Java 从 url 字符串中摆脱获取参数的最佳方法是什么？
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What's the best way to get rid of get parameters from url string?
提问by Maksym
I have URL string like:
我有如下 URL 字符串：
"http://www.xyz/path1/path2/path3?param1=value1¶m2=value2".
“ http://www.xyz/path1/path2/path3?param1=value1¶m2=value2”。
I need to get this url without parameters, so the result should be:
我需要在没有参数的情况下获取这个 url，所以结果应该是：
"http://www.xyz/path1/path2/path3".
“ http://www.xyz/path1/path2/path3”。
I have done it this way:
我是这样做的：
private String getUrlWithoutParameters(String url)
{
return url.substring(0,url.lastIndexOf('?'));
}
Are there any better ways to do it?
有没有更好的方法来做到这一点？
采纳答案by Eran
Probably not the most efficient way, but more type safe :
可能不是最有效的方法，但更安全：
private String getUrlWithoutParameters(String url) throws URISyntaxException {
URI uri = new URI(url);
return new URI(uri.getScheme(),
uri.getAuthority(),
uri.getPath(),
null, // Ignore the query part of the input url
uri.getFragment()).toString();
}
回答by PbxMan
I normally use
我通常使用
url.split("\?")[0]
回答by SMA
Try using substring and indexOf method in String:
尝试在 String 中使用 substring 和 indexOf 方法：
String str = "http://www.xyz/path1/path2/path3?param1=value1¶m2=value2";
int index = str.indexOf("?");
if (index != -1) {
System.out.println(str.substring(0, str.indexOf("?")));
} else {
System.out.println("You dont have question mark in your url");
}
回答by Joop Eggen
With an URL it could be done by methods. With a String:
使用 URL 可以通过方法来完成。使用字符串：
url = url.replaceFirst("\?.*$", "");
This attempts to replace all starting with a question mark. When no question mark, the original string is kept.
这试图替换所有以问号开头的内容。当没有问号时，保留原始字符串。
回答by dchar
You could use something like the following, that removes the query part from the URL.
您可以使用类似以下内容，从 URL 中删除查询部分。
private String getUrlWithoutParameters(String url) throws MalformedURLException {
return url.replace(new URL(url).getQuery(), "");
}
Alternatively, you might want to check if url rewrite covers your needs: http://tuckey.org/urlrewrite/
或者，您可能想检查 url rewrite 是否满足您的需求：http: //tuckey.org/urlrewrite/
回答by Adrian Baker
Using javax.ws.rs.core.UriBuilderfrom JAX-RS 2.0:
使用JAX-RS 2.0 中的javax.ws.rs.core.UriBuilder：
UriBuilder.fromUri("https://www.google.co.nz/search?q=test").replaceQuery(null).build();
Using the very similar org.springframework.web.util.UriBuilderfrom Spring:
使用Spring 中非常相似的org.springframework.web.util.UriBuilder：
UriComponentsBuilder.fromUriString("https://www.google.co.nz/search?q=test").replaceQuery(null).build(Collections.emptyMap());
回答by sendon1982
You can use org.apache.http.client.utils.URIBuilderfrom httpclient jar
您可以org.apache.http.client.utils.URIBuilder从 httpclient jar 中使用
@Test
void test_removeQueryParameters() throws Exception {
String url = "http://www.google.com?query=Shanghai&foo=bar";
String expectedUrl = "http://www.google.com";
String result = removeQueryParameters(url);
Assert.assertThat(result, equalTo(expectedUrl));
}
public String removeQueryParameters(String url) throws URISyntaxException {
URIBuilder uriBuilder = new URIBuilder(url);
uriBuilder.removeQuery();
return uriBuilder.build().toString();
}
