这是我的代码:
String stringToSearch = "https://example.com/excludethis123456/moretext";
Pattern p = Pattern.compile("(?<=.com\\/excludethis).*\\/"); //search for this pattern
Matcher m = p.matcher(stringToSearch); //match pattern in StringToSearch
String store= "";
// print match and store match in String Store
if (m.find())
{
String theGroup = m.group(0);
System.out.format("'%s'\n", theGroup);
store = theGroup;
}
//repeat the process
Pattern p1 = Pattern.compile("(.*)[^\\/]");
Matcher m1 = p1.matcher(store);
if (m1.find())
{
String theGroup = m1.group(0);
System.out.format("'%s'\n", theGroup);
}
我想匹配excludethis
之后和之后的/
之前的所有内容。
使用"(?<=.com\\/excludethis).*\\/"
正则表达式,我将匹配123456/
并将其存储在String store
中。之后,使用"(.*)[^\\/]"
,我将排除/
并得到123456
。
我可以一行执行此操作,即将这两个正则表达式结合起来吗?我不知道如何合并它们。
答案 0 :(得分:1)
这可能对您有用:)
String stringToSearch = "https://example.com/excludethis123456/moretext";
Pattern pattern = Pattern.compile("excludethis([\\d\\D]+?)/");
Matcher matcher = pattern.matcher(stringToSearch);
if (matcher.find()) {
String result = matcher.group(1);
System.out.println(result);
}
答案 1 :(得分:1)
就像在正面使用正面表情一样,您可以在正面使用正面表情并将正则表达式更改为此,
(?<=.com/excludethis).*(?=/)
此外,在Java中,您无需转义/
您修改的代码,
String stringToSearch = "https://example.com/excludethis123456/moretext";
Pattern p = Pattern.compile("(?<=.com/excludethis).*(?=/)"); // search for this pattern
Matcher m = p.matcher(stringToSearch); // match pattern in StringToSearch
String store = "";
// print match and store match in String Store
if (m.find()) {
String theGroup = m.group(0);
System.out.format("'%s'\n", theGroup);
store = theGroup;
}
System.out.println("Store: " + store);
打印
'123456'
Store: 123456
就像您想获取价值一样。
答案 2 :(得分:1)
如果您不想使用regex
,则可以尝试使用String::substring
*
String stringToSearch = "https://example.com/excludethis123456/moretext";
String exclusion = "excludethis";
System.out.println(stringToSearch.substring(stringToSearch.indexOf(exclusion)).substring(exclusion.length(), stringToSearch.substring(stringToSearch.indexOf(exclusion)).indexOf("/")));
输出:
123456
* 绝对不实际使用此