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:
- String str = editor.getText().toString().trim();
- int port = Integer.parseInt(str);
- 08-15 07:13:48.294: ERROR/AndroidRuntime(12281): java.lang.NumberFormatException: unable to parse 'dd' as integer
- /**
- * 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);
- }
那么,如何解决这个问题?
简单办法,就是捕捉异常信息:
- String str = editor.getText().toString().trim();
- try {
- int port = Integer.parseInt(str);
- } catch(NumberFormatException nfe) {
- // to do
- }
- <EditText
- android:id="@+id/edit"
- android:layout_width="fill_parent"
- android:layout_height="wrap_content"
- android:inputType="number"
- android:digits="0123456789"/>

类似的,只接受大小写字母:
- android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"只可以输入52个大小写字母
- android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
关于 parseInt() 用法,下面再示一例:
- // 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