在一个函数调用中得到多个论证，而在另一个函数调用中没有得到多个论证。

Question

我正在开发机器人框架，我的基本方法之一是用python编写的，用于构建一个包含n列和多个where条件的SQL查询。这个函数看起来像，

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('"', '')

在我的机器人关键字中，我将此方法称为：

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

这个函数在另外两个关键字中调用。首先，它工作得很好。另一方面，它一直抛出错误，如

关键字“带有筛选条件的queryBuilderUtility.Get查询”获得了参数“table_name”的多个值.

工作良好的关键字看起来如下：

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}

有人能帮我纠正一下吗?我在这里犯的错误是什么？

Answer 1

这个问题是由于字典中的键值对造成的。字典中的关键之一

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

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，并且工作正常。由于函数采用可以指定为命名参数的位置参数，所以python与我传递的table_name参数和字典中键table_name的参数混淆了。