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Q&A Escape both reserved characters and curly braces in a URI in Spring Boot

We ended up using the old java.net.URLEncoder . More specifically, we used URLEncoder.encode(string, Charsets.UTF_8). We use it to encode the values for the query parameter, and then build a URL ...

posted 1y ago by FractionalRadix‭  ·  edited 1y ago by FractionalRadix‭

Answer
#2: Post edited by user avatar FractionalRadix‭ · 2023-10-26T08:44:40Z (about 1 year ago)
  • We ended up using the old [java.net.URLEncoder](https://docs.oracle.com/javase/8/docs/api/java/net/URLEncoder.html) .
  • We use it to encode the values for the query parameter, and then build a URL from it using String concatenation.
  • This solves most of the problems, if you're willing to overlook that space is encoded to `+` rather than `%20`, and that you should not apply it to a complete URI or URL.
  • If you apply `URLEncoder.encode(...)` to "https://example.com/something", it will return `https%3A%2F%2Fexample.com%2Fsomething`... so, you have to apply it to the individual components of your URI or URL, and then concatenate them.
  • It's not a particularly elegant solution, but it works.
  • We ended up using the old [java.net.URLEncoder](https://docs.oracle.com/javase/8/docs/api/java/net/URLEncoder.html) .
  • More specifically, we used `URLEncoder.encode(string, Charsets.UTF_8)`.
  • We use it to encode the values for the query parameter, and then build a URL from it using String concatenation.
  • This solves most of the problems, if you're willing to overlook that space is encoded to `+` rather than `%20`, and that you should not apply it to a complete URI or URL.
  • If you apply `URLEncoder.encode(...)` to "https://example.com/something", it will return `https%3A%2F%2Fexample.com%2Fsomething`... so, you have to apply it to the individual components of your URI or URL, and then concatenate them.
  • It's not a particularly elegant solution, but it works.
#1: Initial revision by user avatar FractionalRadix‭ · 2023-10-26T08:42:56Z (about 1 year ago)
We ended up using the old [java.net.URLEncoder](https://docs.oracle.com/javase/8/docs/api/java/net/URLEncoder.html) .  

We use it to encode the values for the query parameter, and then build a URL from it using String concatenation.  

This solves most of the problems, if you're willing to overlook that space is encoded to `+` rather than `%20`, and that you should not apply it to a complete URI or URL.  

If you apply `URLEncoder.encode(...)` to "https://example.com/something", it will return `https%3A%2F%2Fexample.com%2Fsomething`... so, you have to apply it to the individual components of your URI or URL, and then concatenate them.  

It's not a particularly elegant solution, but it works.