Question

我从客户端获取我的Java应用程序中的以下字符串。

[["UIButton","Submit","15","30","80","80"],["UILabel","User name","15","75","80","80"],["UITextField","Jonathan","15","75","80","80"]]ˇ˛END

我期望解析这个并分别取出3个字符串，如下所示：

"UIButton","Submit","15","30","80","80"
"UIButton","Submit","15","30","80","80"
"UIButton","Submit","15","30","80","80"

我试过如下，

    // resultStr = [["UIButton","Submit","15","30","80","80"],["UILabel","User name","15","75","80","80"],["UITextField","Jonathan","15","75","80","80"]]ˇ˛END
    if ( resultStr.length()>0 && resultStr!=null ) {

        for (int i=0; i<resultStr.length(); i++) {
            int startInd = resultStr.indexOf('[');
            int endInd = resultStr.indexOf(']');
            if ( startInd>=0 && endInd>0 ) {
                String resStr =  resultStr.substring(startInd , endInd);
                if ( resStr!=null )
                    System.out.println("Applet resStr: " + resStr);
            }
            else
                System.out.println("Applet startindex, endindex failed");
        }
    }

但是，这不起作用，这不正确地解析我所期望的。有人可以建议如何按预期分别解析字符串吗？

Answer 1

您可以使用String.split()方法分割此字符串：

String[] t = myString.substring(2, s.length()-7).split("\\],\\[");

所以你的代码看起来像这样：

//resultStr = [["UIButton","Submit","15","30","80","80"],["UILabel","User name","15","75","80","80"],["UITextField","Jonathan","15","75","80","80"]]ˇ˛END
if ( resultStr.length()>7 && resultStr!=null ) {

    String[] resStrings = resultStr.substring(2, resultStr.length()-7).split("\\],\\[");    
    for (String resString: resStrings) {
        System.out.println("Applet resStr: " + resStr);
    }
}

因此，您应该有以下输出：

Applet resStr: "UIButton","Submit","15","30","80","80"
Applet resStr: "UILabel","User name","15","75","80","80"
Applet resStr: "UITextField","Jonathan","15","75","80","80"

Answer 2

使用 Lazy 方式和Lookaround

尝试使用模式和匹配器

(?<=\[).*?(?=\])

online demo

OR possessive量词

(?<=\[)[^]]*+(?=\])

online demo

示例代码：

Matcher matcher = Pattern.compile("(?<=\\[).*?(?=\\])").matcher(strring);
while (matcher.find()) {
    System.out.println(matcher.group());
}

在Java中解析冗长的字符串

2 个答案: