java singleresult方法_JPA getSingleResult()返回值报异常的解决方案

private AppParameter appParameter = new AppParameter();

@Inject

private Logger logger;

@Inject

private EntityManager entityManager;

private String setTime;

private String showTime;

@PostConstruct//进页面就会加载这个方法

public void dataCall() {

appParameter = findSetTime();

if (appParameter == null) {

showTime = "0";

} else {

showTime = appParameter.getParamValue();

}

}

/**

* 设置settime

*/

@Transactional

public void timerTask() {

if (setTime != null && setTime != "") {

appParameter = new AppParameter();//不new AppParameter,赋值会报空指针因为查出来的结果为空赋值给appParameter了

appParameter.setParamCode(AppParamCode.SET_TIME);

appParameter.setParamValue(setTime);

entityManager.merge(appParameter);

dataCall();

}

}

/**

* .getSingleResult()抛出异常,用findSetTime去初始化 showTime

*

* @return

*/

public AppParameter findSetTime() {

List resultList = entityManager.createQuery("select a from AppParameter a where paramcode = 'SET_TIME'", AppParameter.class)

.getResultList();

if (resultList.size() > 0) {

return resultList.get(0);

} else {

return null;

}

}


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