android 使用InputFilter 对金额和长度进行限制

public class MaxTextLengthFilter implements InputFilter {

    private int mMaxLength;
    private boolean isNum;
    private String note;
    private String message = null;

    /**
     *
     * @param max 支持的最大长度
     * @param isNum 是不是数字
     * @param note 超过后的提示
     */
    public MaxTextLengthFilter(int max, boolean isNum, String note) {
        mMaxLength = max;
        this.isNum = isNum;
        this.note = note;
    }

    public MaxTextLengthFilter setMessage(String message) {
        this.message = message;
        return this;
    }

    public CharSequence filter(CharSequence source, int start, int end,
                               Spanned dest, int dstart, int dend) {
        int keep = mMaxLength - (dest.length() - (dend - dstart));
        if (keep < (end - start)) {
            if (TextUtils.isEmpty(message))
                ToastUtils.showShortToast(String.format("%s最多%s位", note, mMaxLength));
            else {
                ToastUtils.showShortToast(message);
            }
        }
        if (keep <= 0) {
            return "";
        } else if (keep >= end - start) {
            if (isNum) {
                int posDot = dest.toString().indexOf(".");
                if (start < end && posDot > 0 && (dest.length() - posDot) > 2) {//小数点后要保留两位小数
                    ToastUtils.showShortToast("小数点后只能输两位");
                    return "";
                }
                if (start < end && posDot > 1 && dest.toString().equals("0.") && source.equals("0")) {//确保不会出现不是小数的第一位为0
                    ToastUtils.showShortToast("请输入正确格式的金额");
                    return "";
                } else if (start < end && posDot < 0 && dest.toString().equals("0") && !source.equals(".")) {
                    ToastUtils.showShortToast("请输入正确格式的金额");
                    return "";
                }
            } else
                return null;
        } else
            return source.subSequence(start, start + keep);
        return null;
    }
}

你可能感兴趣的:(android)