Java 如何检查字符串数组中的元素是否为空？
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How to check if an element in a string array is empty?
提问by Jakub Pomyka?a
How can I check if an element in a string array is empty? This is my example array:
如何检查字符串数组中的元素是否为空？这是我的示例数组：
private static String correct[] = new String[5];
static {
correct[1] = "some texT";
correct[2] = "some texT3";
correct[4] = "some texT2";
}
I can assign nullto rest of elements, but I want to find another, better way to do this.
I found isEmpty, but it is available only on API 9 and above.
我可以分配null给其余的元素，但我想找到另一种更好的方法来做到这一点。我找到了isEmpty，但它仅适用于 API 9 及更高版本。
if(correct[0].length() > 0)gives me a NPE.
if(correct[0] != nullalso.
if(correct[0].length() > 0)给我一个 NPE。
if(correct[0] != null还。
采纳答案by rolfl
Have you tried the 'obvious':
您是否尝试过“显而易见”：
if(correct[0] != null && correct[0].length() > 0) {
//it is not null and it is not empty
}
回答by rgettman
Just compare it to null, which is the default value for an array of some object type.
只需将它与 比较null，这是某种对象类型数组的默认值。
if (correct[0] != null && correct[0].length() > 0)
The &&operator will only evaluate the right side if the left side is true, i.e. correct[0]isn't nulland won't throw a NullPointerException.
该&&运营商将仅分析右边如果左边是true，即correct[0]是不是null，也不会抛出NullPointerException。
回答by Mike
Well if you can't use isEmpty try to use Array.Contains(null) or try this small loop
那么如果你不能使用 isEmpty 尝试使用 Array.Contains(null) 或尝试这个小循环
for(int i = 0;i<=Array.length;i++){
if(array[i] > 0 && array[i] != null){
//yay its not null
}
}
回答by bosco
If you're trying to check if a specific element in a specific position is empty (not only different from null, but from ""too), the presented solutions won't be enough.
如果您试图检查特定位置的特定元素是否为空（不仅与 不同null，而且来自""），所提供的解决方案是不够的。
Java has a lot of ways to do this. Let's see some of them:
Java 有很多方法可以做到这一点。让我们看看其中的一些：
You can convert your array into a
List(usingArrays's asListmethod) and check the conditions throughcontains()method:import java.util.*; public class Test1 { private static boolean method1_UsingArrays(String[] array) { return Arrays.asList(array).contains(null) || Arrays.asList(array).contains(""); } }It's also possible to use
Collections's disjointmethod, to check whether two collections have elements in common or not:import java.util.*; public class Test2 { private static String[] nullAndEmpty = {"", null}; private static boolean method2_UsingCollectionsDisjoint(String[] array) { // disjoint returns true if the two specified collections have no elements in common. return !Collections.disjoint( // Arrays.asList(array), // Arrays.asList(nullAndEmpty) // ); } }If you're able to use Java 8(my favorite option), it's possible to make use of the new Streams API.
import java.util.stream.*; import java.util.*; public class Test3 { private static boolean method3_UsingJava8Streams(String[] array) { return Stream.of(array).anyMatch(x -> x == null || "".equals(x)); } }Compared to the other options, this is even easier to handle in case you need to
trimeach of your strings (or call any of the otherString's methods):Stream.of(array).anyMatch(x -> x == null || "".equals(x.trim()));
您可以将数组转换为
List（使用Arrays's asList方法）并通过contains()方法检查条件：import java.util.*; public class Test1 { private static boolean method1_UsingArrays(String[] array) { return Arrays.asList(array).contains(null) || Arrays.asList(array).contains(""); } }也可以使用
Collections's disjoint方法来检查两个集合是否有共同的元素：import java.util.*; public class Test2 { private static String[] nullAndEmpty = {"", null}; private static boolean method2_UsingCollectionsDisjoint(String[] array) { // disjoint returns true if the two specified collections have no elements in common. return !Collections.disjoint( // Arrays.asList(array), // Arrays.asList(nullAndEmpty) // ); } }如果您能够使用Java 8（我最喜欢的选项），则可以使用新的Streams API。
import java.util.stream.*; import java.util.*; public class Test3 { private static boolean method3_UsingJava8Streams(String[] array) { return Stream.of(array).anyMatch(x -> x == null || "".equals(x)); } }与其他选项相比，这更容易处理，以防您需要
trim每个字符串（或调用其他任何String方法）：Stream.of(array).anyMatch(x -> x == null || "".equals(x.trim()));
Then, you can easily call and test them:
然后，您可以轻松调用并测试它们：
public static void main(String[] args) {
// Ok tests
String[] ok = {"a", "b", "c"};
System.out.println("Method 1 (Arrays - contains): " + method1_UsingArrays(ok)); // false
System.out.println("Method 2 (Disjoint): " + method2_UsingCollectionsDisjoint(ok)); // false
System.out.println("Method 3 (Java 8 Streams): " + method3_UsingJava8Streams(ok)); // false
System.out.println("-------------------------------");
// Nok tests
String[] nok = {"a", null, "c"};
System.out.println("Method 1 (Arrays - contains): " + method1_UsingArrays(nok)); // true
System.out.println("Method 2 (Disjoint): " + method2_UsingCollectionsDisjoint(nok)); // true
System.out.println("Method 3 (Java 8 Streams): " + method3_UsingJava8Streams(nok)); // true
// Nok tests
String[] nok2 = {"a", "", "c"};
System.out.println("Method 1 (Arrays - contains): " + method1_UsingArrays(nok2)); // true
System.out.println("Method 2 (Disjoint): " + method2_UsingCollectionsDisjoint(nok2)); // true
System.out.println("Method 3 (Java 8 Streams): " + method3_UsingJava8Streams(nok2)); // true
}
回答by Mircea Stanciu
Had similar version where number of arguments could be bigger
有类似的版本，其中参数的数量可能更大
if (Arrays.asList(new String[] {productId, productTitle, imageName, productPrice}).contains(null)) {
return null;
}
回答by Woojin Ahn
Just Use isAnyEmpty
只需使用 isAnyEmpty
org.apache.commons.lang3
StringUtils.isAnyEmpty(final CharSequence... css)
