如何从url中提取id？谷歌表

Question

我有以下网址。

https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY

Foreach url，我需要将工作表 id: 提取1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY到一个 java 字符串中。

我正在考虑使用split但它不能用于所有测试用例：

String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
String[] parts = string.split("/");
String res = parts[parts.length-2];
Log.d("hello res",res );

我怎么可能呢？

Answer 1

您可以使用 regex \/d\/(.*?)(\/|$)( regex demo ) 来解决您的问题，如果您仔细观察，您可以看到 ID 之间存在d/，/或者end of line为此，您可以获得这两者之间的所有内容，请查看此代码演示：

String[] urls = new String[]{
    "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258",
    "https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258",
    "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY"
};

String regex = "\\/d\\/(.*?)(\\/|$)";
Pattern pattern = Pattern.compile(regex);

for (String url : urls) {
    Matcher matcher = pattern.matcher(url);
    while (matcher.find()) {
        System.out.println(matcher.group(1));
    }
}

输出

1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY

Answer 2

看起来您要查找的 id 始终遵循“/spreadsheets/d/”，如果是这种情况，您可以将代码更新为该 ID

        String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
        String[] parts = string.split("spreadsheets/d/");
        String result;
        if(parts[1].contains("/")){
            String[] parts2 = parts[1].split("/");
            result = parts2[0];
        }
        else{
            result=parts[1];
        }
        System.out.println("hello "+ result);