Space 点赞 MySQL 搜索 AWS python
Space in like in MySQL search AWS python
正在尝试查询 AWS 数据库（新手）。查询如下：
dfHealthData = pd.read_sql("SELECT YearStart, LocationDesc, Topic, Question, DataValue "
"FROM HealthData WHERE YearStart = '2018' AND DataValueType = 'Crude Prevalence' AND Question LIKE '%18 years'" , connection_string)
想要获取以“18 年”结尾的字符串（或者理想情况下是“>=18 年”，但前者可以）。
错误如下。
从 AWS 检索时出错：索引 161
处不支持的格式字符 ' ' (0x20)
%在python中有特殊含义。尝试逃脱它：
dfHealthData = pd.read_sql("SELECT YearStart, LocationDesc, Topic, Question, DataValue FROM HealthData WHERE YearStart = '2018' AND DataValueType = 'Crude Prevalence' AND Question LIKE '%%18 years'" , connection_string)
正在尝试查询 AWS 数据库（新手）。查询如下：
dfHealthData = pd.read_sql("SELECT YearStart, LocationDesc, Topic, Question, DataValue "
"FROM HealthData WHERE YearStart = '2018' AND DataValueType = 'Crude Prevalence' AND Question LIKE '%18 years'" , connection_string)
想要获取以“18 年”结尾的字符串（或者理想情况下是“>=18 年”，但前者可以）。
错误如下。 从 AWS 检索时出错：索引 161
处不支持的格式字符 ' ' (0x20)%在python中有特殊含义。尝试逃脱它：
dfHealthData = pd.read_sql("SELECT YearStart, LocationDesc, Topic, Question, DataValue FROM HealthData WHERE YearStart = '2018' AND DataValueType = 'Crude Prevalence' AND Question LIKE '%%18 years'" , connection_string)