java 如何知道ArrayList中指定的对象是否为空？
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How to know if specified object in ArrayList is null?
提问by Dominik
I want to know if an Object in an ArrayList is null. If it's null, then it shouldn't do anything. Example:
我想知道 ArrayList 中的对象是否为空。如果它为空，那么它不应该做任何事情。例子：
if(!(theList.get(theIndexofObject) == null)){
do something...
}
else{
do nothing...
}
This doesn't work, because it throws an exception cause of the '.get()'-method. Any ideas to solve this problem?
这不起作用，因为它引发了 '.get()' 方法的异常原因。有什么想法可以解决这个问题吗？
回答by Alex
回答by Adriaan Koster
You are probably confused about how to use the API. Here is a simple example of how it works:
您可能对如何使用 API 感到困惑。这是它如何工作的一个简单示例：
import java.util.ArrayList;
import java.util.List;
public class NullItems {
public static void main(String[] args) {
List<Object> items = new ArrayList<Object>();
items.add("foo");
items.add(null);
items.add(25);
for (int i = 0; i < items.size(); i++) {
Object item = items.get(i);
if (item != null) {
System.out.println(item);
}
}
// or shorter:
for (Object item : items) {
if (item != null) {
System.out.println(item);
}
}
}
}
回答by adamjmarkham
You are using the getmethod wrong. You need to pass the index an item is at to the getmethod. You could use the containsmethod to see if the object is in the ArrayList.
你使用的get方法不对。您需要将项目所在的索引传递给该get方法。您可以使用该contains方法查看对象是否在 ArrayList 中。
Example:
例子：
if(theList.contains(theObject))
//do something
Otherwise you could use a try and catch which seems confusing and hard to read so I would strongly not recommend doing the following but have included it to show you:
否则，您可以使用看起来令人困惑且难以阅读的 try 和 catch，因此我强烈不建议您执行以下操作，但已将其包含在内以向您展示：
for(int i=0; i<theList.size(); i++)
{
try
{
if(!(theList.get(i) == null))
{
//do something
}
else
{
//do nothing
}
}
catch(NullPointerException npe)
{
//do something else
}
}
Alternatively use a for-each loop.
或者使用 for-each 循环。
回答by Warrior
In javaScript itemArray.length, for java u have to use ARRAY.size() insted of length function
在 javaScript itemArray.length 中，对于 java，你必须使用长度函数的 ARRAY.size() 插入
var itemArray=//Assign some list of value;
for (var i = 0; i < itemArray.length; i++){
if(itemArray[i].value == null){
Do nothing
}else{
Do something
}
}
回答by Sina
i think that your arraylist is null chang first condition to:
我认为你的 arraylist 是 null 改变第一个条件：
if(theList!=null && !(theList.get(theIndexofObject) == null)){
// do something...
}
else{
// do nothing...
}
回答by Muhammad Aamir Ali
The method arrayList.size() returns the number of items in the list - so if the index is greater than or equal to the size(), it doesn't exist.
方法 arrayList.size() 返回列表中的项目数 - 因此，如果索引大于或等于 size()，则它不存在。
回答by user3571695
if(!(theList.get(theIndexofObject) == null)){
do something...
}
else{
do nothing...
}
instead of writing this code.Try in the below format,I think you will get answer
而不是写这个代码。试试下面的格式，我想你会得到答案
if(theList.get(theIndexofObject)!= null)){
do something...
}
else{
do nothing...
}
