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.

"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. — CommonsWare, May 26 '16 at 19:16
The RemoteMessage object is parcelable, you could just send that directly in the intent — morepork, Oct 03 '17 at 21:17

Marcin Koziński · Accepted Answer · 2016-05-26T21:33:18.283

59

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());
    }

edited May 26 '16 at 21:33

answered May 26 '16 at 20:26

Marcin Koziński

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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. – Amir Ziarati Aug 23 '17 at 11:20
2

@AmirZiarati I don't think that is true anymore, `RemoteMessage#getData()` now returns a `Map` excluding a bunch of the internal values. – Mikael Ohlson Sep 11 '18 at 09:26

bgplaya · Answer 2 · 2020-07-27T18:38:26.223

27

Neat Kotlin solution using spread operator will look like that:

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

edited Jul 27 '20 at 18:38

answered Jul 07 '20 at 17:35

bgplaya

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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: https://detekt.github.io/detekt/performance.html#spreadoperator – patrick.elmquist Dec 22 '21 at 07:57
worth mentioning that this won't create nested bundles for nested maps, in case you're using something like FIrebase Analytics that requires that – kassim Nov 17 '22 at 11:04

score 16 · Answer 3 · answered Jan 07 '17 at 13:58

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()
    }
  }
}

never heard of kotlin before today; interesting find – CCJ Apr 13 '17 at 22:34 — CCJ, Apr 13 '17 at 22:34

score 12 · Answer 4 · answered Oct 03 '19 at 12:54

12

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

answered Oct 03 '19 at 12:54

bakua

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score 1 · Answer 5 · edited Dec 13 '16 at 10:46

1

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.

edited Dec 13 '16 at 10:46

Pierre-Luc Paour

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answered Sep 07 '16 at 05:28

SQLiteNoob

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`Parcel` is a superclass of `Bundle`, so while it can be similarly useful, they can't be used interchangeably. – Pierre-Luc Paour Dec 13 '16 at 09:48
Yep, I did mention that it's the parent class of it. – SQLiteNoob Mar 06 '17 at 04:41
1

Your answer maybe is better, but the lack of explanation or example is making it worst. – Stoycho Andreev Feb 08 '19 at 00:00

score 1 · Answer 6 · answered Feb 21 '19 at 14:25

1

here is how I did it in Kotlin

val bundle = Bundle()

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

answered Feb 21 '19 at 14:25

Dan Alboteanu

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score 0 · Answer 7 · answered May 12 '22 at 09:20

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())

score 0 · Answer 8 · answered Feb 22 '23 at 16:43

0

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)
}

answered Feb 22 '23 at 16:43

LaurentY

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score -1 · Answer 9 · answered Mar 29 '21 at 13:42

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())
        }
}

score -10 · Answer 10 · edited Sep 01 '16 at 14:39

-10

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

edited Sep 01 '16 at 14:39

Nikolay Mihaylov

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answered Sep 01 '16 at 14:19

Chetan Sankhla

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Convert Map to Bundle in android

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