blocks|key|1368332|text|抛出异常是getSingleResult()指示找不到它的方式。就我个人而言，我不能忍受这种API。它强制进行虚假的异常处理，但没有真正的好处。您只需将代码包装在try-catch块中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1368333|或者，您也可以查询一个列表，看看它是否为空。这不会抛出异常。实际上，由于从技术上讲，您不是在执行主键查找，因此可能会有多个结果(即使只有一个、两个，或者外键或约束的组合在实践中不可能实现这一点)，所以这可能是更合适的解决方案。|1368334|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javaee/7/api/javax/persistence/Query.html#getSingleResult%2528%2529^0|5|H|5|H|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Throwing an exception is how <a href="http://docs.oracle.com/javaee/7/api/javax/persistence/Query.html#getSingleResult%28%29" rel="noreferrer"><code>getSingleResult()</code></a> indicates it can't be found. Personally I can't stand this kind of API. It forces spurious exception handling for no real benefit. You just have to wrap the code in a try-catch block.

Alternatively you can query for a list and see if its empty. That doesn't throw an exception. Actually since you're not doing a primary key lookup technically there could be multiple results (even if one, both or the combination of your foreign keys or constraints makes this impossible in practice) so this is probably the more appropriate solution.

blocks|key|1368369|text|我将逻辑封装在下面的助手方法中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368370|public+class+JpaResultHelper+{
++++public+static+Object+getSingleResultOrNull(Query+query){
++++++++List+results+=+query.getResultList();
++++++++if+(results.isEmpty())+return+null;
++++++++else+if+(results.size()+==+1)+return+results.get(0);
++++++++throw+new+NonUniqueResultException();
++++}
}|code-block|syntax|javascript|1368371|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I encapsulated the logic in the following helper method.

<pre><code>public class JpaResultHelper {
 public static Object getSingleResultOrNull(Query query){
 List results = query.getResultList();
 if (results.isEmpty()) return null;
 else if (results.size() == 1) return results.get(0);
 throw new NonUniqueResultException();
 }
}
</code></pre>

blocks|key|1368773|text|在Java+8中尝试一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368774|Optional+first+=+query.getResultList().stream().findFirst();|code-block|syntax|javascript|1368775|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this in Java 8:

<pre><code>Optional first = query.getResultList().stream().findFirst();
</code></pre>

blocks|key|1153327|text|这是一个很好的选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153328|public+static+<T>+T+getSingleResult(TypedQuery<T>+query)+{
++++query.setMaxResults(1);
++++List<T>+list+=+query.getResultList();
++++if+(list+==+null+%7C%7C+list.isEmpty())+{
++++++++return+null;
++++}

++++return+list.get(0);
}|code-block|syntax|javascript|1153329|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a good option for doing this:

<pre><code>public static &lt;T&gt; T getSingleResult(TypedQuery&lt;T&gt; query) {
 query.setMaxResults(1);
 List&lt;T&gt; list = query.getResultList();
 if (list == null || list.isEmpty()) {
 return null;
 }

 return list.get(0);
}
</code></pre>

blocks|key|1368590|text|Spring为此提供了a+utility+method：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1368591|TypedQuery<Profile>+query+=+em.createNamedQuery(namedQuery,+Profile.class);
...
return+org.springframework.dao.support.DataAccessUtils.singleResult(query.getResultList());|code-block|syntax|javascript|1368592|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/dao/support/DataAccessUtils.html#singleResult-java.util.Collection-^0|B|G|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Spring has <a href="http://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/dao/support/DataAccessUtils.html#singleResult-java.util.Collection-">a utility method</a> for this:

<pre><code>TypedQuery&lt;Profile&gt; query = em.createNamedQuery(namedQuery, Profile.class);
...
return org.springframework.dao.support.DataAccessUtils.singleResult(query.getResultList());
</code></pre>

blocks|key|1153798|text|我已经做到了(在Java+8中)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153799|query.getResultList().stream().findFirst().orElse(null);|code-block|syntax|javascript|1153800|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I've done (in Java 8):

<pre><code>query.getResultList().stream().findFirst().orElse(null);
</code></pre>

blocks|key|1154005|text|From+JPA+2.2，而不是.getResultList()和检查list是否为空或创建流，您可以返回stream并获取第一个元素。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1154006|.getResultStream()
.findFirst()
.orElse(null);|code-block|syntax|javascript|1154007|entityMap|0|LINK|mutability|MUTABLE|url|https://thoughts-on-java.org/jpa-2-2s-new-stream-method-and-how-you-should-not-use-it/^0|G|G|0|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

<a href="https://thoughts-on-java.org/jpa-2-2s-new-stream-method-and-how-you-should-not-use-it/" rel="noreferrer">From JPA 2.2</a>, instead of <code>.getResultList()</code> and checking if list is empty or creating a stream you can return stream and take first element.

<pre><code>.getResultStream()
.findFirst()
.orElse(null);
</code></pre>

blocks|key|1153453|text|如果您希望使用try/catch机制来处理此问题..然后，它就可以用来充当if/else的角色。当我没有找到现有的记录时，我使用try/catch添加了一个新记录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153454|try+{++//if+part

++++record+=+query.getSingleResult();+++
++++//use+the+record+from+the+fetched+result.
}
catch(NoResultException+e){+//else+part
++++//create+a+new+record.
++++record+=+new+Record();
++++//.........
++++entityManager.persist(record);+
}|code-block|syntax|javascript|1153455|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you wish to use the try/catch mechanism to handle this problem.. then it can be used to act like if/else. I used the try/catch to add a new record when I didn't find an existing one.

<pre><code>try { //if part

 record = query.getSingleResult(); 
 //use the record from the fetched result.
}
catch(NoResultException e){ //else part
 //create a new record.
 record = new Record();
 //.........
 entityManager.persist(record); 
}
</code></pre>

blocks|key|1368546|text|这是一个基于Rodrigo+IronMan实现的类型化/泛型版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368547|+public+static+<T>+T+getSingleResultOrNull(TypedQuery<T>+query)+{
++++query.setMaxResults(1);
++++List<T>+list+=+query.getResultList();
++++if+(list.isEmpty())+{
++++++++return+null;
++++}
++++return+list.get(0);
}|code-block|syntax|javascript|1368548|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a typed/generics version, based on Rodrigo IronMan's implementation:

<pre><code> public static &lt;T&gt; T getSingleResultOrNull(TypedQuery&lt;T&gt; query) {
 query.setMaxResults(1);
 List&lt;T&gt; list = query.getResultList();
 if (list.isEmpty()) {
 return null;
 }
 return list.get(0);
}
</code></pre>

blocks|key|1368480|text|有一个我推荐的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368481|Query+query+=+em.createQuery("your+query");
List<Element>+elementList+=+query.getResultList();
return+CollectionUtils.isEmpty(elementList+)+?+null+:+elementList.get(0);|code-block|syntax|javascript|1368482|这可以防止空指针异常，保证只返回1个结果。|1368483|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

There is an alternative which I would recommend: 

<pre><code>Query query = em.createQuery("your query");
List&lt;Element&gt; elementList = query.getResultList();
return CollectionUtils.isEmpty(elementList ) ? null : elementList.get(0);
</code></pre>

This safeguards against Null Pointer Exception, guarantees only 1 result is returned.

blocks|key|1153648|text|结合现有答案的有用部分(限制结果的数量，检查结果的唯一性)并使用estabilshed方法名(Hibernate)，我们得到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153649|/**
+*+Return+a+single+instance+that+matches+the+query,+or+null+if+the+query+returns+no+results.
+*
+*+@param+query+query+(required)
+*+@param+<T>+result+record+type
+*+@return+record+or+null
+*/
public+static+<T>+T+uniqueResult(@NotNull+TypedQuery<T>+query)+{
++++List<T>+results+=+query.setMaxResults(2).getResultList();
++++if+(results.size()+>+1)+throw+new+NonUniqueResultException();
++++return+results.isEmpty()+?+null+:+results.get(0);
}|code-block|syntax|javascript|1153650|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Combining the useful bits of the existing answers (limiting the number of results, checking that the result is unique) and using the estabilshed method name (Hibernate), we get:

<pre><code>/**
 * Return a single instance that matches the query, or null if the query returns no results.
 *
 * @param query query (required)
 * @param &lt;T&gt; result record type
 * @return record or null
 */
public static &lt;T&gt; T uniqueResult(@NotNull TypedQuery&lt;T&gt; query) {
 List&lt;T&gt; results = query.setMaxResults(2).getResultList();
 if (results.size() &gt; 1) throw new NonUniqueResultException();
 return results.isEmpty() ? null : results.get(0);
}
</code></pre>

blocks|key|1368825|text|org.hibernate.query.Query中未记录的方法uniqueResultOptional应该可以做到这一点。您可以直接调用query.uniqueResultOptional().orElse(null)，而不必捕获NoResultException。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1368826|entityMap^0|W|K|1X|15|38|H|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

The undocumented method <code>uniqueResultOptional</code> in org.hibernate.query.Query should do the trick. Instead of having to catch a <code>NoResultException</code> you can just call <code>query.uniqueResultOptional().orElse(null)</code>.

blocks|key|1153851|text|我通过使用List<?>+myList+=+query.getResultList();并检查myList.size()是否等于零解决了这个问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1153852|entityMap^0|5|13|1B|D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

I solved this by using <code>List&lt;?&gt; myList = query.getResultList();</code> and checking if <code>myList.size()</code> equals to zero.

blocks|key|1153949|text|看看这段代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153950|return+query.getResultList().stream().findFirst().orElse(null);|offset|length|style|CODE|1153951|当调用findFirst()时，可能会抛出一个NullPointerException。|1153952|最好的方法是：|1153953|return+query.getResultList().stream().filter(Objects::nonNull).findFirst().orElse(null);|1153954|entityMap^0|0|0|1R|0|3|B|0|0|0|2G|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|O|$]]

Look this code :

<code>return query.getResultList().stream().findFirst().orElse(null);</code>

When <code>findFirst()</code> is called maybe can be throwed a NullPointerException.

the best aproach is:

<code>return query.getResultList().stream().filter(Objects::nonNull).findFirst().orElse(null);</code>

blocks|key|1368424|text|这里的逻辑与其他人的建议相同(获取resultList，返回其唯一的元素或null)，使用Google+Guava和TypedQuery。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368425|public+static+<T>+getSingleResultOrNull(final+TypedQuery<T>+query)+{
++++return+Iterables.getOnlyElement(query.getResultList(),+null);+
}|code-block|syntax|javascript|1368426|请注意，如果结果集有多个结果，则Guava将返回不直观的IllegalArgumentException。(这个异常对于getOnlyElement()的客户机来说是有意义的，因为它将结果列表作为它的参数，但是对于getSingleResultOrNull()的客户机来说就不那么容易理解了。)|1368427|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here's the same logic as others suggested (get the resultList, return its only element or null), using Google Guava and a TypedQuery.

<pre><code>public static &lt;T&gt; getSingleResultOrNull(final TypedQuery&lt;T&gt; query) {
 return Iterables.getOnlyElement(query.getResultList(), null); 
}
</code></pre>

Note that Guava will return the unintuitive IllegalArgumentException if the result set has more than one result. (The exception makes sense to clients of getOnlyElement(), as it takes the result list as its argument, but is less understandable to clients of getSingleResultOrNull().)

blocks|key|1368511|text|这是另一个扩展，这次是在Scala中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368512|customerQuery.getSingleOrNone+match+{
++case+Some(c)+=>+//+...
++case+None++++=>+//+...
}|code-block|syntax|javascript|1368513|用这个皮条客：|1368514|import+javax.persistence.{NonUniqueResultException,+TypedQuery}
import+scala.collection.JavaConversions._

object+Implicits+{

++class+RichTypedQuery[T](q:+TypedQuery[T])+{

++++def+getSingleOrNone+:+Option[T]+=+{

++++++val+results+=+q.setMaxResults(2).getResultList

++++++if+(results.isEmpty)
++++++++None
++++++else+if+(results.size+==+1)
++++++++Some(results.head)
++++++else
++++++++throw+new+NonUniqueResultException()
++++}
++}

++implicit+def+query2RichQuery[T](q:+TypedQuery[T])+=+new+RichTypedQuery[T](q)
}|1368515|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here's another extension, this time in Scala.

<pre><code>customerQuery.getSingleOrNone match {
 case Some(c) =&gt; // ...
 case None =&gt; // ...
}
</code></pre>

With this pimp:

<pre><code>import javax.persistence.{NonUniqueResultException, TypedQuery}
import scala.collection.JavaConversions._

object Implicits {

 class RichTypedQuery[T](q: TypedQuery[T]) {

 def getSingleOrNone : Option[T] = {

 val results = q.setMaxResults(2).getResultList

 if (results.isEmpty)
 None
 else if (results.size == 1)
 Some(results.head)
 else
 throw new NonUniqueResultException()
 }
 }

 implicit def query2RichQuery[T](q: TypedQuery[T]) = new RichTypedQuery[T](q)
}
</code></pre>

blocks|key|1153564|text|所以这个页面中的所有“尝试无异常重写”解决方案都有一个小问题。它既不会抛出NonUnique异常，也不会在某些错误的情况下抛出它(见下文)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153565|我认为正确的解决方案是(也许)这样：|1153566|public+static+<L>+L+getSingleResultOrNull(TypedQuery<L>+query)+{
++++List<L>+results+=+query.getResultList();
++++L+foundEntity+=+null;
++++if(!results.isEmpty())+{
++++++++foundEntity+=+results.get(0);
++++}
++++if(results.size()+>+1)+{
++++++++for(L+result+:+results)+{
++++++++++++if(result+!=+foundEntity)+{
++++++++++++++++throw+new+NonUniqueResultException();
++++++++++++}
++++++++}
++++}
++++return+foundEntity;
}|code-block|syntax|javascript|1153567|如果列表中有0个元素，则返回null；如果列表中有不同的元素，则返回nonunique；但如果其中一个select设计不正确，并且多次返回相同的对象，则不返回nonunique。|1153568|请随时发表评论。|1153569|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

So all of the "try to rewrite without an exception" solution in this page has a minor problem. Either its not throwing NonUnique exception, nor throw it in some wrong cases too (see below).

I think the proper solution is (maybe) this:

<pre><code>public static &lt;L&gt; L getSingleResultOrNull(TypedQuery&lt;L&gt; query) {
 List&lt;L&gt; results = query.getResultList();
 L foundEntity = null;
 if(!results.isEmpty()) {
 foundEntity = results.get(0);
 }
 if(results.size() &gt; 1) {
 for(L result : results) {
 if(result != foundEntity) {
 throw new NonUniqueResultException();
 }
 }
 }
 return foundEntity;
}
</code></pre>

Its returning with null if there is 0 element in the list, returning nonunique if there are different elements in the list, but not returning nonunique when one of your select is not properly designed and returns the same object more then one times.

Feel free to comment.

blocks|key|1369098|text|如果你可以使用新的JPA特性，我更喜欢@Serafins的答案，但这是一个相当简单的方法，我很惊讶以前没有在这里提到过：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369099|++++try+{
++++++++return+(Profile)+query.getSingleResult();
++++}+catch+(NoResultException+ignore)+{
++++++++return+null;
++++}|code-block|syntax|javascript|1369100|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I prefer @Serafins answer if you can use the new JPA features, but this is one fairly straight forward way to do it which I'm surprised hasn't been mentioned here before:
<pre><code> try {
 return (Profile) query.getSingleResult();
 } catch (NoResultException ignore) {
 return null;
 }
</code></pre>

blocks|key|1153892|text|这对我来说很管用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1153893|Optional<Object>+opt+=+Optional.ofNullable(nativeQuery.getSingleResult());
return+opt.isPresent()+?+opt.get()+:+null;|code-block|syntax|javascript|1153894|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Thats works to me:

<pre><code>Optional&lt;Object&gt; opt = Optional.ofNullable(nativeQuery.getSingleResult());
return opt.isPresent() ? opt.get() : null;
</code></pre>

I have an <code>insertOrUpdate</code> method which inserts an <code>Entity</code> when it doesn't exist or update it if it does. To enable this, I have to <code>findByIdAndForeignKey</code>, if it returned <code>null</code> insert if not then update. The problem is how do I check if it exists? So I tried <code>getSingleResult</code>. But it throws an exception if the

<pre><code>public Profile findByUserNameAndPropertyName(String userName, String propertyName) {
 String namedQuery = Profile.class.getSimpleName() + ".findByUserNameAndPropertyName";
 Query query = entityManager.createNamedQuery(namedQuery);
 query.setParameter("name", userName);
 query.setParameter("propName", propertyName);
 Object result = query.getSingleResult();
 if (result == null) return null;
 return (Profile) result;
}
</code></pre>

but <code>getSingleResult</code> throws an <code>Exception</code>. 

Thanks

JPA getSingleResult() or null

我有一个insertOrUpdate方法，它在Entity不存在时插入它，或者在它存在时更新它。要启用此功能，我必须执行findByIdAndForeignKey，如果返回null，则插入，如果没有，则执行更新。问题是我如何检查它是否存在？所以我尝试了getSingleResult。但它会抛出异常public Profile findByUserNameAndPropertyName(String

问JPA getSingleResult()或null
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回答 19

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回答 19

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问JPA getSingleResult()或null
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