java ArrayList<String> 类型中的 add(String) 方法不适用于参数 (String[])
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The method add(String) in the type ArrayList<String> is not applicable for the arguments (String[])
提问by user3756933
I'm a beginner in JAVA and programming in general, so please be patient as I may not use the correct terms to correctly describe my doubts. I'll do my best, nevertheless. So, I have this ArrayList that I'm going to use regular expressions on, to split it on the commas. I really needed some help solving this problem, even if I have to change the way that I do the process. It's not important that it stays this way, it's the final result that matters the most to me. Thank you.
我是 JAVA 和一般编程的初学者，所以请耐心等待，因为我可能不会使用正确的术语来正确描述我的疑虑。尽管如此，我会尽力而为。所以，我有这个 ArrayList，我将在上面使用正则表达式，以逗号分隔它。我真的需要一些帮助来解决这个问题，即使我必须改变我做这个过程的方式。保持这种状态并不重要，对我来说最重要的是最终结果。谢谢你。
String temp;
String temp2;
ArrayList <String> tempsplit = new ArrayList<String> ();
ArrayList <String> dominios = new ArrayList<String> (); {
for (int h = 0; h < 191; h++){
temp = features.get(h);
**temp2.add(temp.split(","));
tempsplit.add(temp.split(","));**
//in these last couple lines I get the error "The method add(String) in the type ArrayList<String> is not applicable for the arguments (String[])"
for(int oi = 0; oi < tempsplit.size(); oi++){
for (int t = 0; t < dominios.size() ; t++){
int conf = 0;
if (tempsplit.get(oi) == dominios.get(t)){
conf = 0;
}
else{ conf = 1;
}
if (conf == 1){
dominios.add (tempsplit.get(oi));
}
}
}
回答by Joop Eggen
Collections.addAll(temp2, temp.split(","));
This uses the help class Collections to add a String[]by item.
这使用帮助类 Collections 添加一个String[]按项。
回答by Jean Logeart
temp.split(",")returns a String[].
temp.split(",")返回一个String[].
List<String>.addtakes a Stringas a parameter, not an array of strings.
List<String>.add将 aString作为参数，而不是字符串数组。
Yet you can use Collections.addAll, that takes an array as second argument:
但是您可以使用Collections.addAll, 将数组作为第二个参数：
Collections.addAll(temp2, temp.split(","));
You can alternatively use the addAll(Collection<String> c)method from the tempArrayList<String>but then you have to convert the array to a Collection:
您也可以使用 中的addAll(Collection<String> c)方法，tempArrayList<String>但是您必须将数组转换为 a Collection：
temp2.addAll(Arrays.asList(temp.split(",")));
回答by Bohemian
The code with the problem is essentially:
有问题的代码本质上是：
ArrayList <String> tempsplit = new ArrayList<String>();
tempsplit.add(temp.split(",")); // compile error
The problem is that split()returns a String[], but the list will only accept a String.
问题是split()返回 a String[]，但列表只接受 a String。
To fix, convert the array to a List and pass it to addAll():
要修复，请将数组转换为 List 并将其传递给addAll()：
tempsplit.addAll(Arrays.asList(temp.split(",")));
or use the utility addAll()method:
或使用实用addAll()方法：
Collections.addAll(tempsplit, temp.split(","));
回答by silentprogrammer
Your Arraylist is of type Stringnot String[]and you are passing string array in add()because string.split()will give you array.Instead you can do
你的 Arraylist 的类型String不是String[]，你正在传入字符串数组，add()因为string.split()会给你数组。相反，你可以做
for(String a:temp.split(",")){
tempsplit.add(a);
}
回答by Jesper
The method add(String) in the type ArrayList is not applicable for the arguments (String[])
ArrayList 类型中的 add(String) 方法不适用于参数 (String[])
This means that you have an ArrayList<String>, and you are trying to call the addmethod with an arrayof strings (String[]) instead of with a single String. The method needs a single String, not an array.
这意味着您有一个ArrayList<String>，并且您正在尝试add使用字符串数组( String[]) 而不是单个String. 该方法需要单个String，而不是数组。
