fix: 修复添加用户失败的BUG(因为telephone字段拼写错误)
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9774070d93
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@ -185,6 +185,8 @@ public class UserServiceImpl implements UserService {
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userAddParam.setTelephone(userAddParam.getTelephone()==null?"":userAddParam.getTelephone());
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userAddParam.setTelephone(userAddParam.getTelephone()==null?"":userAddParam.getTelephone());
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userAddParam.setAvatar(userAddParam.getAvatar()==null?"":userAddParam.getAvatar());
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userAddParam.setAvatar(userAddParam.getAvatar()==null?"":userAddParam.getAvatar());
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userAddParam.setCode(userAddParam.getCode()==null?"":userAddParam.getCode());
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userAddParam.setCode(userAddParam.getCode()==null?"":userAddParam.getCode());
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log.info("开始创建新用户");
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log.info(userAddParam.toString());
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int result = userMapper.addUser(userAddParam);
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int result = userMapper.addUser(userAddParam);
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if (result == 1) {
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if (result == 1) {
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return ResponseResult.success(userAddParam);
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return ResponseResult.success(userAddParam);
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@ -246,6 +248,7 @@ public class UserServiceImpl implements UserService {
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// 设置不需要更新的字段为 null
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// 设置不需要更新的字段为 null
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userInfo.setPassword(null);
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userInfo.setPassword(null);
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userInfo.setCreateTime(null);
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userInfo.setCreateTime(null);
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userInfo.setStatus(String.valueOf(userAddParam.getStatus()));
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// 使用 LambdaUpdateWrapper 构建更新条件
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// 使用 LambdaUpdateWrapper 构建更新条件
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LambdaUpdateWrapper<UserInfo> updateWrapper = new LambdaUpdateWrapper<>();
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LambdaUpdateWrapper<UserInfo> updateWrapper = new LambdaUpdateWrapper<>();
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@ -75,7 +75,7 @@
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INSERT INTO user_info (
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INSERT INTO user_info (
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id, name, username, password, avatar, status, telephone, email, create_time, merchant_code, deleted, role_id, creator_id
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id, name, username, password, avatar, status, telephone, email, create_time, merchant_code, deleted, role_id, creator_id
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) VALUES (
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) VALUES (
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#{id}, #{name}, #{username}, #{password}, #{avatar}, #{status}, #{telphone}, #{email}, #{createTime},
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#{id}, #{name}, #{username}, #{password}, #{avatar}, #{status}, #{telephone}, #{email}, #{createTime},
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#{merchantCode},
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#{merchantCode},
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0, #{roleId}, #{creatorId}
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0, #{roleId}, #{creatorId}
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)
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)
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