!513 优化代码生成498 解决解决 Set access token expire time to 0 报错问题和邮件发送用户编号为空问题
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@ -18,6 +18,7 @@ import javax.validation.Valid;
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import java.util.List;
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import static cn.iocoder.yudao.framework.common.pojo.CommonResult.success;
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import static cn.iocoder.yudao.framework.security.core.util.SecurityFrameworkUtils.getLoginUserId;
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@Tag(name = "管理后台 - 邮件模版")
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@RestController
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@ -81,7 +82,7 @@ public class MailTemplateController {
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@Operation(summary = "发送短信")
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@PreAuthorize("@ss.hasPermission('system:mail-template:send-mail')")
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public CommonResult<Long> sendMail(@Valid @RequestBody MailTemplateSendReqVO sendReqVO) {
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return success(mailSendService.sendSingleMailToAdmin(sendReqVO.getMail(), null,
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return success(mailSendService.sendSingleMailToAdmin(sendReqVO.getMail(), getLoginUserId(),
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sendReqVO.getTemplateCode(), sendReqVO.getTemplateParams()));
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}
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@ -37,7 +37,9 @@ public class OAuth2AccessTokenRedisDAO {
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// 清理多余字段,避免缓存
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accessTokenDO.setUpdater(null).setUpdateTime(null).setCreateTime(null).setCreator(null).setDeleted(null);
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long time = LocalDateTimeUtil.between(LocalDateTime.now(), accessTokenDO.getExpiresTime(), ChronoUnit.SECONDS);
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stringRedisTemplate.opsForValue().set(redisKey, JsonUtils.toJsonString(accessTokenDO), time, TimeUnit.SECONDS);
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if (time > 0) {
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stringRedisTemplate.opsForValue().set(redisKey, JsonUtils.toJsonString(accessTokenDO), time, TimeUnit.SECONDS);
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}
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}
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public void delete(String accessToken) {
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