[英]Do not execute function until another function's ajax call finishes
我已經嘗試搜索了一段時間,但沒有找到解決這種情況的方法。 Say I have a function that executes an ajax call, but the result of the ajax is not returned, and I cannot modify this function, I can only call it. I need to call this function and then execute another function once the ajax call in the former function is complete.
這是我嘗試的眾多解決方案之一的示例:
function doAjaxCall() {
$.ajax({
url: "https://api.publicapis.org/entries"
}).done(function(response) {
console.log("I want this message to log first");
});
}
function anotherCall() {
console.log("and this message to log second");
}
promise = new Promise(() => {
doAjaxCall();
});
$.when(
promise,
).then(
(r1) => { anotherCall(); }
);
這不起作用; 在 ajax 調用完成之前, anotherCall()
仍在執行。 同樣,在此示例中,我無法修改doAjaxCall()
的內容。
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