I am executing the below javascript using selenium java script exceuter, I want to return the response fromt the fetch call and want to store it in java variable in code. But below code is showing script time out, any suggestions how can I achive above requirements ???
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);
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Answer
I’m not how to escape the quotes but the javascript should be:
let url = "https://..." fetch(url).then(r => r.blob()).then(arguments[0])
arguments[0] is the callback, it needs to call that within 30 seconds or the timeout happens