FAQ_10_EditText与parseInt方法

FROM（http://blog.csdn.net/androidbluetooth/article/details/6689726）

 

 

需求：从android的EditText中获取输入的数字，作为端口号。

基本思路 ，见下面代码示意：

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1. String str = editor.getText().toString().trim();
2. int port = Integer.parseInt(str);

String str = editor.getText().toString().trim();int port = Integer.parseInt(str);

如果用户输入的字符包括非数字如mark123,那么，会报异常：

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1. 08-15 07:13:48.294: ERROR/AndroidRuntime(12281): java.lang.NumberFormatException: unable to parse 'dd' as integer

08-15 07:13:48.294: ERROR/AndroidRuntime(12281): java.lang.NumberFormatException: unable to parse 'dd' as integer

parseInt()源码：

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1. /**
2. * Parses the specified string as a signed decimal integer value. The ASCII
3. * character \u002d ('-') is recognized as the minus sign.
4. *
5. * @param string
6. * the string representation of an integer value.
7. * @return the primitive integer value represented by {@code string}.
8. * @throws NumberFormatException
9. * if {@code string} is {@code null}, has a length of zero or
10. * can not be parsed as an integer value.
11. */
12. public staticint parseInt(String string) throws NumberFormatException {
13. return parseInt(string,10);
14. }

 /**     * Parses the specified string as a signed decimal integer value. The ASCII     * character \u002d ('-') is recognized as the minus sign.     *     * @param string     *            the string representation of an integer value.     * @return the primitive integer value represented by {@code string}.     * @throws NumberFormatException     *             if {@code string} is {@code null}, has a length of zero or     *             can not be parsed as an integer value.     */    public static int parseInt(String string) throws NumberFormatException {        return parseInt(string, 10);    }

可以看出，参数字符串为null、或者长度为0以及非数字的字符时，报NumberFormatException异常。

那么，如何解决这个问题？

简单办法，就是捕捉异常信息：

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1. String str = editor.getText().toString().trim();
2. try {
3. int port = Integer.parseInt(str);
4. } catch(NumberFormatException nfe) {
5. // to do
6. }

String str = editor.getText().toString().trim();try {int port = Integer.parseInt(str);} catch(NumberFormatException nfe) {// to do}

还有一个办法，利用EditText的xml属性（android:digits）。

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1. <EditText
2. android:id="@+id/edit"
3. android:layout_width="fill_parent"
4. android:layout_height="wrap_content"
5. android:inputType="number"
6. android:digits="0123456789"/>

<EditText    android:id="@+id/edit"      android:layout_width="fill_parent"     android:layout_height="wrap_content"    android:inputType="number"     android:digits="0123456789"/>

这样以来，在该EditText中只可以输入数字，其它任何字符都不会被接受。如回车、空格等。

类似的，只接受大小写字母：

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1. android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"只可以输入52个大小写字母

android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"只可以输入52个大小写字母

或者只接受数字与字母：

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1. android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"

android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"

关于 parseInt() 用法，下面再示一例：

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1. // 将16进制转换为10进制
2. int decimalNum = Integer.parseInt("101a",16);
3. System.out.println("decimalNum= " + decimalNum);
5. // 将10进制转换为16进制
6. String str = Integer.toHexString(100);
7. System.out.println("str= " + str);
8. int hexNum = Integer.parseInt(str);
9. System.out.println("hexNum= " + hexNum);
11. // Integer类还有toBinaryString()、toOctalString()将int转换为2进制、8进制

// 将16进制转换为10进制int decimalNum = Integer.parseInt("101a", 16);System.out.println("decimalNum= " + decimalNum);// 将10进制转换为16进制String str = Integer.toHexString(100);System.out.println("str= " + str);int hexNum = Integer.parseInt(str);System.out.println("hexNum= " + hexNum);// Integer类还有toBinaryString()、toOctalString()将int转换为2进制、8进制