在一个函数调用中获得多个参数，但在另一个函数调用中没有

Question

我正在研究 Robot Framework，我的基本方法之一是在 python 中编写的，用于构建具有 n 列和多个 where 条件的 SQL 查询。该函数看起来像，

from pypika import Query, Table, Field
def get_query_with_filter_conditions(table_name, *column, **where):
    table_name_with_no_lock = table_name + ' with (nolock)'
    table = Table(table_name_with_no_lock)
    where_condition = get_where_condition(**where)
    sql_query = Query.from_(table).select(
        *column
    ).where(
        Field(where_condition)
    )
    return str(sql_query).replace('"', '')

我在我的 Robot 关键字中调用此方法为：

Get Query With Filter Conditions    ${tableName}    ${column}    &{tableFilter}

此函数在另外两个关键字中被调用。一方面，它工作正常。另一方面，它不断抛出错误

Keyword 'queryBuilderUtility.Get Query With Filter Conditions' got multiple values for argument 'table_name'.

效果不错的关键字如下：

Verify the ${element} in ${grid} is fetched from ${column} column in ${tableName} table from DB
    [Documentation]    Verifies Monetary values in the View Sale Grid
    ${feature}=    Get Variable Value    ${FEATURE_NAME}
    ${filterValue}=    Get Variable value    ${FILTER_VALUE}
    ${queryFilter}=    Get the Test Data    valid    ${filterValue}    ${feature}
    &{tableFilter}=    Create Dictionary
    Set To Dictionary    ${tableFilter}    ${filterValue}=${queryFilter}
    Set To Dictionary    ${tableFilter}    form_of_payment_type=${element}
    ${tableName}=    Catenate    SEPARATOR=.    SmartPRASales    ${tableName}
    ${query}=    Get query with Filter Conditions    ${tableName}    ${column}    &{tableFilter}
    Log    ${query}
    @{queryResult}=    CommonPage.Get a Column values from DB    ${query}

总是抛出错误的函数如下所示：

Verify ${element} drop down contains all values from ${column} column in ${tableName} table
    [Documentation]    To verify the drop down has all values from DB
    ${feature}=    Get Variable Value    ${FEATURE_NAME}
    ${filterElement}=    Run Keyword If    '${element}'=='batch_type'    Set Variable    transaction_type
    ...    ELSE IF    '${element}'=='channel'    Set Variable    agency_type
    ...    ELSE    Set Variable    ${element}
    &{tableFilter}=    Create Dictionary
    Set To Dictionary    ${tableFilter}    table_name=GENERAL
    Set To Dictionary    ${tableFilter}    column_name=${filterElement}
    Set To Dictionary    ${tableFilter}    client_id=QR
    Log    ${tableFilter}
    Log    ${tableName}
    Log    ${column}
    ${tableName}=    Catenate    SEPARATOR=.    SmartPRAMaster    ${tableName}
    ${query}=    Get Query With Filter Conditions    ${tableName}    ${column}    &{tableFilter}
    Log    ${query}
    @{expectedvalues}=    CommonPage.Get a Column values from DB    ${query}

有人可以帮我改正我在这里做错了什么吗？

Answer 1

问题出在字典中的键值对。字典中的关键字之一

&{tableFilter}=    Create Dictionary
    Set To Dictionary    ${tableFilter}    table_name=GENERAL

与

中的参数之一相同

def get_query_with_filter_conditions(table_name, *column, **where):

将 get_query_with_filter_conditions 函数中的参数从 table_name 更改为 p_table_name 并且有效。由于该函数采用可以指定为命名参数的位置参数，因此 python 与我通过字典中的键 table_name 传递的 table_name 参数混淆了。

在一个函数调用中获得多个参数，但在另一个函数调用中没有

Got multiple Arguements in one function call but not in another function call

python-3.x

parameter-passing

robotframework

keyword-argument

positional-parameter