java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String android.net.Uri.toString()' on a null object reference
时间: 2023-10-13 15:05:02 浏览: 88
这个错误通常表示你在调用一个空对象的方法,需要检查你的代码中是否存在对象为空的情况。具体来说,这里的错误可能是因为你在使用一个 Uri 对象时,该对象为 null,导致在调用 toString() 方法时出现了空指针异常。可以使用条件语句或者空对象判断来避免这个问题,例如:
```
if(uri != null){
String uriString = uri.toString();
//其他操作
} else {
//对象为空的处理
}
```
相关问题
java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String java.lang.Object.toString()' on a null object reference
This error occurs when you are trying to invoke the 'toString()' method on a null object reference. The 'toString()' method is used to convert an object to a string representation. However, if the object is null, then there is no object to convert and this error is thrown.
To fix this error, you need to make sure that the object you are calling the 'toString()' method on is not null. You can do this by checking for null before calling the method, like this:
if (object != null) {
String stringRepresentation = object.toString();
// Do something with the string representation
}
Alternatively, you can use the null-safe operator '?' to avoid a NullPointerException like this:
String stringRepresentation = object?.toString();
java.lang.NullPointerException: Attempt to invoke virtual method 'int java.lang.String.length()' on a null object reference
当你尝试在一个空对象上调用方法时,就会出现java.lang.NullPointerException异常。在这个特定的引用中,你尝试在一个空字符串上调用length()方法,因此出现了异常。要解决这个问题,你需要确保在调用方法之前,对象不是空的。你可以使用if语句来检查对象是否为空,或者使用try-catch块来捕获异常并处理它们。
以下是一个使用if语句检查字符串是否为空的例子:
```java
String str = null;
if (str != null) {
int length = str.length();
System.out.println("Length of string is: " + length);
} else {
System.out.println("String is null.");
}
```
以下是一个使用try-catch块处理空字符串的例子:
```java
String str = null;
try {
int length = str.length();
System.out.println("Length of string is: " + length);
} catch (NullPointerException e) {
System.out.println("String is null.");
}
```