the requirement is:
ajax
I can"t write well no matter how I write it. Look at the Internet, most of them add an id, to the form tag and then new FormData ($("- sharpid") [0]), and then submit it at once, so my three fields are useless, aren"t they?
ask the elder brother to teach me.
let formdata = new FormData();
formdata.append("img1", $('-sharpimg1')[0]);
formdata.append("img2", $('-sharpimg2')[0]);
formdata.append("img3", $('-sharpimg3')[0]);
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