Question

当我正在使我的应用程序在线版本即时获取来自json的数据，如下面的代码，现在我从数据库获取数据，请参阅下面的代码我将如何使用相同的textview和linearlayout并显示数据，就像我在应用程序的在线版本中显示的那样来自json解析??? http://imgur.com/RVrKJ98

final LinearLayout table3 = (LinearLayout) findViewById(R.id.table3);

JSONArray school5 = json2.getJSONArray("dish_ingredient");
String concatProduct = "";

for (int i = 0; i < school5.length(); i++) {
String name = school5.getJSONObject(i).getString("name");
String subIngredient = school5.getJSONObject(i).getString("sub_ingredients");
if (i == 0) {

if (subIngredient.equals("")) {
concatProduct = " " + "<b>" + name + "</b>"+ ", " + subIngredient;
} else {
concatProduct = " " + "<b>" + name + "</b>"+ ": " + subIngredient;
}

} else {
if (subIngredient.equals("")) {
if (i == school5.length() - 1) {

concatProduct += " " + "<b>" + name + "</b>" + ". " + subIngredient;
} else {

concatProduct += " " + "<b>" + name+ "</b>" + ", " + subIngredient;
}
} else {

concatProduct += " " + "<b>" + name + "</b>"+ ": " + subIngredient;
}
}

ingredient.setText(Html.fromHtml(concatProduct));





     now i m geting data from database with array how i will show on same 
    layout????check below code which used to get array from database


     static ArrayList<String> Ingredient_name = new ArrayList<String>();
static ArrayList<String> Ingredient_sub = new ArrayList<String>();



        Cursor mCursor2 =  db.selectQuery("SELECT i.name, i.sub_ingredients FROM  
    uss_dish_to_ingredient ");


    if (mCursor2.moveToFirst()) {
        do {

   Ingredient_name.add(mCursor2.getString(mCursor2.getColumnIndex("name")));

 Ingredient_sub.add(mCursor2.getString(mCursor2.getColumnIndex("sub_ingredients")));

} while (mCursor2.moveToNext());
}

    mCursor2.close();














              FATAL EXCEPTION: main
     java.lang.NullPointerException
at com.schoollunchapp.fifthscreen.getFormattedIngredientList(fifthscreen.java:681)
at com.schoollunchapp.fifthscreen.displayData(fifthscreen.java:561)
at com.schoollunchapp.fifthscreen.access$0(fifthscreen.java:517)
at com.schoollunchapp.fifthscreen$TheTask.onPostExecute(fifthscreen.java:274)
at com.schoollunchapp.fifthscreen$TheTask.onPostExecute(fifthscreen.java:1)
at android.os.AsyncTask.finish(AsyncTask.java:417)
at android.os.AsyncTask.access$300(AsyncTask.java:127)
at android.os.AsyncTask$InternalHandler.handleMessage(AsyncTask.java:429)
at android.os.Handler.dispatchMessage(Handler.java:99)
at android.os.Looper.loop(Looper.java:123)
at android.app.ActivityThread.main(ActivityThread.java:4627)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:521)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:868)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:626)
at dalvik.system.NativeStart.main(Native Method)








                Cursor mCursor2 =  db.selectQuery("SELECT xyz");


    if (mCursor2.moveToFirst()) {
        do {
            String ingredientName =    
       mCursor2.getString(mCursor2.getColumnIndex("name"));
            String ingredientSub = 
    mCursor2.getString(mCursor2.getColumnIndex("sub_ingredients"));

            // create an Ingredient object in your loop and fill its
            // values
            Ingredient ingredientItem = new Ingredient();
            ingredientItem.setName(ingredientName);
            ingredientItem.setSub(ingredientSub);

            // Add your ingredient to your ingredientList
            ingredientList.add(ingredientItem);

        } while (mCursor2.moveToNext());
    }

    // Now you got a list of ingredients.
    String formattedAsString = getFormattedIngredientList(ingredientList);
    ingredient.setText(Html.fromHtml(formattedAsString));

    mCursor2.close();

Answer 1

你最好定义一个成分类：

public class Ingredient {
    private String name, sub;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getSub() {
        return sub;
    }

    public void setSub(String sub) {
        this.sub = sub;
    }   
}

在你的json和db结果中使用这个包装类。

        ArrayList<Ingredient> ingredientList = new ArrayList<Ingredient>();

        if (mCursor2.moveToFirst()) {
            do {
                String ingredientName = mCursor2.getString(mCursor2.getColumnIndex("name"));
                String ingredientSub = mCursor2.getString(mCursor2.getColumnIndex("sub_ingredients"));

                // create an Ingredient object in your loop and fill its
                // values
                Ingredient ingredientItem = new Ingredient();
                ingredientItem.setName(ingredientName);
                ingredientItem.setSub(ingredientSub);

                // Add your ingredient to your ingredientList
                ingredientList.add(ingredientItem);

            } while (mCursor2.moveToNext());
        }

        // Now you got a list of ingredients.
        String formattedAsString = getFormattedIngredientList(ingredientList);
        ingredientTextView.setText(Html.fromHtml(formattedAsString));

在json和db结果中使用的

getFormattedIngredientList方法

private String getFormattedIngredientList(ArrayList<Ingredient> ingredientList) {

    String concatProduct = "";
    for (int i = 0; i < ingredientList.size(); i++) {
        Ingredient currentIngredient = ingredientList.get(i);
                    if(currentIngredient.getSub() == null) {
                       currentIngredient.setSub("");  
                    }
        if (i == 0) {
            if ((currentIngredient.getSub() == null || currentIngredient.getSub().equals("")) {
                concatProduct = " " + "<b>" + currentIngredient.getName() + "</b>" + ", " + currentIngredient.getSub();
            } else {
                concatProduct = " " + "<b>" + currentIngredient.getName() + "</b>" + ": " + currentIngredient.getSub();
            }

        } else {
            if ((currentIngredient.getSub() == null || currentIngredient.getSub().equals("")) {
                if (i == ingredientList.size() - 1) {
                    concatProduct += " " + "<b>" + currentIngredient.getName() + "</b>" + ". " + currentIngredient.getSub();
                } else {
                    concatProduct += " " + "<b>" + currentIngredient.getName() + "</b>" + ", " + currentIngredient.getSub();
                }
            } else {
                concatProduct += " " + "<b>" + currentIngredient.getName() + "</b>" + ": " + currentIngredient.getSub();
            }
        }
    }

    return concatProduct;
}

还要将json逻辑改为：

        ArrayList<Ingredient> ingredientList = new ArrayList<Ingredient>();
        for (int i = 0; i < school5.length(); i++) {
            String ingredientName = school5.getJSONObject(i).getString("name");
            String ingredientSub = school5.getJSONObject(i).getString("sub_ingredients");


            Ingredient ingredientItem = new Ingredient();
            ingredientItem.setName(ingredientName);
            ingredientItem.setSub(ingredientSub);

            // Add your ingredient to your ingredientList
            ingredientList.add(ingredientItem);
        }

        // Now you got a list of ingredients.
        String formattedAsString = getFormattedIngredientList(ingredientList);
        ingredientTextView.setText(Html.fromHtml(formattedAsString));

现在你有一个逻辑来格式化和设置json和db结果的结果。它易于实施和维护。

Answer 2

您可以使用database或assets。

如何在没有json的情况下从脱机显示数据

2 个答案: