如何防止我的搜索栏需要在另一个窗口中打开

如何解决如何防止我的搜索栏需要在另一个窗口中打开

我创建了一个 JS 搜索栏。目前我的代码如下所示：

let results = document.getElementById('searchBar');
console.log(results);
let theWebsites = [];
results.addEventListener('click',(e) => {
  if (window.location.href != "file:///Users/eswizzle/project/searchbarhtml.html?searchBar"){
    window.location.replace("file:///Users/eswizzle/project/searchbarhtml.html?searchBar");
  }
});
results.addEventListener('keypress',(e) => {
    console.log(e);
    if (results.value.length > 1 && e.keyCode === 13){
      e.preventDefault();
      const searchString = results.value.toLowerCase();
      const filteredWebsites = theWebsites.filter(website => {
        return website.name.toLowerCase().includes(searchString) || website.keywords.toLowerCase().includes (searchString);
      });
      console.log(searchString);
      displayWebsites(filteredWebsites);
      e.preventDefault();
      results.value = "";
      console.log(filteredWebsites);
    }
});

const loadWebsites = async () => {
  try{
    const res = await fetch('http://localhost:3000/websites');
    theWebsites = await res.json();
    console.log(theWebsites);
  } catch(err){
    console.error(err);
    console.log("error");
  }
};
const displayWebsites = (websites) => {
  console.log(websites);

  let htmlString = websites.map((character) => {
      return `
      <li class = "normalContainerDiv">
        <h2><a href = "Users/eswizzle/project${character.link}.html">${character.name}</h2></a>
      </li>
      `;

  })
  .join('');

  console.log(htmlString);
  if (htmlString == false){
    htmlString = `
    <li class = "normalContainerDiv">
      <h2>No Results</h2>
    </li>
    `;
  }
  console.log(htmlString);
  websiteList.innerHTML = htmlString;
  waitMoment.innerHTML = "Search Results:";
};
loadWebsites();

搜索栏显示在每个页面的顶部。在第 4 行，我必须为单击创建一个事件侦听器。当用户点击搜索栏时，如果用户不在搜索结果页面上，网站会自动重定向到该页面。我必须创建此代码，因为如果没有此代码，则无法从搜索结果页面之外的其他页面进行搜索，因为不会显示任何结果。

如何允许用户不仅从搜索结果页面而且从其他页面进行搜索？