Java Community | Help. Code. Learn.•9mo ago

Java .properties read Strings with the same name

Ok so I got a .properties file in my Project and now, I would like to read every single property, which starts with "API" in this example and return it as a list or something, how can I do that

31 Replies

JavaBot•9mo ago

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nettyOP•9mo ago

Defective•9mo ago

You can either iterate over all keys in properties and filter all that start with "API", or use streams Depends how you are currently reading your properties

nettyOP•9mo ago

like this

nettyOP•9mo ago

and resources

Defective•9mo ago

So you have a Properties object, right?

nettyOP•9mo ago

yes

Defective•9mo ago

Okay, sec So basically, you want to read all properties that start with specified string and store it in a list right? So in this example you expect to get a list containing just the api links?

nettyOP•9mo ago

yes indeed

Defective•9mo ago

So you have two options. Let me start with the one that is fairly simple to understand.

Properties props = ...

List<String> api_links = new ArrayList<>();

for(Map.Entry<Object, Object> entry : props.entrySet()) {
    if(entry.getKey() instanceof String key && key.startsWith("API")) {
        Object value = entry.getValue();
        if(value instanceof String valueString) {
            api_links.add(valueString);
        }
    }
}

Properties props = ...

List<String> api_links = new ArrayList<>();

for(Map.Entry<Object, Object> entry : props.entrySet()) {
    if(entry.getKey() instanceof String key && key.startsWith("API")) {
        Object value = entry.getValue();
        if(value instanceof String valueString) {
            api_links.add(valueString);
        }
    }
}

So, you can iterate over all entries in your Properties object Properties extends a map, so it also has an entrySet method, which returns a set of entries in this properties object You can then check every key one by one if it starts with desired string ("API" in your case), and if it does, just add a value associated with current entry to a List created before

nettyOP•9mo ago

ok I see

Defective•9mo ago

You should do instanceof checks, because Properties#entrySet() returns a set of <Object, Object> pairs, so it's safer to check if these Objects are Strings for sure

nettyOP•9mo ago

Ok I see, now what do I call or use to uhm write for example the first API thing into a string or something like where will the links be saved

Defective•9mo ago

In this example it will be the api_links List, it should contain all links read from Properties

nettyOP•9mo ago

api_links[] ojk ye I can just iterate through that list using .length right

Defective•9mo ago

size(), but yeah

nettyOP•9mo ago

ok do I need .getProperty also or no No Cuz its already saved in the list right

Defective•9mo ago

Yeah

nettyOP•9mo ago

Defective•9mo ago

This list is completely independent from the Properties object

nettyOP•9mo ago

ok good ok thats it I guess thanks men

JavaBot•9mo ago

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Defective•9mo ago

You are welcome

nettyOP•9mo ago

uh another question rq How do I create such a list, with as much elements as a method for example needs cuz I got a vars.java file where all read properties are saved as static

nettyOP•9mo ago

would this work

Defective•9mo ago

Depends on what you want to achieve, but if your file looks roughly like this:

public static class Vars {
  public static List<String> apis;

  public static void setApis(List<String> api) {
    apis = api
  }
}

public static class Vars {
  public static List<String> apis;

  public static void setApis(List<String> api) {
    apis = api
  }
}

then that should work. You will be able to access the same list by just calling Vars.apis

nettyOP•9mo ago

ok, will Vars.apis return single strings or god ive never used this, that shit is breaking my brain

Defective•9mo ago

It returns the very same List you passed with setApis setApis just assigns your list to a variable in the class

nettyOP•9mo ago

ok it wokred fire ok for now ill try to work w that thanks mate

JavaBot•9mo ago

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Java .properties read Strings with the same name

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