2016-11-27 50 views
0

比方說,當URL包含有效的JSON時,我需要使用RestTemplate作爲GET請求。一個例子URL是:當URL包含JSON時處理RestTemplate的模式匹配

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

的代碼:

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的開始,而不是整個URL 。從實驗中,我知道花括號是問題所在。但是,使用反斜槓轉義它們並不能解決問題。下面的代碼產生「未封閉組」異常:

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) 

我怎樣才能讓RestTemplate正確引用網址是什麼?

+0

爲什麼你需要納入'json' url中,如果它是一個'GET'要求? – NiVeR

+0

不幸的是,API就是這麼設計的...... – Ozilophile

+0

正如David S.指出的那樣,URL編碼是關鍵。然而,質疑API設計。另請注意,瀏覽器/服務器確實有字符限制。使用POST方法的請求體將更有意義 – Hirako

回答

1

您試圖直接在URL中傳遞JSON字符串,從而破壞了url編碼。在使用resttemplate之前,您應該對您的JSON負載進行網址編碼。

轉問這個問題,要學會如何來urlencode幾個方面:

spring resttemplate url encoding