CODEWITHSUNDEEP
javaspring-boot
qccaprospect
qccaprospect

Reputation: 151

How to Pass URL as Path Variable

I am trying to pass a URL as a path variable, but when I input the URL as a parameter and access the route, it returns an error. I want to be able to see the address after it passes into the route as a parameter.

@RequestMapping("/addAddress/{address}")
public String addAddress(@PathVariable("address") String address) {
    System.out.println("Address: "+address)
    return address;
}

For example, if I put into the URL:

localhost:8080/addAddress/http://samplewebsite.com

I should see 

http://samplewebsite.com

printed out in the back end.

Upvotes: 5

Views: 17317

Answers (3)

DCTID
DCTID

Reputation: 1337

The forward slashes are your issue.

You have a few choices.

  1. Use 2 path variables. This still works with a double forward slash. This works with URL: "localhost:8080/addAddress/http://samplewebsite.com"

    @GetMapping("/addAddress/{schema}/{address}")
public String addAddress(@PathVariable("schema") String schema, @PathVariable("address") String address) {
    System.out.println("Address: "+ schema + "//" + address);
    return schema + "//" + address;
}

  2. Use a query (request) param, your url would then be URL: "localhost:8080/addAddress/?address=http://samplewebsite.com"

    @GetMapping("/addAddress2")
public String addAddress2(@RequestParam("address") String address) {
    System.out.println("Address: "+address);
    return address;
}

  3. Encode the slashes in URL: 

    localhost:8080/addAddress/http:%2F%2Fsamplewebsite.com"

    and configure Tomcat or Jetty, whatever you use to allow encoded slashes. Here is an example in Tomcat

Upvotes: 4

buræquete
buræquete

Reputation: 14678

You can do the following;

@RequestMapping("/addAddress/**")
public String addAddress(HttpServletRequest request) {
    String fullUrl = request.getRequestURL().toString();
    String url = fullUrl.split("/addAddress/")[1];
    System.out.println(url);
    return url;
}

with @PathVariable that is not doable due to the / char breaking the behaviour you are looking for, unless you encode/decode, but I feel like this is a simpler way to go for both user of the endpoint, and for the backend.

Also this will not fetch the request query part, e.g. ?input=user, to do that you can add this logic

Upvotes: 3

Milton BO
Milton BO

Reputation: 608

You can use SafeUrl to parse your URL in your path.

Some documentation:

https://www.urlencoder.io/java/

It let you pass parameters safe by the url and you can decode where you need.

Upvotes: 0

Related Questions