilter 的新类
class DecimalDigitsInputFilter(digitsBeforeDecimal: Int, digitsAfterDecimal: Int) : InputFilter {
var mPattern: Pattern = Pattern.compile("[0-9]{0,$digitsBeforeDecimal}+((\\.[0-9]{0,$digitsAfterDecimal})?)||(\\.)?")
override fun filter(
source: CharSequence?,
start: Int,
end: Int,
dest: Spanned?,
dstart: Int,
dend: Int
): CharSequence? {
val matcher: Matcher = mPattern.matcher(
dest?.subSequence(0, dstart).toString() + source?.subSequence(
start,
end
).toString() + dest?.subSequence(dend, dest.length).toString()
)
if (!matcher.matches())
return ""
else
return null
}
}
使用以下行调用此类
et_buy_amount.filters = (arrayOf<InputFilter>(DecimalDigitsInputFilter(8,2)))
有太多相同的答案，但它允许您输入小数点前 8 位数字和小数点后 2 位数字
其他答案只接受 8 位数字
DecimalFormat form = new DecimalFormat("#.##", new DecimalFormatSymbols(Locale.US));
EditText et;
et.setOnEditorActionListener(new TextView.OnEditorActionListener() {
@Override
public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
if (actionId == EditorInfo.IME_ACTION_DONE) {
double a = Double.parseDouble(et.getText().toString());
et.setText(form.format(a));
}
return false;
}
});
这样做是当您退出编辑阶段时，它将字段格式化为正确的格式。目前，它只有 2 个十进制字符。我认为这是很容易做到这一点的方法。
我真的很喜欢 Pinhassi 的答案，但注意到用户在小数点后输入了指定的数字后，您无法再在小数点左侧输入文本。问题在于该解决方案仅测试之前输入的文本，而不测试当前输入的文本。所以这是我的解决方案，它将新字符插入原始文本中进行验证。
package com.test.test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import android.text.InputFilter;
import android.text.Spanned;
import android.util.Log;
public class InputFilterCurrency implements InputFilter {
Pattern moPattern;
public InputFilterCurrency(int aiMinorUnits) {
// http://www.regexplanet.com/advanced/java/index.html
moPattern=Pattern.compile("[0-9]*+((\\.[0-9]{0,"+ aiMinorUnits + "})?)||(\\.)?");
} // InputFilterCurrency
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String lsStart = "";
String lsInsert = "";
String lsEnd = "";
String lsText = "";
Log.d("debug", moPattern.toString());
Log.d("debug", "source: " + source + ", start: " + start + ", end:" + end + ", dest: " + dest + ", dstart: " + dstart + ", dend: " + dend );
lsText = dest.toString();
// If the length is greater then 0, then insert the new character
// into the original text for validation
if (lsText.length() > 0) {
lsStart = lsText.substring(0, dstart);
Log.d("debug", "lsStart : " + lsStart);
// Check to see if they have deleted a character
if (source != "") {
lsInsert = source.toString();
Log.d("debug", "lsInsert: " + lsInsert);
} // if
lsEnd = lsText.substring(dend);
Log.d("debug", "lsEnd : " + lsEnd);
lsText = lsStart + lsInsert + lsEnd;
Log.d("debug", "lsText : " + lsText);
} // if
Matcher loMatcher = moPattern.matcher(lsText);
Log.d("debug", "loMatcher.matches(): " + loMatcher.matches() + ", lsText: " + lsText);
if(!loMatcher.matches()) {
return "";
}
return null;
} // CharSequence
} // InputFilterCurrency
以及设置 editText 过滤器的调用
editText.setFilters(new InputFilter[] {new InputFilterCurrency(2)});
Ouput with two decimal places
05-22 15:25:33.434: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:25:33.434: D/debug(30524): source: 5, start: 0, end:1, dest: 123.4, dstart: 5, dend: 5
05-22 15:25:33.434: D/debug(30524): lsStart : 123.4
05-22 15:25:33.434: D/debug(30524): lsInsert: 5
05-22 15:25:33.434: D/debug(30524): lsEnd :
05-22 15:25:33.434: D/debug(30524): lsText : 123.45
05-22 15:25:33.434: D/debug(30524): loMatcher.matches(): true, lsText: 123.45
Ouput inserting a 5 in the middle
05-22 15:26:17.624: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:26:17.624: D/debug(30524): source: 5, start: 0, end:1, dest: 123.45, dstart: 2, dend: 2
05-22 15:26:17.624: D/debug(30524): lsStart : 12
05-22 15:26:17.624: D/debug(30524): lsInsert: 5
05-22 15:26:17.624: D/debug(30524): lsEnd : 3.45
05-22 15:26:17.624: D/debug(30524): lsText : 1253.45
05-22 15:26:17.624: D/debug(30524): loMatcher.matches(): true, lsText: 1253.45
小智 5
我改进了 Pinhassi 使用正则表达式的解决方案，因此它也可以正确处理边缘情况。在检查输入是否正确之前，首先按照 android 文档的描述构造最终字符串。
public class DecimalDigitsInputFilter implements InputFilter {
private Pattern mPattern;
private static final Pattern mFormatPattern = Pattern.compile("\\d+\\.\\d+");
public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal) {
mPattern = Pattern.compile(
"^\\d{0," + digitsBeforeDecimal + "}([\\.,](\\d{0," + digitsAfterDecimal +
"})?)?$");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest,
int dstart, int dend) {
String newString =
dest.toString().substring(0, dstart) + source.toString().substring(start, end)
+ dest.toString().substring(dend, dest.toString().length());
Matcher matcher = mPattern.matcher(newString);
if (!matcher.matches()) {
return "";
}
return null;
}
}
用法：
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});
我已经更改了答案 \xe2\x84\x966 （由 Favas Kv ），因为你可以将点放在第一个位置。
\n\nfinal InputFilter [] filter = { new InputFilter() {\n\n @Override\n public CharSequence filter(CharSequence source, int start, int end,\n Spanned dest, int dstart, int dend) {\n StringBuilder builder = new StringBuilder(dest);\n builder.replace(dstart, dend, source\n .subSequence(start, end).toString());\n if (!builder.toString().matches(\n "(([1-9]{1})([0-9]{0,4})?(\\\\.)?)?([0-9]{0,2})?"\n\n )) {\n if(source.length()==0)\n return dest.subSequence(dstart, dend);\n return "";\n }\n return null;\n }\n}};\n