blocks|key|704297|text|ArrayList根据equals(Object+obj)方法删除对象。所以你应该正确地实现这个方法。类似于：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|704298|public+boolean+equals(Object+obj)+{
++++if+(obj+==+null)+return+false;
++++if+(obj+==+this)+return+true;
++++if+(!(obj+instanceof+ArrayTest))+return+false;
++++ArrayTest+o+=+(ArrayTest)+obj;
++++return+o.i+==+this.i;
}|code-block|syntax|javascript|704299|或|704300|public+boolean+equals(Object+obj)+{
++++if+(obj+instanceof+ArrayTest)+{
++++++++ArrayTest+o+=+(ArrayTest)+obj;
++++++++return+o.i+==+this.i;
++++}
++++return+false;
}|704301|entityMap^0|0|9|B|I|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

<code>ArrayList</code> removes objects based on the <code>equals(Object obj)</code> method. So you should implement properly this method. Something like:

<pre><code>public boolean equals(Object obj) {
 if (obj == null) return false;
 if (obj == this) return true;
 if (!(obj instanceof ArrayTest)) return false;
 ArrayTest o = (ArrayTest) obj;
 return o.i == this.i;
}
</code></pre>

Or

<pre><code>public boolean equals(Object obj) {
 if (obj instanceof ArrayTest) {
 ArrayTest o = (ArrayTest) obj;
 return o.i == this.i;
 }
 return false;
}
</code></pre>

blocks|key|488958|text|如果您使用的是Java+8：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|488959|test.removeIf(t+->+t.i+==+1);|code-block|syntax|javascript|488960|Java8在集合接口中有一个removeIf方法。对于ArrayList，它有一个高级实现(n阶)。|offset|length|style|CODE|488961|entityMap^0|0|0|E|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you are using Java 8:

<pre><code>test.removeIf(t -&gt; t.i == 1);
</code></pre>

Java 8 has a <code>removeIf</code> method in the collection interface. For the ArrayList, it has an advanced implementation (order of n).

blocks|key|704335|text|通常，可以通过两种方式从ArrayList+(或任何List)中删除对象，即按索引(remove(int)+)和按对象(remove(Object)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|704336|在这个特定的场景中:向ArrayTest类添加一个equals(Object)方法。这将使ArrayList.remove(Object)能够识别正确的对象。|704337|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html|1|http://docs.oracle.com/javase/6/docs/api/java/util/List.html|2|http://docs.oracle.com/javase/6/docs/api/java/util/List.html#remove(int%2529)|3|http://docs.oracle.com/javase/6/docs/api/java/util/List.html#remove(java.lang.Object%2529)|4|http://docs.oracle.com/javase/6/docs/api/java/lang/Object.html#equals(java.lang.Object)|5|http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#remove(java.lang.Object)^0|C|9|Q|4|16|B|1O|E|C|9|0|Q|4|1|16|B|2|1O|E|3|0|B|9|P|E|19|O|P|E|4|19|O|5|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]|$9|1E|A|1F|1|1G]|$9|1H|A|1I|1|1J]]|E|$]]|$1|F|3|G|5|6|7|1K|8|@$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@$9|1R|A|1S|1|1T]|$9|1U|A|1V|1|1W]]|E|$]]|$1|H|3|-4|5|6|7|1X|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]|R|$5|K|L|M|E|$N|S]]|T|$5|K|L|M|E|$N|U]]|V|$5|K|L|M|E|$N|W]]|X|$5|K|L|M|E|$N|Y]]]]

In general an object can be removed in two ways from an <a href="http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html" rel="noreferrer"><code>ArrayList</code></a> (or generally any <a href="http://docs.oracle.com/javase/6/docs/api/java/util/List.html" rel="noreferrer"><code>List</code></a>), by index (<a href="http://docs.oracle.com/javase/6/docs/api/java/util/List.html#remove(int)" rel="noreferrer"><code>remove(int)</code></a>) and by object (<a href="http://docs.oracle.com/javase/6/docs/api/java/util/List.html#remove(java.lang.Object)" rel="noreferrer"><code>remove(Object)</code></a>).

In this particular scenario: Add an <a href="http://docs.oracle.com/javase/6/docs/api/java/lang/Object.html#equals(java.lang.Object)" rel="noreferrer"><code>equals(Object)</code></a> method to your <code>ArrayTest</code> class. That will allow <a href="http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#remove(java.lang.Object)" rel="noreferrer"><code>ArrayList.remove(Object)</code></a> to identify the correct object.

blocks|key|704376|text|使用此代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|704377|test.remove(test.indexOf(obj));

test+is+your+ArrayList+and+obj+is+the+Object,+first+you+find+the+index+of+obj+in+ArrayList+and+then+you+remove+it+from+the+ArrayList.|code-block|syntax|javascript|704378|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|$]]

use this code

<pre><code>test.remove(test.indexOf(obj));

test is your ArrayList and obj is the Object, first you find the index of obj in ArrayList and then you remove it from the ArrayList.
</code></pre>

blocks|key|704743|text|List<Object>+list+=+new+ArrayList();
for+(Iterator<Object>+iterator+=+list.iterator();+iterator.hasNext();)+{
++Object+obj=+iterator.next();
++++if+(obj.getId().equals("1"))+{
+++++++//+Remove+the+current+element+from+the+iterator+and+the+list.
+++++++iterator.remove();
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|704744|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;Object&gt; list = new ArrayList();
for (Iterator&lt;Object&gt; iterator = list.iterator(); iterator.hasNext();) {
 Object obj= iterator.next();
 if (obj.getId().equals("1")) {
 // Remove the current element from the iterator and the list.
 iterator.remove();
 }
}
</code></pre>

blocks|key|489181|text|+private+List<Employee>+list+=+new+ArrayList<>();
+list.removeIf(emp+->+emp.getId()+==+10);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|489182|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> private List&lt;Employee&gt; list = new ArrayList&lt;&gt;();
 list.removeIf(emp -&gt; emp.getId() == 10);
</code></pre>

blocks|key|488774|text|AValchev是对的。一种更快的解决方案是解析所有元素，并通过唯一的属性进行比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|488775|String+property+=+"property+to+delete";

for(int+j+=+0;+j+<+i.size();+j%2B%2B)
{
++++Student+obj+=+i.get(j);

++++if(obj.getProperty().equals(property)){
+++++++//found,+delete.
++++++++i.remove(j);
++++++++break;
++++}

}|code-block|syntax|javascript|488776|THis是一个快速的解决方案。你最好在更大的项目中实现对象比较。|488777|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|$]]

AValchev is right.
A quicker solution would be to parse all elements and compare by an unique property. 

<pre><code>String property = "property to delete";

for(int j = 0; j &lt; i.size(); j++)
{
 Student obj = i.get(j);

 if(obj.getProperty().equals(property)){
 //found, delete.
 i.remove(j);
 break;
 }

}
</code></pre>

THis is a quick solution. You'd better implement object comparison for larger projects.

blocks|key|489011|text|如果您想删除与该属性匹配的多个对象，请尝试执行此操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|489012|我使用下面的代码从对象数组中删除元素，这对我很有帮助。|489013|一般来说，可以通过两种方式从对象(或任何列表)中删除对象，通过index+(remove(int))和object+(remove(+ArrayList+))。|offset|length|style|CODE|489014|对于使用这些行的arrayList.remove(index)或arrayList.remove(obj.get(index))来说，有时可能无法工作，请尝试使用以下代码。|489015|for+(Iterator<DetailInbox>+iter+=+detailInboxArray.iterator();+iter.hasNext();+)+{
++++DetailInbox+element+=+iter.next();
+++if+(element.isSelected())+{
++++++iter.remove();
+++}
}|code-block|syntax|javascript|489016|entityMap^0|0|0|V|5|1F|6|1V|9|0|8|N|W|W|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@$F|V|G|W|H|I]|$F|X|G|Y|H|I]|$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@$F|12|G|13|H|I]|$F|14|G|15|H|I]]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

If you want to remove multiple objects that are matching to the property try this.

I have used following code to remove element from object array it helped me.

In general an object can be removed in two ways from an <code>ArrayList</code> (or generally any List), by <code>index</code> (remove(int)) and by <code>object</code> (remove(Object)).

some time for you <code>arrayList.remove(index)</code>or <code>arrayList.remove(obj.get(index))</code> using these lines may not work try to use following code.

<pre><code>for (Iterator&lt;DetailInbox&gt; iter = detailInboxArray.iterator(); iter.hasNext(); ) {
 DetailInbox element = iter.next();
 if (element.isSelected()) {
 iter.remove();
 }
}
</code></pre>

blocks|key|489076|text|我已经尝试过了，它对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|489077|ArrayList<cartItem>+cartItems=+new+ArrayList<>();

//filling+the+cartItems

cartItem+ci=new+cartItem(itemcode,itemQuantity);//the+one+I+want+to+remove

Iterator<cartItem>+itr+=cartItems.iterator();
while+(itr.hasNext()){
+++cartItem+ci_itr=itr.next();
+++if+(ci_itr.getClass()+==+ci.getClass()){
++++++itr.remove();
++++++return;
++++}
}|code-block|syntax|javascript|489078|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 have tried this and it works for me:

<pre><code>ArrayList&lt;cartItem&gt; cartItems= new ArrayList&lt;&gt;();

//filling the cartItems

cartItem ci=new cartItem(itemcode,itemQuantity);//the one I want to remove

Iterator&lt;cartItem&gt; itr =cartItems.iterator();
while (itr.hasNext()){
 cartItem ci_itr=itr.next();
 if (ci_itr.getClass() == ci.getClass()){
 itr.remove();
 return;
 }
}
</code></pre>

blocks|key|704359|text|++++ArrayTest+obj=new+ArrayTest(1);
++++test.add(obj);
++++ArrayTest+obj1=new+ArrayTest(2);
++++test.add(obj1);
++++ArrayTest+obj2=new+ArrayTest(3);
++++test.add(obj2);

++++test.remove(object+of+ArrayTest);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|704360|您可以指定如何控制每个对象。|unstyled|704361|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> ArrayTest obj=new ArrayTest(1);
 test.add(obj);
 ArrayTest obj1=new ArrayTest(2);
 test.add(obj1);
 ArrayTest obj2=new ArrayTest(3);
 test.add(obj2);

 test.remove(object of ArrayTest);
</code></pre>

you can specify how you control each object.

blocks|key|488843|text|您可以使用Collections.binarySearch查找元素，然后对返回的索引调用remove。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|488844|请参阅此处的Collections.binarySearch文档：http://docs.oracle.com/javase/1.4.2/docs/api/java/util/Collections.html#binarySearch%2528java.util.List,%2520java.lang.Object%2529|offset|length|488845|不过，这需要ArrayTest对象实现.equals。您还需要调用Collections.sort来对列表进行排序。最后，为了让binarySearch正确运行，ArrayTest必须实现可比较的接口。|488846|在Java中，这是一种“恰当”的方式。如果您只是想以一种快速而肮脏的方式解决问题，那么您只需迭代元素并删除具有您要查找的属性的元素即可。|488847|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/1.4.2/docs/api/java/util/Collections.html#binarySearch%2528java.util.List,%2520java.lang.Object%2529^0|0|X|3G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

You can use Collections.binarySearch to find the element, then call remove on the returned index.

See the documentation for Collections.binarySearch here:
<a href="http://docs.oracle.com/javase/1.4.2/docs/api/java/util/Collections.html#binarySearch%28java.util.List,%20java.lang.Object%29" rel="nofollow">http://docs.oracle.com/javase/1.4.2/docs/api/java/util/Collections.html#binarySearch%28java.util.List,%20java.lang.Object%29</a>

This would require the ArrayTest object to have .equals implemented though. You would also need to call Collections.sort to sort the list. Finally, ArrayTest would have to implement the Comparable interface, so that binarySearch would run correctly.

This is the "proper" way to do it in Java. If you are just looking to solve the problem in a quick and dirty fashion, then you can just iterate over the elements and remove the one with the attribute you are looking for.

blocks|key|488882|text|这对我很有帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|488883|++++++++card+temperaryCardFour+=+theDeck.get(theDeck.size()+-+1);
++++++++theDeck.remove(temperaryCardFour);++++|code-block|syntax|javascript|488884|而不是|488885|theDeck.remove(numberNeededRemoved);|488886|我在第一个代码片段上得到了一个删除确认，在第二个代码片段上得到了一个未删除确认。|488887|尝试使用第一个代码片段切换您的代码，我认为这是您的问题所在。|488888|内森·纳尔逊|488889|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

This helped me:

<pre><code> card temperaryCardFour = theDeck.get(theDeck.size() - 1);
 theDeck.remove(temperaryCardFour); 
</code></pre>

instead of 

<pre><code>theDeck.remove(numberNeededRemoved);
</code></pre>

I got a removal conformation on the first snippet of code and an un removal conformation on the second.

Try switching your code with the first snippet I think that is your problem.

Nathan Nelson

blocks|key|489088|text|简单使用remove()函数。并将对象作为您想要删除的参数进行传递。ur+arraylist.remove(obj)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|489089|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

simple use remove() function. and pass object as param u want to remove.
ur arraylist.remove(obj)

blocks|key|704884|text|或者您可以使用java+8+lambda。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|704885|test.removeIf(i+->+i==2)；|704886|它将简单地删除所有满足条件的对象|704887|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

or you can use java 8 lambda
test.removeIf(i -&gt; i==2);
it will simply remove all object that meet the condition

blocks|key|489291|text|简单字符串列表中的示例，如果有人想要的话：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|489292|+public+ArrayList<String>+listAfterRemoved(ArrayList<String>+arrayList,+String+toRemove)+{

++++++++for+(int+i+=+0;+i+<+arrayList.size();+i%2B%2B)+{
++++++++++++if+(arrayList.get(i).equals(toRemove))+{
++++++++++++++++arrayList.remove(toRemove);
++++++++++++}
++++++++}
++++++++return+arrayList;
++++}|code-block|syntax|javascript|489293|调用是这样的：|489294|ArrayList<String>+arrayList+=+new+ArrayList<>();
++++++++arrayList.add("1");
++++++++arrayList.add("2");
++++++++arrayList.add("3");
++++++++arrayList.add("4");

++++++++System.out.println("Array+List+before:+"+%2B+arrayList.toString());
++++++++arrayList+=+listAfterRemoved(arrayList,+"2");
++++++++System.out.println("Array+List+after+:+"+%2B+arrayList.toString());|489295|​|489296|📷|atomic|offset|length|489297|489298|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/7b390f044317d75140a79e5b60faef1d.png|imageAlt^0|0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|13|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|15|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|16|8|@]|9|@$P|17|Q|18|1|19]]|A|$]]|$1|R|3|L|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z|10|-4]]]]

Example within a simple String List, if anyone wants :
<pre><code> public ArrayList&lt;String&gt; listAfterRemoved(ArrayList&lt;String&gt; arrayList, String toRemove) {

 for (int i = 0; i &lt; arrayList.size(); i++) {
 if (arrayList.get(i).equals(toRemove)) {
 arrayList.remove(toRemove);
 }
 }
 return arrayList;
 }
</code></pre>
And the call is :
<pre><code>ArrayList&lt;String&gt; arrayList = new ArrayList&lt;&gt;();
 arrayList.add(&quot;1&quot;);
 arrayList.add(&quot;2&quot;);
 arrayList.add(&quot;3&quot;);
 arrayList.add(&quot;4&quot;);

 System.out.println(&quot;Array List before: &quot; + arrayList.toString());
 arrayList = listAfterRemoved(arrayList, &quot;2&quot;);
 System.out.println(&quot;Array List after : &quot; + arrayList.toString());
</code></pre>
<a href="https://i.stack.imgur.com/Vf69v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vf69v.png" alt="enter image description here" /></a>

How can I remove specific object from ArrayList?
Suppose I have a class as below:

<pre><code>import java.util.ArrayList; 
public class ArrayTest {
 int i;

 public static void main(String args[]){
 ArrayList&lt;ArrayTest&gt; test=new ArrayList&lt;ArrayTest&gt;();
 ArrayTest obj;
 obj=new ArrayTest(1);
 test.add(obj);
 obj=new ArrayTest(2);
 test.add(obj);
 obj=new ArrayTest(3);
 test.add(obj);
 }
 public ArrayTest(int i){
 this.i=i;
 }
}
</code></pre>

How can I remove object with <code>new ArrayTest(1)</code> from my <code>ArrayList&lt;ArrayList&gt;</code>

How to remove specific object from ArrayList in Java?

如何从ArrayList中删除特定对象？假设我有一个如下的类：import java.util.ArrayList; public class ArrayTest { int i; public static void main(String args[]){ ArrayList<ArrayTest> test=new ArrayList<ArrayTest>();

问在Java中如何从ArrayList中删除特定对象？
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