如何根据我的情况进行正确的SQL查询

debugcn 发表于 Dev

阿特姆·图马斯（Artem Tumas）

想法在这里，但它没有用，想纠正它很热，我是一个完全的初学者）

String temperature = editText5.getText().toString();
String percent = editText7.getText().toString();

Cursor c1 = db.query("mytable", null, "temperature, percent = ?", new String[]{temperature, percent}, null, null, null);

if (c1.moveToFirst()) {
    int ckoef = c1.getColumnIndex("ckoef");
    int density = c1.getColumnIndex("density");

    do {
        textView9.setText(c1.getString(ckoef));
        textView11.setText(c1.getString(density));
    } while (c1.moveToNext());

} else

c1.close();

桌子看起来像这样

ID = 1, temperature = 0, percent = 10, ckoef = 4.025, density = 1012.5
ID = 2, temperature = 10, percent = 10, ckoef = 4.034, density = 1012.5
ID = 3, temperature = 20, percent = 10, ckoef = 4.043, density = 1012.5
ID = 4, temperature = 30, percent = 10, ckoef = 4.057, density = 1012.5
ID = 5, temperature = 40, percent = 10, ckoef = 4.075, density = 1012.5
ID = 6, temperature = 50, percent = 10, ckoef = 4.085, density = 1012.5
ID = 7, temperature = 60, percent = 10, ckoef = 4.094, density = 1012.5
ID = 8, temperature = 70, percent = 10, ckoef = 4.103, density = 1012.5
ID = 9, temperature = 80, percent = 10, ckoef = 4.113, density = 1012.5
ID = 10, temperature = 90, percent = 10, ckoef = 4.122, density = 1012.5
ID = 11, temperature = 100, percent = 10, ckoef = 4.132, density = 1012.5

以此类推百分比10-50，并带有第10步

所以请帮忙。

伊甸园

游标c1 = db.query（“ mytable”，null，“ temperature =？和percent =？”，temperature，percent，null，null，null）;

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