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jQuery.ajaxFileUpload({
            url: '<?php echo base_url(); ?>profile_pic/prof_photo_upload',
            secureuri: true,
            fileElementId: 'photoimg',
            dataType: 'html',
            success: function(resp){
                $('.right-register-container-left-rows-fields').html(resp);
            }
        });
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1 回答 1

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您正在将一个对象传递给 ajaxFileUpload,要添加更多参数,只需向该对象添加一个新属性,例如:

jQuery.ajaxFileUpload({
    url: '<?php echo base_url(); ?>profile_pic/prof_photo_upload',
    type: "POST",
    secureuri: true,
    fileElementId: 'photoimg',
    dataType: 'html',
    data: {
        somedata: "XXX",
        moredata: "YYY"
    },
    success: function(resp) {
        $('.right-register-container-left-rows-fields').html(resp);
    }
});
于 2013-09-08T08:41:01.713 回答