dousigan0499 2018-06-29 04:21
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通过ajax将输入文件类型传递给php无法正常工作

I have this function to get values from a form

$("button[name='createimport']").click(function() {
var fd = new FormData();
var files = $('#xsfile')[0].files[0];
fd.append('file',files);

var batchid = $("input[name='batchid']").val();
var yrstart = $("input[name='yrstart']").val();
var yrend = $("input[name='yrend']").val();

$.ajax({
    url:"fetch_import.php",
    method:"POST",
    data: { batchid : batchid, yrstart: yrstart, yrend: yrend, fd},
    success: function (data) {
        //show success result div
        if(data)
        {
            showSuccess();
        }
        else
        {
            showFailure();
        }

    },
    error: function () {
        //show failure result div
        showFailure();
        }
}); 
});

and a php code like this:

enter code here$bcid =  $_POST['batchid'];
$yrs =  $_POST['yrstart'];
$yrg =  $_POST['yrend'];

/* Getting file name */
$filename = $_FILES['file']['name'];

/* Location */
$location = "upload/".$filename;
$FileType = pathinfo($location,PATHINFO_EXTENSION);
move_uploaded_file($_FILES['file']['tmp_name'],$location);

passing the file doesn't work. I've searched for this but still not working for me, i think i'll understand how it will work. Any idea? Tyia

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1条回答 默认 最新

  • dpbfb7119 2018-06-29 04:28
    关注

    You should append your fields to fd and simply use that as data-parameter in $.ajax:

    $("button[name='createimport']").click(function() {
        var fd = new FormData();
        var files = $('#xsfile')[0].files[0];
        fd.append('file',files);
    
        fd.append('batchid', $("input[name='batchid']").val());
        fd.append('yrstart', $("input[name='yrstart']").val());
        fd.append('yrend', $("input[name='yrend']").val());
    
        $.ajax({
            url:"fetch_import.php",
            method:"POST",
            data: fd,
            success: function (data) {
                //show success result div
                if(data)
                {
                    showSuccess();
                }
                else
                {
                    showFailure();
                }
    
            },
            error: function () {
                //show failure result div
                showFailure();
                }
        }); 
    });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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