2016-11-27 62 views
0

比方说,当URL包含有效的JSON时,我需要使用RestTemplate作为GET请求。一个例子URL是:当URL包含JSON时处理RestTemplate的模式匹配

http://example.com/?json={"property1":{"property2":0},"property3":{"property4":0.0,"property5":0.0}} 

的代码:

String url = "http://example.com/?json={\"property1\":{\"property2\":0},\"property3\":{\"property4\":0.0,\"property5\":0.0}}"; 
RestTemplate template = new RestTemplate(); 
ResponseEntity<String> response = template.getForEntity(url, String.class); 

除外:

Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition near index 32 
\Qhttp://example.com/?json=\E({"property2":0)\Q,"property3":\E(0.0,"property5":0.0)\Q}\E 
          ^
at java.util.regex.Pattern.error(Pattern.java:1955) 
at java.util.regex.Pattern.closure(Pattern.java:3157) 
at java.util.regex.Pattern.sequence(Pattern.java:2134) 
at java.util.regex.Pattern.expr(Pattern.java:1996) 
at java.util.regex.Pattern.group0(Pattern.java:2905) 
at java.util.regex.Pattern.sequence(Pattern.java:2051) 
at java.util.regex.Pattern.expr(Pattern.java:1996) 
at java.util.regex.Pattern.compile(Pattern.java:1696) 
at java.util.regex.Pattern.<init>(Pattern.java:1351) 
at java.util.regex.Pattern.compile(Pattern.java:1028) 
at org.springframework.web.util.UriTemplate$Parser.getMatchPattern(UriTemplate.java:247) 
at org.springframework.web.util.UriTemplate$Parser.access$200(UriTemplate.java:196) 
at org.springframework.web.util.UriTemplate.<init>(UriTemplate.java:70) 
at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:471) 
at org.springframework.web.client.RestTemplate.getForEntity(RestTemplate.java:260) 

RestTemplate被引用的URL到JSON的开始,而不是整个URL 。从实验中,我知道花括号是问题所在。但是,使用反斜杠转义它们并不能解决问题。下面的代码产生“未封闭组”异常:

String url = "http://example.com/?json=\\{\"property1\":\\{\"property2\":0\\}\\}"; 
RestTemplate template = new RestTemplate(); 
ResponseEntity<String> response = template.getForEntity(url, String.class); 

例外:

Exception in thread "main" java.util.regex.PatternSyntaxException: Unclosed group near index 56 
\Qhttp://example.com/?json=\\E(\{"property2":0\)\Q\}\E 
                ^
at java.util.regex.Pattern.error(Pattern.java:1955) 
at java.util.regex.Pattern.accept(Pattern.java:1813) 
at java.util.regex.Pattern.group0(Pattern.java:2908) 
at java.util.regex.Pattern.sequence(Pattern.java:2051) 
at java.util.regex.Pattern.expr(Pattern.java:1996) 
at java.util.regex.Pattern.compile(Pattern.java:1696) 
at java.util.regex.Pattern.<init>(Pattern.java:1351) 
at java.util.regex.Pattern.compile(Pattern.java:1028) 
at org.springframework.web.util.UriTemplate$Parser.getMatchPattern(UriTemplate.java:247) 
at org.springframework.web.util.UriTemplate$Parser.access$200(UriTemplate.java:196) 
at org.springframework.web.util.UriTemplate.<init>(UriTemplate.java:70) 
at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:471) 
at org.springframework.web.client.RestTemplate.getForEntity(RestTemplate.java:260) 
at com.ebay.sdo.misc.ThemisTest.main(ThemisTest.java:55) 

我怎样才能让RestTemplate正确引用网址是什么?

+0

为什么你需要纳入'json' url中,如果它是一个'GET'要求? – NiVeR

+0

不幸的是,API就是这么设计的...... – Ozilophile

+0

正如David S.指出的那样,URL编码是关键。然而,质疑API设计。另请注意,浏览器/服务器确实有字符限制。使用POST方法的请求体将更有意义 – Hirako

回答

1

您试图直接在URL中传递JSON字符串,从而破坏了url编码。在使用resttemplate之前,您应该对您的JSON负载进行网址编码。

转问这个问题,要学会如何来urlencode几个方面:

spring resttemplate url encoding