blocks|key|4464403|text|您只需要过滤具有null名称的汽车：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4464404|requiredCars+=+cars.stream()
+++++++++++++++++++.filter(c+->+c.getName()+!=+null)
+++++++++++++++++++.filter(c+->+c.getName().startsWith("M"));|code-block|syntax|javascript|4464405|entityMap^0|8|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You just need to filter the cars that have a <code>null</code> name:

<pre><code>requiredCars = cars.stream()
 .filter(c -&gt; c.getName() != null)
 .filter(c -&gt; c.getName().startsWith("M"));
</code></pre>

blocks|key|4464441|text|您可以在单个筛选器步骤中完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4464442|requiredCars+=+cars.stream().filter(c+->+c.getName()+!=+null+&&+c.getName().startsWith("M"));|code-block|syntax|javascript|4464443|如果您不想多次调用getName()+(例如，调用成本很高)，您可以这样做：|offset|length|style|CODE|4464444|requiredCars+=+cars.stream().filter(c+->+{
++++String+name+=+c.getName();
++++return+name+!=+null+&&+name.startsWith("M");
});|4464445|或者以更复杂的方式：|4464446|requiredCars+=+cars.stream().filter(c+->+
++++Optional.ofNullable(c.getName()).filter(name+->+name.startsWith("M")).isPresent());|4464447|entityMap^0|0|0|9|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You can do this in single filter step:

<pre><code>requiredCars = cars.stream().filter(c -&gt; c.getName() != null &amp;&amp; c.getName().startsWith("M"));
</code></pre>

If you don't want to call <code>getName()</code> several times (for example, it's expensive call), you can do this:

<pre><code>requiredCars = cars.stream().filter(c -&gt; {
 String name = c.getName();
 return name != null &amp;&amp; name.startsWith("M");
});
</code></pre>

Or in more sophisticated way:

<pre><code>requiredCars = cars.stream().filter(c -&gt; 
 Optional.ofNullable(c.getName()).filter(name -&gt; name.startsWith("M")).isPresent());
</code></pre>

blocks|key|4464512|text|在这个特殊的例子中，我认为@Tagir是100%25正确的，把它放到一个过滤器中，然后做两个检查。我不会使用Optional.ofNullable，可选的东西实际上是返回类型，而不是逻辑。但实际上不是这里也不是那里。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4464513|我想指出的是，在广泛的情况下，java.util.Objects有一个很好的方法来解决这个问题，所以您可以这样做：|4464514|cars.stream()
++++.filter(Objects::nonNull)|code-block|syntax|javascript|4464515|这将清除您的空对象。对于任何不熟悉的人来说，这是以下内容的速记：|4464516|cars.stream()
++++.filter(car+->+Objects.nonNull(car))|4464517|要部分回答手头的问题，请返回以"M"开头的汽车名称列表|4464518|cars.stream()
++++.filter(car+->+Objects.nonNull(car))
++++.map(car+->+car.getName())
++++.filter(carName+->+Objects.nonNull(carName))
++++.filter(carName+->+carName.startsWith("M"))
++++.collect(Collectors.toList());|4464519|一旦你习惯了速记的lambdas，你也可以这样做：|4464520|cars.stream()
++++.filter(Objects::nonNull)
++++.map(Car::getName)++++++++//+Assume+the+class+name+for+car+is+Car
++++.filter(Objects::nonNull)
++++.filter(carName+->+carName.startsWith("M"))
++++.collect(Collectors.toList());|4464521|不幸的是，一旦你.map(Car::getName)，你将只返回名单的名字，而不是汽车。所以不那么漂亮，但完全回答了这个问题：|4464522|cars.stream()
++++.filter(car+->+Objects.nonNull(car))
++++.filter(car+->+Objects.nonNull(car.getName()))
++++.filter(car+->+car.getName().startsWith("M"))
++++.collect(Collectors.toList());|4464523|entityMap^0|1G|J|0|F|H|0|0|0|0|F|3|0|0|0|0|8|I|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1H|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1I|8|@]|D|@]|E|$K|L]]|$1|Y|3|Z|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|10|3|11|5|J|7|1M|8|@]|D|@]|E|$K|L]]|$1|12|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|13|$]]

In this particular example I think @Tagir is 100% correct get it into one filter and do the two checks. I wouldn't use <code>Optional.ofNullable</code> the Optional stuff is really for return types not to be doing logic... but really neither here nor there.

I wanted to point out that <code>java.util.Objects</code> has a nice method for this in a broad case, so you can do this:

<pre><code>cars.stream()
 .filter(Objects::nonNull)
</code></pre>

Which will clear out your null objects. For anyone not familiar, that's the short-hand for the following:

<pre><code>cars.stream()
 .filter(car -&gt; Objects.nonNull(car))
</code></pre>

To partially answer the question at hand to return the list of car names that starts with <code>"M"</code>:

<pre><code>cars.stream()
 .filter(car -&gt; Objects.nonNull(car))
 .map(car -&gt; car.getName())
 .filter(carName -&gt; Objects.nonNull(carName))
 .filter(carName -&gt; carName.startsWith("M"))
 .collect(Collectors.toList());
</code></pre>

Once you get used to the shorthand lambdas you could also do this:

<pre><code>cars.stream()
 .filter(Objects::nonNull)
 .map(Car::getName) // Assume the class name for car is Car
 .filter(Objects::nonNull)
 .filter(carName -&gt; carName.startsWith("M"))
 .collect(Collectors.toList());
</code></pre>

Unfortunately once you <code>.map(Car::getName)</code> you'll only be returning the list of names, not the cars. So less beautiful but fully answers the question:

<pre><code>cars.stream()
 .filter(car -&gt; Objects.nonNull(car))
 .filter(car -&gt; Objects.nonNull(car.getName()))
 .filter(car -&gt; car.getName().startsWith("M"))
 .collect(Collectors.toList());
</code></pre>

blocks|key|4464330|text|提出的答案很好。我只想提出一个改进方案，以处理使用Optional.ofNullable、new+feature+in+Java+8|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4464331|+List<String>+carsFiltered+=+Optional.ofNullable(cars)
++++++++++++++++.orElseGet(Collections::emptyList)
++++++++++++++++.stream()
++++++++++++++++.filter(Objects::nonNull)
++++++++++++++++.collect(Collectors.toList());|code-block|syntax|javascript|4464332|因此，完整的答案将是：|4464333|+List<String>+carsFiltered+=+Optional.ofNullable(cars)
++++++++++++++++.orElseGet(Collections::emptyList)
++++++++++++++++.stream()
++++++++++++++++.filter(Objects::nonNull)+//filtering+car+object+that+are+null
++++++++++++++++.map(Car::getName)+//now+it's+a+stream+of+Strings
++++++++++++++++.filter(Objects::nonNull)+//filtering+null+in+Strings
++++++++++++++++.filter(name+->+name.startsWith("M"))
++++++++++++++++.collect(Collectors.toList());+//back+to+List+of+Strings|4464334|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/Optional.html^0|P|J|19|L|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

The proposed answers are great. Just would like to suggest an improvement to handle the case of null list using <code>Optional.ofNullable</code>, <a href="https://docs.oracle.com/javase/8/docs/api/java/util/Optional.html" rel="noreferrer">new feature in Java 8</a>:

<pre><code> List&lt;String&gt; carsFiltered = Optional.ofNullable(cars)
 .orElseGet(Collections::emptyList)
 .stream()
 .filter(Objects::nonNull)
 .collect(Collectors.toList());
</code></pre>

So, the full answer will be:

<pre><code> List&lt;String&gt; carsFiltered = Optional.ofNullable(cars)
 .orElseGet(Collections::emptyList)
 .stream()
 .filter(Objects::nonNull) //filtering car object that are null
 .map(Car::getName) //now it's a stream of Strings
 .filter(Objects::nonNull) //filtering null in Strings
 .filter(name -&gt; name.startsWith("M"))
 .collect(Collectors.toList()); //back to List of Strings
</code></pre>

blocks|key|4464591|text|你可以使用这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4464592|List<Car>+requiredCars+=+cars.stream()
++++.filter+(t->++t!=+null+&&+StringUtils.startsWith(t.getName(),"M"))
++++.collect(Collectors.toList());|code-block|syntax|javascript|4464593|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|$]]

you can use this

<pre><code>List&lt;Car&gt; requiredCars = cars.stream()
 .filter (t-&gt; t!= null &amp;&amp; StringUtils.startsWith(t.getName(),"M"))
 .collect(Collectors.toList());

</code></pre>

blocks|key|4464611|text|利用java.util.Optional#map()的强大功能|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4464612|List<Car>+requiredCars+=+cars.stream()
++.filter+(car+->+
++++Optional.ofNullable(car)
++++++.map(Car::getName)
++++++.map(name+->+name.startsWith("M"))
++++++.orElse(false)+//+what+to+do+if+either+car+or+getName()+yields+null?+false+will+filter+out+the+element
++++)
++.collect(Collectors.toList())
;|code-block|syntax|javascript|4464613|entityMap^0|2|O|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Leveraging the power of <code>java.util.Optional#map()</code>:

<pre><code>List&lt;Car&gt; requiredCars = cars.stream()
 .filter (car -&gt; 
 Optional.ofNullable(car)
 .map(Car::getName)
 .map(name -&gt; name.startsWith("M"))
 .orElse(false) // what to do if either car or getName() yields null? false will filter out the element
 )
 .collect(Collectors.toList())
;
</code></pre>

I have a list of objects say <code>car</code>. I want to filter this list based on some parameter using Java 8. But if the parameter is <code>null</code>, it throws <code>NullPointerException</code>. How to filter out null values?

Current code is as follows

<pre><code>requiredCars = cars.stream().filter(c -&gt; c.getName().startsWith("M"));
</code></pre>

This throws <code>NullPointerException</code> if <code>getName()</code> returns <code>null</code>.

Filter values only if not null using lambda in Java8

我有一个对象列表，上面写着car。我想使用Java8根据一些参数来过滤这个列表，但是如果参数是null，它就会抛出NullPointerException。如何过滤掉null值？当前代码如下requiredCars = cars.stream().filter(c -> c.getName().startsWith("...

问仅当不为空时使用Java8中的lambda筛选值
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