java 从电话号码字符串条目中删除所有特殊字符，除了 + 仅出现在第一位
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Remove all special characters from a phone number string entry except + occurring only at first place
提问by S P
I want to remove all special characters from a phone number string entry except + symbol. That too, if it is occurring only at first place. Example : +911234567890 should be valid but +91+1234#1234 should be invalid.
我想从电话号码字符串条目中删除除 + 符号之外的所有特殊字符。这也是，如果它只发生在第一位。例如：+911234567890 应该是有效的，但 +91+1234#1234 应该是无效的。
回答by Nargis
You can use something like:
你可以使用类似的东西：
String number = "+91+1234#1234"
number=number.replaceAll("[\D]", "")
This will replace all non digit characters with space but then for your additional "+" in the beginning ,you may need to add it as a prefix to the result.
这将用空格替换所有非数字字符，但是对于开头的附加“+”，您可能需要将其添加为结果的前缀。
Hope this helps!
希望这可以帮助！
回答by Yakiv Mospan
The best way is to use regular expression:
最好的方法是使用正则表达式：
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main
{
public static void main(String[] args)
{
String sPhoneNumber = "+911234567890";
Pattern pattern = Pattern.compile("^[+]\d*");
Matcher matcher = pattern.matcher(sPhoneNumber);
if (matcher.matches()) {
System.out.println("Phone Number Valid");
} else {
System.out.println("Phone Number must start from + ");
}
}
}
Best wishes.
最好的祝愿。
回答by user2515154
Scanner scan=new Scanner(System.in);
String input=scan.next();
String onlyDigits = input.replaceAll("[^0-9]+","");
System.out.println(onlyDigits);
回答by Dulanga
Try this
试试这个
public static void main(String arg[]){
String num="+45*#545454*j";
String edited="";
for (int i=0;i<num.length();i++){
char c=num.charAt(i);
if (i==0&&c=='+'){
edited+=c;
}
else if (Character.isDigit(c)){
edited+=c;
}
}
System.out.println(edited);
}
}
回答by NoJoshua
Use a MaskFormatterto constrain the input, something like this
使用MaskFormatter来约束输入，像这样
MaskFormatter formatter = null;
try {
formatter = new MaskFormatter("+###");
} catch (ParseException e) {
e.printStackTrace();
}
JFormattedTextField tf = new JFormattedTextField(formatter);
The string "+###" in the constructor defines what can be typed in the textfield; in this case the first character needs to be '+' and the following three needs to be digits. See the APIfor details.
构造函数中的字符串“+###”定义了可以在文本字段中输入的内容；在这种情况下，第一个字符需要是“+”，接下来的三个字符需要是数字。有关详细信息，请参阅API。
回答by Sanjaya Liyanage
Here is an answer for you.But please give a try and then ask the question so that you can improve yourself. At least give a try to understand the code without directly use it
这是给您的答案。但请尝试一下，然后提出问题，以便您可以提高自己。至少尝试理解代码而不直接使用它
public static boolean isValid(String number) {
if(number.startsWith("+")){
number=number.substring(1);
}
try {
Integer.parseInt(number);
} catch (Exception e) {
return false;
}
return true;
}
