Question

我从JSON转移到字符串的一个字符串如下：

"FIELDLIST": [      "Insurance Num",      "Insurance PersonName",      "InsurancePayDate",      "InsuranceFee",      "InsuranceInvType"    ]

我试图剥离空间，我希望结果是：

"FIELDLIST":["Insurance Num","Insurance PersonName","InsurancePayDate","InsuranceFee","InsuranceInvType"]

我在c＃中编写代码如下：

string[] rearrange_sign = { ",", "[", "]", "{", "}", ":" };
string rtnStr = _prepareStr;

for (int i = 0; i < rearrange_sign.Length; i++)
{
    while (true)
    {
        rtnStr = rtnStr.Replace(@rearrange_sign[i] + " ", rearrange_sign[i]);
        rtnStr = rtnStr.Replace(" " + @rearrange_sign[i], rearrange_sign[i]);
        if (rtnStr.IndexOf(@rearrange_sign[i] + " ").Equals(-1) && rtnStr.IndexOf(" " + @rearrange_sign[i]).Equals(-1))
        {
            break;
        }
    }
}

但它根本不起作用，似乎我必须使用Regex来替换，我该如何使用？

Answer 1

使用正则表达式(\s(?=")|\s(?=\])|\s(?=\[))+

这是代码：

string str = "\"FIELDLIST\": [ \"Insurance Num\", \"Insurance PersonName\", \"InsurancePayDate\", \"InsuranceFee\", \"InsuranceInvType\" ] ";
string result = System.Text.RegularExpressions.Regex.Replace(str, "(\\s(?=\")|\\s(?=\\])|\\s(?=\\[))+", string.Empty);

Answer 2

只匹配两个非单词字符之间存在的空格，然后用空字符串替换匹配的空格。

@"(?<=\W)\s+(?=\W)"

DEMO

<强>代码：

string str = @"""FIELDLIST"": [      ""Insurance Num"",      ""Insurance PersonName"",      ""InsurancePayDate"",      ""InsuranceFee"",      ""InsuranceInvType""    ]";
string result = Regex.Replace(str, @"(?<=\W)\s+(?=\W)", "");
Console.WriteLine(result);

<强>输出：

"FIELDLIST":["Insurance Num","Insurance PersonName","InsurancePayDate","InsuranceFee","InsuranceInvType"]

IDEONE

如何使用Regex替换此字符串

2 个答案: