Cannot find appropriate system libraries for WITH_SSL=system. Make sure you have specified a supported SSL version. Valid options are : system (use the OS openssl library), yes (synonym for system), </path/to/custom/openssl/installation> CMake Error at cmake/ssl.cmake:66 (MESSAGE): Please install the appropriate openssl developer package. Call Stack (most recent call first): cmake/ssl.cmake:260 (FATAL_SSL_NOT_FOUND_ERROR) CMakeLists.txt:483 (MYSQL_CHECK_SSL)

这个错误是由于缺少 OpenSSL 库或版本不支持导致的。解决方法如下：

1. 确保 OpenSSL 库已经安装。在 Linux 上，您可以使用包管理器安装 OpenSSL 库，例如 apt-get、yum 等。在 Windows 上，您可以从 OpenSSL 官网下载适合您的版本并安装。

2. 如果您已经安装了 OpenSSL 库，但是仍然遇到此错误，请确保您使用的是支持的版本。MySQL 可以使用 OpenSSL 1.0.x 或 1.1.x 版本。检查您的 OpenSSL 版本并确保它与 MySQL 兼容。

3. 如果您使用的是自定义 OpenSSL 安装路径，请在 WITH_SSL 选项中指定路径，例如：-DWITH_SSL=/path/to/openssl。

4. 如果您在构建 MySQL 时使用了系统 OpenSSL 库，并且仍然遇到此错误，请确保您的系统 OpenSSL 库版本与 MySQL 兼容。您可以尝试使用自定义 OpenSSL 安装路径或升级您的系统 OpenSSL 库。

如果您仍然无法解决此问题，请提供更多上下文信息，例如您正在尝试构建什么，使用的操作系统和 CMake 版本等。

相关问题

优化代码 def cluster_format(self, start_time, end_time, save_on=True, data_clean=False, data_name=None): """ local format function is to format data from beihang. :param start_time: :param end_time: :return: """ # 户用簇级数据清洗 if data_clean: unused_index_col = [i for i in self.df.columns if 'Unnamed' in i] self.df.drop(columns=unused_index_col, inplace=True) self.df.drop_duplicates(inplace=True, ignore_index=True) self.df.reset_index(drop=True, inplace=True) dupli_header_lines = np.where(self.df['sendtime'] == 'sendtime')[0] self.df.drop(index=dupli_header_lines, inplace=True) self.df = self.df.apply(pd.to_numeric, errors='ignore') self.df['sendtime'] = pd.to_datetime(self.df['sendtime']) self.df.sort_values(by='sendtime', inplace=True, ignore_index=True) self.df.to_csv(data_name, index=False) # 调用基本格式化处理 self.df = super().format(start_time, end_time) module_number_register = np.unique(self.df['bat_module_num']) # if registered m_num is 0 and not changed, there is no module data if not np.any(module_number_register): logger.logger.warning("No module data!") sys.exit() if 'bat_module_voltage_00' in self.df.columns: volt_ref = 'bat_module_voltage_00' elif 'bat_module_voltage_01' in self.df.columns: volt_ref = 'bat_module_voltage_01' elif 'bat_module_voltage_02' in self.df.columns: volt_ref = 'bat_module_voltage_02' else: logger.logger.warning("No module data!") sys.exit() self.df.dropna(axis=0, subset=[volt_ref], inplace=True) self.df.reset_index(drop=True, inplace=True) self.headers = list(self.df.columns) # time duration of a cluster self.length = len(self.df) if self.length == 0: logger.logger.warning("After cluster data clean, no effective data!") raise ValueError("No effective data after cluster data clean.") self.cluster_stats(save_on) for m in range(self.mod_num): print(self.clusterid, self.mod_num) self.module_list.append(np.unique(self.df[f'bat_module_sn_{str(m).zfill(2)}'].dropna())[0])

Here are some possible optimizations for the given code:

1. Instead of using a list comprehension to find columns with 'Unnamed' in their names, you can use the `filter()` function along with a lambda function to achieve the same result in a more concise way:

   unused_index_col = list(filter(lambda x: 'Unnamed' in x, self.df.columns))

2. Instead of dropping duplicates and resetting the index separately, you can use the `drop_duplicates()` function with the `ignore_index` parameter set to `True` to achieve both in one step:

   self.df.drop_duplicates(inplace=True, ignore_index=True)

3. Instead of using `sys.exit()` to terminate the program when there is no module data, you can raise a `ValueError` with an appropriate error message:

   raise ValueError("No module data!")

4. Instead of using a series of `if` statements to find the voltage reference column, you can use the `loc` accessor with a boolean mask to select the first column that starts with 'bat_module_voltage':

   volt_ref_col = self.df.columns[self.df.columns.str.startswith('bat_module_voltage')][0]

5. Instead of using a loop to append a single item to a list, you can use the `append()` method directly:

   self.module_list.append(np.unique(self.df[f'bat_module_sn_{str(m).zfill(2)}'].dropna())[0])

By applying these optimizations, the code can become more concise and efficient.

Write a function is_bst, which takes a Tree t and returns True if, and only if, t is a valid binary search tree, which means that: Each node has at most two children (a leaf is automatically a valid binary search tree). The children are valid binary search trees. For every node, the entries in that node's left child are less than or equal to the label of the node. For every node, the entries in that node's right child are greater than the label of the node.其中class Tree: def init(self, label, branches=[]): for b in branches: assert isinstance(b, Tree) self.label = label self.branches = list(branches) def is_leaf(self): return not self.branches def map(self, fn): self.label = fn(self.label) for b in self.branches: b.map(fn) def contains(self, e): if self.label == e: return True for b in self.branches: if e in b: return True return False def repr(self): if self.branches: branch_str = ', ' + repr(self.branches) else: branch_str = '' return 'Tree({0}{1})'.format(self.label, branch_str) def str(self): def print_tree(t, indent=0): tree_str = ' ' * indent + str(t.label) + "\n" for b in t.branches: tree_str += print_tree(b, indent + 1) return tree_str return print_tree(self).rstrip()

Here's one possible implementation of the `is_bst` function using recursion to check the validity of each node in the binary search tree:

def is_bst(t):
    def is_bst_helper(t, min_val, max_val):
        if t.is_leaf():
            return True
        elif len(t.branches) == 1:
            return False
        else:
            left_bst = is_bst_helper(t.branches[0], min_val, t.label)
            right_bst = is_bst_helper(t.branches[1], t.label, max_val)
            return left_bst and right_bst and min_val <= t.label and t.label <= max_val
    
    return is_bst_helper(t, float('-inf'), float('inf'))

The `is_bst_helper` function takes in a tree `t`, a minimum value `min_val`, and a maximum value `max_val`, and returns `True` if `t` is a valid binary search tree within the range `[min_val, max_val]`. If `t` is a leaf, it is automatically a valid binary search tree, so `True` is returned. If `t` has only one child, it cannot be a valid binary search tree, so `False` is returned. Otherwise, the function recursively checks that the left and right subtrees are valid binary search trees within the appropriate ranges, and that the label of `t` is between `min_val` and `max_val`. Finally, the `is_bst` function calls `is_bst_helper` with the root of the input tree `t` and the initial range `[-inf, inf]`.