How to send JSON as a parameter in url using Spring RestTemplate?

问题

I am trying to achieve same thing as this: How to use query parameter represented as JSON with Spring RestTemplate?, sending JSON string as a URL parameter in restTemplate.exchange().

The accepted answer mentions that sending JSON object as request parameter is generally not a good idea since you will probably face problem with curly brackets inside JSON. That is exactly what is happening when I am trying to make a GET call to an API. Since this is an API from another system, I cannot ask them to change the format and will have to call the GET endpoint, passing JSON as parameter. How can I achieve this in restTemplate.exchange() call?

Note: The mentioned related question does not guide on how to overcome this problem and I do not have enough reputation to comment on it to ask the author of the answer.

回答1:

Answering my own question. While it is a bad idea to pass JSON like this in a query/url parameter, there is a workaround as suggested here: https://jira.spring.io/browse/SPR-9220?focusedCommentId=76760&page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel#comment-76760.
Replicating the code here in case this link goes dead:

String url = "http://localhost:8983/solr/select?wt=json&indent=true&fl=*&q=*:*&fq={!geofilt}&sfield=venue_location&pt=28.0674,-80.5595&d=25";
URI uri = UriComponentsBuilder.fromUriString(url).build().encode().toUri();

System.out.println(uri);
// http://localhost:8983/solr/select?wt=json&indent=true&fl=*&q=*:*&fq=%7B!geofilt%7D&sfield=venue_location&pt=28.0674,-80.5595&d=25

Basically, instead of passing url having JSON query/url parameters as a string, pass it as a URI. Then call exchange method as before, but with URI instead of String:

restTemplate.exchange(uri, HttpMethod.GET, requestEntity, String.class)

回答2:

You can use URLEncoder class to encode the URL in exchange method, e.g.:

String url = "http://www.yoursite.com/api?param={\"some_key\":\"some_value\"}";
System.out.println(URLEncoder.encode(url, StandardCharsets.UTF_8.name()));

This will encode the characters (like braces and double quotes) and server then will decode it back to json.

回答3:

If this is 3rd party API and you cannot control or change JSON processing on backend side - there is no solution. Even if you will encode with URLEncoder - there is no guarantee that API backend would process such request correctly.

来源：https://stackoverflow.com/questions/43252750/how-to-send-json-as-a-parameter-in-url-using-spring-resttemplate

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