Question

我需要从EditText传递一个参数，当我点击按钮时，它会启动另一个活动并获取该参数并将其传递给查询。它如下：

final Button ara = (Button) findViewById(R.id.maddebutton);
    ara.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub
            String maddeno = madde.getText().toString(); //madde is my EditText
            Intent intent = new Intent(Anayasa.this, MaddeBul.class);
            intent.putExtra("maddeler", maddeno);
            startActivity(intent);
            }
    });

我的第二堂课如下：

Intent intent = getIntent();
    int maddebul = intent.getIntExtra("maddeler", 0); //I don't want to set a default value but it pushes me

try {
        Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = maddebul", null, null, null, null));
        ShowRecord(cursor);
    } finally {
        database.close();
    }

我的FetchRecord(Cursor c)和ShowRecord(Cursor cursor)功能正常，因为我在其他课程中使用它们。有＆＃34;不＆＃34;在我的＆＃34; anayasa＆＃34;保存整数值的数据库。

在LogCat上，它说＆＃34;没有列像maddebul＆＃34;。确实，没有。它假设是：

SELECT * FROM anayasa WHERE no = maddebul; //as sql command

任何帮助？

Answer 1

您在此处添加额外的字符串：

intent.putExtra("maddeler", maddeno);

但是当您尝试检索额外内容时，您将其检索为int：

int maddebul = intent.getIntExtra("maddeler", 0);

尝试使用此insteald

String maddebul = intent.getStringExtra("maddeler", "");

此处参考的是getStringExtra() method的文档。

Answer 2

On LogCat, it says "no column such maddebul". It is true, there isn't.

由于其中的条款“no = maddebul”，maddebul不是变量，它是字符串部分所以更改whereClause以获取它的值

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                    "no = maddebul", null, null, null, null));

应该是

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = "+maddebul, null, null, null, null));

Answer 3

你正在双引号中写下maddebul。

替换

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = maddebul", null, null, null, null));

通过

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = "+maddebul, null, null, null, null));

更好地使用

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = ?",new String[]{String.valueOf(maddebul)}, null, null, null));

Answer 4

你把Extra作为字符串并将其作为整数 - 也许这就是问题？

Answer 5

存在多个问题：

你把maddebul作为字符串放在这里，所以需要像这样：

String maddebul = intent.getStringExtra("maddeler", "");

此外，您正在构建错误的SQL查询，maddebul作为字符串传递给数据库，而您实际上想要传递该变量的值。所以它可能是这样的：

Cursor cursor = FetchRecord(db.query("anayasa", SELECT,
                "no = ?",new String[]{maddebul}, null, null, null));

通过意图将参数传递给查询

5 个答案: