[英]On success reponse from Async Await API call,how to do one moreAPI call with request params
const activeStepOne = async () => {
const formData = new FormData();
formData.append("file", csvData);
formData.append("filetype", fileType);
let postData = {
filetype: fileType,
file: csvData,
};
let getData = {
filetype: fileType,
file: csvData.name,
};
sessionStorage.setItem("uploadDetails", JSON.stringify(getData));
const requestOptions = {
method: "POST",
body: formData,
};
let responseData = await callAPI(`${API_URL}/fileupload`, requestOptions);
if (responseData?.status) {
setDisable(true);
setActiveStep((prevActiveStep) => prevActiveStep + 1);
setIsLoading(false);
setShowSnackbar(false);
setSnackbarType("success");
setSnackbarMessage("File uploaded successfully!");
setShowSnackbar(true);
} else {
setStatusCode(true);
setIsLoading(false);
setSnackbarMessage(
"Error uploading file. Please try again with proper data."
);
setSnackbarType("error");
setShowSnackbar(true);
}
console.log(responseData);
};
我在這里做 API 調用,一旦我得到成功響應,我想再做一個 API 調用,請求參數如下
{“上傳”:“True”,“名稱”:“Bolzen”,“工作流程名稱”:“file1.csv”,“UserId”:1}
不確定,如果我了解您所面臨的挑戰。 您可以像這樣發起另一個請求:
let responseData = await callAPI(`${API_URL}/fileupload`, requestOptions);
if (responseData?.status) {
let responseData = await callAPI(`${API_URL}/endpoint`, { method: "POST", body: {"Upload":"True", "Name":"Bolzen", "WorkflowName":"file1.csv", "UserId":1} };
}
我試圖推斷callAPI
的用法。
不過,如果值不會改變,我建議使用const
而不是let
。
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