Android ListPreference：使用字符串数组作为条目和整数数组作为条目值不起作用

ListPreference: use string-array as Entry and integer-array as Entry Values doesn't work

提问by Santacrab

I'm using in a settings.xml file a ListPreference. I want to show the user a list of 3 options to select. When the user chooses one of the options in the Settings, I get this error:

我在 settings.xml 文件中使用了 ListPreference。我想向用户显示要选择的 3 个选项的列表。当用户选择设置中的选项之一时，我收到此错误：

java.lang.NullPointerException
    at android.preference.ListPreference.onDialogClosed(ListPreference.java:264)
    at android.preference.DialogPreference.onDismiss(DialogPreference.java:381)
    at android.app.Dialog$ListenersHandler.handleMessage(Dialog.java:1228)
    at android.os.Handler.dispatchMessage(Handler.java:99)
    at android.os.Looper.loop(Looper.java:137)
    at android.app.ActivityThread.main(ActivityThread.java:4424)
    at java.lang.reflect.Method.invokeNative(Native Method)
    at java.lang.reflect.Method.invoke(Method.java:511)
    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:784)
    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:551)
    at dalvik.system.NativeStart.main(Native Method)

This is the code of the ListPreference:

这是 ListPreference 的代码：

<ListPreference
    android:entries="@array/date_alignement"
    android:entryValues="@array/date_alignement_values"
    android:key="settings_date_alignement"
    android:summary="@string/settings_date_alignement_summary"
    android:title="@string/settings_date_alignement_title" />

And here are the arrays I use to populate the entries:

这是我用来填充条目的数组：

<string-array name="date_alignement">
    <item>"Top"</item>
    <item>"Center"</item>
    <item>"Bottom"</item>
</string-array>
<integer-array name="date_alignement_values">
    <item >0</item>
    <item >1</item>
    <item >2</item>
</integer-array>

In my onSharedPreferenceChanged I use those values in this way:

在我的 onSharedPreferenceChanged 中，我以这种方式使用这些值：

@Override
public void onSharedPreferenceChanged(
            SharedPreferences sharedPreferences, String key) {          

        //Text
        mAlignment =  mPrefs.getInt(PREF_ALIGNMENT, 1);
        switch (mAlignment) {
        case 0:
            offsetY = mHeight/3.0f;
            break;

        case 2:
            offsetY = mHeight*2.0f/3.0f;
            break;

        default:
            offsetY = mHeight/2.0f;
            break;
        }
}

If I use another string-array for the entryValuesit works. For example if I use the same string array as values and entries:

如果我为entryValues使用另一个字符串数组，它会起作用。例如，如果我使用相同的字符串数组作为值和条目：

    android:entries="@array/date_alignement"
    android:entryValues="@array/date_alignement"

then I have to change my code a little but it works:

然后我必须稍微更改我的代码，但它有效：

        if(mAlignment.equalsIgnoreCase("center")) {
            offsetY = mHeight/2.0f;
        } else if(mAlignment.equalsIgnoreCase("top")) {
            offsetY = mHeight/3.0f;
        } else if(mAlignment.equalsIgnoreCase("bottom")) {
            offsetY = mHeight*2.0f/3.0f;
        }

Why I can't use a string-array and an integer-array for a ListPreference entries and values?

为什么我不能对 ListPreference 条目和值使用字符串数组和整数数组？

回答by Egor

The answer is simple: because the Android is designed this way. It just uses Stringarrays for both entries and entry values and that's all. And I can't see any problem in this fact, since you can easily convert a Stringto an intusing the Integer.parseInt()method. Hope this helps.

答案很简单：因为 Android 就是这样设计的。它只是将String数组用于条目和条目值，仅此而已。而且我看不出有任何问题，因为您可以使用该方法轻松地将 a 转换String为 an 。希望这可以帮助。intInteger.parseInt()

回答by kasvith

The answer is listed in List Preference Documentation.

答案列在List Preference Documentation 中。

int findIndexOfValue (String value)

will return the index for given value and the argument is taken as a String, so the entryValues array should be a string array to get this work.

将返回给定值的索引，并且参数被视为字符串，因此 entryValues 数组应该是一个字符串数组才能完成这项工作。

回答by Sam

You can convert your entry values to strings to keep ListPreferencehappy and then convert them to ints when talking to the persistent data store.

您可以将条目值转换为字符串以保持ListPreference满意，然后int在与持久数据存储交谈时将它们转换为s。

1. When setting the entry values, use strings instead of ints: "1", "2", "3"etc
2. Make a custom IntListPreferencethat persists the values as ints
3. In your preferences.xmlfile, change the <ListPreference>into a <your.app.package.IntListPreference>

1. 设置条目值时，使用字符串而不是整数："1", "2", "3"等
2. 自定义IntListPreference将值保留为整数
3. 在你的preferences.xml文件，更改<ListPreference>成<your.app.package.IntListPreference>

IntListPreference.java

IntListPreference.java

Here's a sample implementation. Tested and working on AndroidX Preference 1.0.0.

这是一个示例实现。在 AndroidX Preference 1.0.0 上测试和工作。

public class IntListPreference extends ListPreference {

    public IntListPreference(Context context, AttributeSet attrs, int defStyleAttr, int defStyleRes) {
        super(context, attrs, defStyleAttr, defStyleRes);
    }

    public IntListPreference(Context context, AttributeSet attrs, int defStyleAttr) {
        super(context, attrs, defStyleAttr);
    }

    public IntListPreference(Context context, AttributeSet attrs) {
        super(context, attrs);
    }

    public IntListPreference(Context context) {
        super(context);
    }

    @Override
    protected boolean persistString(String value) {
        int intValue = Integer.parseInt(value);
        return persistInt(intValue);
    }

    @Override
    protected String getPersistedString(String defaultReturnValue) {
        int intValue;

        if (defaultReturnValue != null) {
            int intDefaultReturnValue = Integer.parseInt(defaultReturnValue);
            intValue = getPersistedInt(intDefaultReturnValue);
        } else {
            // We haven't been given a default return value, but we need to specify one when retrieving the value

            if (getPersistedInt(0) == getPersistedInt(1)) {
                // The default value is being ignored, so we're good to go
                intValue = getPersistedInt(0);
            } else {
                throw new IllegalArgumentException("Cannot get an int without a default return value");
            }
        }

        return Integer.toString(intValue);
    }

}

回答by Александр

The following worked for me:

以下对我有用：

String objectName = prefs.getString("listPrefMelodyYd1", "");
switch (objectName.toUpperCase()) {
    case "1":
        playSound(catSound);
        break;
    case "2":
        playSound(dogSound);
        break;
    case "3":
        playSound(cowSound);
        break;
}