Question

尝试使用Javascript，我有一个很长的字符串，我需要使用在字符串中重复的单词将其拆分为数组。

示例：

long_string = "THS& | Willam | Baker | 1234 Corker St| Jacksonville, TX 75074| THS& Steve | James | 4312 Corker St | Jacksonville, TX 75074| THS& | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|"

我尝试过拆分和匹配，但始终忽略了THS＆

split_string = [];
split_string = long_string.split(/THS&/);
console.log(split_string);

放入数组：

[THS& | Willam | Baker | 1234 Corker St| Jacksonville, TX 75074|, THS& Steve | James | 4312 Corker St | Jacksonville, TX 75074|, THS& | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|]

但是我得到的是类似的东西

[| Willam | Baker | 1234 Corker St| Jacksonville, TX 75074|, Steve | James | 4312 Corker St | Jacksonville, TX 75074|, | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|]

Answer 1

您在split中匹配的任何内容（例如THS&）均不包括在结果中。解决方案是使用look-ahead，它实际上不会捕获字符串：

var long_string = "THS& | Willam | Baker | 1234 Corker St| Jacksonville, TX 75074| THS& Steve | James | 4312 Corker St | Jacksonville, TX 75074| THS& | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|"

var split_string = long_string.split(/(?=THS&)/);
console.log(split_string);

Answer 2

split()方法在结果子字符串中不包含分隔符。如果分隔符总是相同，请考虑在每个子字符串的开头将其串联。（您可以遍历split_string并在数组的每个字符串的开头添加串联的"THS& "）

Answer 3

分隔符不会包含在结果中，因此您必须将其重新添加。而且，第一项将为空，因此请使用Array.prototype.shift()将其从结果数组中剥离。 EG：

var long_string = "THS& | Willam | Baker | 1234 Corker St| Jacksonville, TX 75074| THS& Steve | James | 4312 Corker St | Jacksonville, TX 75074| THS& | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|";

var split_string = long_string.split("THS&").map(function(item) {
  return "THS&"+item;
});
split_string.shift();

console.log(split_string);

/* 
[
  "THS& | Willam | Baker | 1234 Corker St| Jacksonville, TX 75074| ",
  "THS& Steve | James | 4312 Corker St | Jacksonville, TX 75074| ",
  "THS& | Samuel | Cade | 1257 Corker St | Jacksonville, TX 75074|"
]
*/

Answer 4

尝试一下

//create adapter class implementing the on click listener
public class WordListAdapter extends RecyclerView.Adapter<WordListAdapter.WordViewHolder> implements View.OnClickListener {
    private final LinkedList<String> mWordList;
    private LayoutInflater mInFlater;
    private final RecyclerView recyclerView;

    //in constructor, pass in RecyclerView created in MainActivity 
    public WordListAdapter(Context context, LinkedList<String> wordList, RecyclerView rView) {
        recyclerView = rView; <----
        mInFlater = LayoutInflater.from(context);
        this.mWordList = wordList;
    }

    //implementation of View.OnClickListener makes
    @Override 
    public void onClick(View view) {
        //get the position of the item that was clicked
        int mPosition = recyclerView.getChildAdapterPosition(view); <----

        //other code to do what you want with the list(eg. String element = mWordList.get(mPosition);)
    }

    //And you set your pointer in the adapter class
    @Override
    public WordListAdapter.WordViewHolder onCreateViewHolder( @NonNull ViewGroup parent, int viewType) {
        View mItemView = mInFlater.inflate(R.layout.wordlist_item, parent, false);
        mItemView.setOnClickListener(this); <----
        return new WordViewHolder(mItemView, this);
    }

THS之前的空格用于修剪最后一个空格

我需要将重复单词上的长字符串拆分为数组

4 个答案: