1: 修改 relation_type的字段类型为integer

dev_3.1.0
xueqingkun 10 months ago
parent 62dbc764fc
commit 49c365a866

@ -82,7 +82,7 @@ public class DiseaseAncillary implements Serializable {
@Schema(description = "关系类型 0:关联疾病 1:关联病历")
private String relationType;
private Integer relationType;
/**
* ID

@ -87,7 +87,7 @@ public class DiseasePhysical implements Serializable {
private String trait;
@Schema(description = "关系类型 0:关联疾病 1:关联病历")
private String relationType;
private Integer relationType;
/**
* ID

@ -19,7 +19,7 @@
<result property="result" column="result" jdbcType="VARCHAR"/>
<result property="normalResult" column="normal_result" jdbcType="VARCHAR"/>
<result property="description" column="description" jdbcType="VARCHAR"/>
<result property="relationType" column="relation_type" jdbcType="VARCHAR"/>
<result property="relationType" column="relation_type" jdbcType="INTEGER"/>
<result property="createUserId" column="create_user_id" jdbcType="VARCHAR"/>
<result property="createTime" column="create_time" jdbcType="TIMESTAMP"/>
<result property="updateUserId" column="update_user_id" jdbcType="VARCHAR"/>

@ -20,7 +20,7 @@
<result property="result" column="result" jdbcType="VARCHAR"/>
<result property="normalResult" column="normal_result" jdbcType="VARCHAR"/>
<result property="trait" column="trait" jdbcType="VARCHAR"/>
<result property="relationType" column="relation_type" jdbcType="VARCHAR"/>
<result property="relationType" column="relation_type" jdbcType="INTEGER"/>
<result property="createUserId" column="create_user_id" jdbcType="VARCHAR"/>
<result property="createTime" column="create_time" jdbcType="TIMESTAMP"/>
<result property="updateUserId" column="update_user_id" jdbcType="VARCHAR"/>

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