Convert Map to Bundle in android

Question

Is there an easy way to convert a Map to a Bundle in android without explicit iteration?

Why?

Firebase returns a Map for Notification getData(). I need to pass the data to an intent. Formerly GCM gave me a bundle, so I didn't need to worry about this.

Organon

"Is there an easy way to convert a Map to a Bundle in android without explicit iteration?" -- note that a Map does not necessarily equate to a Bundle. A Map can have keys that are not strings. A Map can have values that are not one of the supported data types. – Angary

Yes, but in my case they are all primitives and String – Organon

You could serialize it. – Rhizo

The RemoteMessage object is parcelable, you could just send that directly in the intent – Demivolt

Answer 1

I guess a good old fashioned for loop is the easiest way:

    Bundle bundle = new Bundle();
    for (Map.Entry<String, String> entry : getData().entrySet()) {
        bundle.putString(entry.getKey(), entry.getValue());
    }

Lingonberry

this will probably throw an exception because first parameter "google_sent_time" has a Long type and you are using putString. you better use the intent to handle type variation for you and use intent.putExtra(entry.getKey(), entry.getValue()) so different data types dont make problem for you. – Spectrum

@AmirZiarati I don't think that is true anymore, RemoteMessage#getData() now returns a Map<String, String> excluding a bunch of the internal values. – Raye

Answer 2

Neat Kotlin solution using spread operator will look like that:

fun Map<String, Any?>.toBundle(): Bundle = bundleOf(*this.toList().toTypedArray())

Belgravia

Worth noting is that the spread operator is not the best for performance and the above solution will generate a performance warning in some static analysis tools like for example Detekt: detekt.github.io/detekt/performance.html#spreadoperator – Fillender

worth mentioning that this won't create nested bundles for nested maps, in case you're using something like FIrebase Analytics that requires that – Sleeping

Answer 3

Came across this same issue with firebase messaging and created a kotlin extension function for it. The gist is here, code below. Although I am using this method there are some caveats:

• it doesn't cover all of the types that can be put into a bundle
• it is still under development and hasn't been fully tested

With this in mind, please use it as a guide not a definitive solution. I will keep the gist up to date as it evolves.

import android.os.Bundle 
import android.os.IBinder
import android.os.Parcelable
import java.io.Serializable

fun <V> Map<String, V>.toBundle(bundle: Bundle = Bundle()): Bundle = bundle.apply {
  forEach {
    val k = it.key
    val v = it.value
    when (v) {
      is IBinder -> putBinder(k, v)
      is Bundle -> putBundle(k, v)
      is Byte -> putByte(k, v)
      is ByteArray -> putByteArray(k, v)
      is Char -> putChar(k, v)
      is CharArray -> putCharArray(k, v)
      is CharSequence -> putCharSequence(k, v)
      is Float -> putFloat(k, v)
      is FloatArray -> putFloatArray(k, v)
      is Parcelable -> putParcelable(k, v)
      is Serializable -> putSerializable(k, v)
      is Short -> putShort(k, v)
      is ShortArray -> putShortArray(k, v)

//      is Size -> putSize(k, v) //api 21
//      is SizeF -> putSizeF(k, v) //api 21

      else -> throw IllegalArgumentException("$v is of a type that is not currently supported")
//      is Array<*> -> TODO()
//      is List<*> -> TODO()
    }
  }
}

Orfinger

never heard of kotlin before today; interesting find – Turin

Answer 4

Nowadays you can use fun bundleOf(vararg pairs: Pair<String, Any?>)

Intermarriage

Answer 5

You can use writeToParcel(Parcel out, int flags)to generate a Parcel that could be similarly useful, since it's a parent class of Bundle, and it's handily built into the Firebase framework as part of the RemoteMessage class. Documentation is here.

Trimester

Parcel is a superclass of Bundle, so while it can be similarly useful, they can't be used interchangeably. – Orthostichy

Yep, I did mention that it's the parent class of it. – Trimester

Your answer maybe is better, but the lack of explanation or example is making it worst. – Assumption

Answer 6

here is how I did it in Kotlin

val bundle = Bundle()

for (entry in data.entries)
    bundle.putString(entry.key, entry.value)

Nagoya

Answer 7

In a kotlin way:

private fun Map<String, Any>.toBundle(): Bundle {
    val pairs = this.map { entry ->
        Pair(entry.key, entry.value)
    }.toTypedArray()
    return bundleOf(*pairs)
}

Wolter

Answer 8

Here is an extention method in Kotlin:

fun Map<String, Any>.toBundle() = bundleOf(*toList().toTypedArray())

Bartonbartosch

Answer 9

In Kotlin you can build a function extension to convert Map to Bundle without explicit iteration like so:

override fun onMessageReceived(remoteMessage: RemoteMessage) {
        val payload = remoteMessage.data.toBundle()
        Log.d(TAG, "Notification payload: $payload")
        // ...
}

where toBundle() is defined like so:

inline fun <reified T> Map<String, T>.toBundle(): Bundle = bundleOf(*toList<String, T?>().toTypedArray())

or again:

inline fun <reified T> Map<String, T>.toBundle(): Bundle = bundleOf(*map { it.key to it.value }.toTypedArray())

Pleistocene

Answer 10

For future reference 👇

Convert Map to String

  fun Map<String, String>.convertExtraMapToString(): String? {
    return keys.stream()
        .map { key: String -> "$key=" + this[key] }.collect(Collectors.joining(", ", "{", "}"))
}

Convert String to Map then to Bundle

fun String.convertExtraMapToBundle(): Bundle {
    return Arrays.stream(split(",".toRegex()).toTypedArray())
        .map { entry: String -> entry.split("=".toRegex()).toTypedArray() }
        .collect(Collectors.toMap(
            { entry: Array<String> -> entry[0] }) { entry: Array<String> -> entry[1] }).let {
            bundleOf(*it.toList().toTypedArray())
        }
}

Zealous

Answer 11

 private ArrayList<Bundle> convertMapToBundleList(ArrayList<HashMap<String, String>> mapList)

Pulpiteer