如何处理javascript执行器的脚本超时以获取调用

Question

我正在使用selenium java script exceuter执行下面的javascript，我想从fetch调用返回响应，并想将其存储在代码中的java变量中。但下面的代码显示了脚本超时，有什么建议吗？我怎样才能达到以上要求？？？

 String location = "!async function(){\n" +
                "let data = await fetch(\"https://raw.githubusercontent.com/IbrahimTanyalcin/LEXICON/master/lexiconLogo.png\")\n" +
                "    .then((response) => response.blob())\n" +
                "    .then(data => {\n" +
                "        return data;\n" +
                "    })\n" +
                "    .catch(error => {\n" +
                "        console.error(error);\n" +
                "    });\n" +
                "\n" +
                "console.log(data);\n" +
                "return data;\n" +
                "}();\n";


        Object str = js.executeAsyncScript(location);

Answer 1

使用setTimeout（）

function resolveAfter2Seconds() {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve('resolved');
    }, 2000);
  });
}

async function asyncCall() {
  console.log('calling');
  const result = await resolveAfter2Seconds();
  console.log(result);
  // expected output: "resolved"
}

异步函数