postgresql 错误：运算符不存在：字符变化 = bytea
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ERROR: operator does not exist: character varying = bytea
提问by rajshekhar
I have a table in postgres with the following structure
我在 postgres 中有一个表，其结构如下
CREATE TABLE rpaul."HK_LOGIN_DETAILS"
(
"HK_LOGIN_DETAILS_ID" bigint NOT NULL,
"HK_LOGIN_DETAILS_USERNAME" character varying(10) NOT NULL,
"HK_LOGIN_DETAILS_PASSWORD" character varying(50) NOT NULL,
CONSTRAINT "HK_LOGIN_DETAILS_PK" PRIMARY KEY ("HK_LOGIN_DETAILS_ID" ),
CONSTRAINT "HK_LOGIN_DETAILS_UK" UNIQUE ("HK_LOGIN_DETAILS_USERNAME" )
)
And hibernate mapping for this table is as mentioned below
该表的休眠映射如下所述
<hibernate-mapping package="net.rpaul.projects.homekeeping.domain.login">
<class name="LoginDetails" table="`HK_LOGIN_DETAILS`">
<id name="id" column="`HK_LOGIN_DETAILS_ID`" type="long">
<generator class="assigned" />
</id>
<property name="userName" type="string" column="`HK_LOGIN_DETAILS_USERNAME`" not-null="true" />
<property name="password" type="string" column="`HK_LOGIN_DETAILS_PASSWORD`" not-null="true" />
</class>
</hibernate-mapping>
In the LoginDetails.java, I have declared id field as long, userName and password fields as String. Still when I try to execute the following
在 LoginDetails.java 中，我已将 id 字段声明为字符串，将用户名和密码字段声明为字符串。仍然当我尝试执行以下操作时
List list = getHibernateTemplate().find("from LoginDetails ld where ld.userName = ?", userName);
I get
我得到
ERROR: operator does not exist: character varying = bytea
错误：运算符不存在：字符变化 = bytea
I am not getting what has went wrong. Any help would be appreciated.
我不明白出了什么问题。任何帮助，将不胜感激。
回答by chavocarlos
I think you should check that your variable "userName" is not null. I experienced this message in cases like that.
我认为您应该检查您的变量“userName”是否为空。在这样的情况下，我经历了这条消息。
回答by Sergey Ponomarev
It seems that Hibernate is for some reason sending the type-parameter as bytea (or rather, probably java.sql.Types.BLOB), instead of leaving it for the server to infer or setting it to text (java.sql.Types.STRING). Here is similar issue with solution JPA lower() function on parameter
似乎 Hibernate 出于某种原因将类型参数作为 bytea（或者更确切地说，可能是 java.sql.Types.BLOB）发送，而不是将其留给服务器来推断或将其设置为文本 (java.sql.Types.细绳）。这是与参数上的解决方案JPA lower() 函数类似的问题
