java - 如何将灵活的搜索查询正确转换为 Java Final String 查询？

Question

我正在尝试在 Hybris Commerce 上创建一个 Web 服务以从模型返回数据。我的查询适用于灵活的搜索控制台，但我的 Java 方法存在语法问题。

我的灵活搜索查询:

select * from {address as a join customerenvasado as c on {a:owner} = {c:pk}} where {c:rut} like '1754%' 

我的错误:

{"errors": [{
    "message": "type code 'a: owner' invalid",
    "type": "FlexibleSearchError"
}]}

我的功能:

public AddressModel getCustomerEnvasadoForRut(String rut) {
    validateParameterNotNull(rut, "Rut must not be null!");
    final String querys = "SELECT * FROM {a: " + AddressModel._TYPECODE + " as a join " + CustomerEnvasadoModel._TYPECODE + " as c on {a:owner} = {c:pk}} where {c:rut} like ?ParamRut";

    final FlexibleSearchQuery query = new FlexibleSearchQuery(querys);
    query.addQueryParameter("ParamRut", rut);
    List<AddressModel> result = getFlexibleSearchService().<AddressModel>search(querys).getResult();
    if (result != null && !result.isEmpty()) {
        return result.get(0);
    }
    return null;
}

Answer 1

只需按如下方式更改您的查询即可，它应该可以工作:

SELECT {pk} FROM {Address AS a JOIN customerenvasado AS c ON {a:owner} = {c:pk}} WHERE {c:rut} LIKE '1754%'

您还可以使用以下查询来代替上面给出的查询:

SELECT {pk} FROM {Address AS a}, {customerenvasado AS c} WHERE {a:owner} = {c:pk} AND {c:rut} LIKE '1754%'