如何处理 javascript 执行程序的脚本超时以获取调用

Question

我正在使用 selenium java 脚本执行器执行下面的 java 脚本，我想 return 获取调用的响应并将其存储在 java 代码中的变量。但是下面的代码显示脚本超时，有什么建议可以满足上述要求吗？？？

 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"
}

async function

How to useexecuteAsyncScriptmethodinorg.openqa.selenium.JavascriptExecutor

Answer 2

我不知道如何转义引号，但 javascript 应该是：

let url = "https://..."
fetch(url).then(r => r.blob()).then(arguments[0])

arguments[0]是回调，需要在30秒内调用否则超时

如何处理 javascript 执行程序的脚本超时以获取调用

how to handle script time out for javascript executer for fetch call

javascript

java

selenium

selenium-webdriver

fetch-api