Question

我在Android中遇到Switch的问题。这是我处理Switch的OnCheckedChangeListener的代码：

public static SparseVector combineSparseVectors(SparseVector... svs) {
    int size = 0;
    int nonzeros = 0;
    for (SparseVector sv : svs) {
        size += sv.size();
        nonzeros += sv.indices().length;
    }

    if (nonzeros != 0) {
        int[] indices = new int[nonzeros];
        double[] values = new double[nonzeros];

        int pointer_D = 0;
        int totalPt_D = 0;
        int pointer_V = 0;
        for (SparseVector sv : svs) {
            int[] indicesSV = sv.indices();
            for (int i : indicesSV) {
                indices[pointer_D++] = i + totalPt_D;
            }
            totalPt_D += sv.size();

            double[] valuesSV = sv.values();
            for (double d : valuesSV) {
                values[pointer_V++] = d;
            }

        }
        return new SparseVector(size, indices, values);
    } else {
        System.out.println("all zeroes");
        return null;
    }

}

条件 if（compoundButton.isPressed（））以避免自动调用onCheckedChanged。但当我滑动切换时， myCodeHere（）不会触发。

请给我一些提示来解决这个问题。或者以任何方式禁用Switch的滑块？非常感谢

Answer 1

<?php
  while ( $row = mysqli_fetch_array($result) ) {
      echo '<td><input type="checkbox" value ="" name="todelete[]" /> '. $row['id'] . '</td><td>' . $row['title'] . '</td><td>' . $row['author_last'] . '</td><td>' . $row['author_first'] . '</td><td>' . $row['format'] . '</td><td>' . $row['price'] . '</td><td>' . $row['isbncode'] . '</td>';    
      echo '</tr>';
 }
?>

复合按钮的选中状态发生变化时调用。您可以使用布尔onCheckedChanged(CompoundButton buttonView, boolean isChecked)在开关打开/关闭时执行操作

isChecked

Android中的滑块开关

1 个答案: