Java正则表达式：用于不带前导空格或0或+；

在从电话号码中开始实际数字字符之前，我想删除前导0、+或任何空格（\s）。也许这并不难，但因为我对雷吉克斯还不熟悉，所以我一直在寻求帮助。我试着做我自己的，但没有成功

这是一个相同的链接，但它也添加了leading+，但我不希望这样

因此，我尝试了这一点，但它也删除了内部0

(^\s*|0*|[+]*)

我也尝试过这个，但实际上它在java中不起作用，而只在Php中起作用，所以我需要基于java的regix的帮助

^(?:\s*0+|[+]0*|(\d+)0*)(?!$)

示例输入

+490232345678
0049032345678
+1 (555) 234-5078
+7 (23) 45/6789+10
(0123) 345/5678, ext. 666

期望输出

490232345678
49032345678
15552345678
72345678910
1233455678666

我只需要regix，因为我知道如何在java中使用regix。 我有一段代码需要regix

    String value = "+490232345678";
    Pattern p = Pattern.compile("(^\s*|0*|[+]*)");      
    Matcher m = p.matcher(value);
    value = m.replaceAll("");   

试试这个：

public static void main(String[] args) {
    String [] testResult = {"+490232345678", 
                            "0049032345678", 
                            "+1 (555) 234-5078", 
                            "+7 (23) 45/6789+10", 
                            "(0123) 345/5678, ext. 666"};
    String reg = "^([\\(+ ]0| +|0+|\\(\\)|\\+)| +|[^\\d]+|/$";
    for (String phone : testResult) {
        System.out.println(phone.replaceAll(reg, ""));
    }
}

输出将是：

490232345678
49032345678
15552345078
72345678910
1233455678666

更简单的方法是分两步进行：

 .replaceAll("[^\\d]+", "").replaceAll("^0+", "")

删除所有非数字，然后删除前导零

正则表达式解释

^([\(+ ]0| +|0+|\(\)|\+)| +|[^\d]+|\/$

 1st Alternative ^([\(+ ]0| +|0+|\(\)|\+)
     ^ asserts position at start of the string
   1st Capturing Group ([\(+ ]0| +|0+|\(\)|\+)
       1st Alternative [\(+ ]0
           Match a single character present in the list below [\(+ ]
           \( matches the character ( literally (case sensitive)
           +  matches a single character in the list +  (case sensitive)
           0 matches the character 0 literally (case sensitive)
       2nd Alternative  +
           + matches the character   literally (case sensitive)
           + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
       3rd Alternative 0+
           0+ matches the character 0 literally (case sensitive)
           + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
       4th Alternative \(\)
           \( matches the character ( literally (case sensitive)
           \) matches the character ) literally (case sensitive)
       5th Alternative \+
           \+ matches the character + literally (case sensitive)
 2nd Alternative  +
    + matches the character   literally (case sensitive)
    + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
 3rd Alternative [^\d]+
    Match a single character not present in the list below [^\d]+
    + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
    \d matches a digit (equal to [0-9])
 4th Alternative \/$
    \/ matches the character / literally (case sensitive)
 $ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

解释来源：

这是一个来自

试试这个：

public static void main(String[] args) {
    String [] testResult = {"+490232345678", 
                            "0049032345678", 
                            "+1 (555) 234-5078", 
                            "+7 (23) 45/6789+10", 
                            "(0123) 345/5678, ext. 666"};
    String reg = "^([\\(+ ]0| +|0+|\\(\\)|\\+)| +|[^\\d]+|/$";
    for (String phone : testResult) {
        System.out.println(phone.replaceAll(reg, ""));
    }
}

输出将是：

490232345678
49032345678
15552345078
72345678910
1233455678666

更简单的方法是分两步进行：

 .replaceAll("[^\\d]+", "").replaceAll("^0+", "")

删除所有非数字，然后删除前导零

正则表达式解释

^([\(+ ]0| +|0+|\(\)|\+)| +|[^\d]+|\/$

 1st Alternative ^([\(+ ]0| +|0+|\(\)|\+)
     ^ asserts position at start of the string
   1st Capturing Group ([\(+ ]0| +|0+|\(\)|\+)
       1st Alternative [\(+ ]0
           Match a single character present in the list below [\(+ ]
           \( matches the character ( literally (case sensitive)
           +  matches a single character in the list +  (case sensitive)
           0 matches the character 0 literally (case sensitive)
       2nd Alternative  +
           + matches the character   literally (case sensitive)
           + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
       3rd Alternative 0+
           0+ matches the character 0 literally (case sensitive)
           + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
       4th Alternative \(\)
           \( matches the character ( literally (case sensitive)
           \) matches the character ) literally (case sensitive)
       5th Alternative \+
           \+ matches the character + literally (case sensitive)
 2nd Alternative  +
    + matches the character   literally (case sensitive)
    + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
 3rd Alternative [^\d]+
    Match a single character not present in the list below [^\d]+
    + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
    \d matches a digit (equal to [0-9])
 4th Alternative \/$
    \/ matches the character / literally (case sensitive)
 $ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

解释来源：

这是一个来自

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伙计，我不需要任何代码。我只是在搜索正则表达式。我知道如何编写java代码。但是我正在寻找一个可以与我的input@Mubasher看看我的答案。伙计，我不需要任何代码。我只是在搜索正则表达式。我知道如何编写java代码。但是我正在寻找一个可以与我的input@Mubasher看看我的答案，谢谢你，乔治。Jorge，在我的应用程序中不可能有两个步骤，在我们的应用程序中，我们有一个通用验证器，它接受regix输入并将捕获组替换为替换字符串，所以我将使用您的第一个regix。这个validtor是通用的，只需更改regix和replacement，就可以应用于任何类型的数据。jorge，如果你能解释一下它是如何工作的，那将非常有帮助。因为我正在学习regix和regix中的新蜜蜂，所以这将有助于我的理解。如果你有时间的话。好的，我会的，等一下！奇怪的是，我也在尝试regex101.com，但没能成功。非常感谢，谢谢你，乔治。Jorge，在我的应用程序中不可能有两个步骤，在我们的应用程序中，我们有一个通用验证器，它接受regix输入并将捕获组替换为替换字符串，所以我将使用您的第一个regix。这个validtor是通用的，只需更改regix和replacement，就可以应用于任何类型的数据。jorge，如果你能解释一下它是如何工作的，那将非常有帮助。因为我正在学习regix和regix中的新蜜蜂，所以这将有助于我的理解。如果你有时间的话。好的，我会的，等一下！奇怪的是，我也在尝试regex101.com，但没能成功。非常感谢你。