Get Key, not value, of the ListPreference selection - Possible?

问题

Getting the value of the currently selected item in a ListPreference is straightforward:

String selected = sharedPrefs.getString(
    getString(R.string.list_preference_array),
    "default string"
);

But now I need to get the key of the currently selected item, instead. Is this possible?

To clarify, a typical ListPreference definition in the XML file has the following components:

<ListPreference 
    android:key="@string/list_preference_array"
    android:title="Title of ENTIRE list (not seen by user?)"
    android:summary="this is what the user sees in small fonts" 
    android:defaultValue="just in case"
    android:entries="@array/user_friendly_labels" 
    android:entryValues="@array/code_meaningful_strings"
    android:dialogTitle="User Prompt(big font)" 
    android:showDefault="true"
    android:showSilent="true" 
/>

What sharedPrefs.getString() returns is the current selection from android:entryValues. What I am interested in getting is the current selection from android:entries. I mistakenly called it "key" but really it is a "corresponding label", which must be different than actual content.

回答1:

A bit of a guess:

int index = mylistpreference.findIndexOfValue(selected)  // <- selected taken from your code above
String entry = mylistpreference.getEntries()[index];

回答2:

just use:

mylistpreference.getEntry()

to get mylistpreference use:

mylistpreference= (ListPreference) getPreferenceScreen().findPreference(key);

key is the android:key you defined in preference.xml inside < ListPreference> tag.

回答3:

The function is defined as

SharedPreferences.getString(String key, String defaultValue);

So in your example code, getString(R.string.select_string) would return the key.

When you add the SharedPreference, you need to specify the key, so the key would be the same one you used to set the value.

Edit:

using SharedPreferences.getString() with the ListPreference key as the key will return the value the user selected from the list. You don't need to create keys for each option in the ListPreference array, and these keys aren't created automatically. Use case statements that correlate to the ListPreference's entryValues.

回答4:

@shai Your method lags to last selected entry

listPreference.getEntry()

Rather @Erik 's method works nicely

int index = listPreference.findIndexOfValue((String) value)
String entry = listpreference.getEntries()[index];

回答5:

To properly update the summary of a ListPreference (using the label, instead of the key).

   @Override
    public boolean onPreferenceChange(Preference preference, Object value) {
        String stringValue = value.toString();

        if (preference instanceof ListPreference) {
            ListPreference listPreference = (ListPreference) preference;
            int prefIndex = listPreference.findIndexOfValue(stringValue);
            if (prefIndex >= 0) {
                CharSequence[] labels = listPreference.getEntries();
                preference.setSummary(labels[prefIndex]);
            }
        } else {
            preference.setSummary(stringValue);
        }
        return true;

    }

来源：https://stackoverflow.com/questions/7261141/get-key-not-value-of-the-listpreference-selection-possible

标签

android

android-preferences