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怎么通过ajax一起传递formData和另一个对象数据
data:formData+"&type="+$("#1").val(), 这样不行吧?
ringa_lee2017-05-16 13:14:31
var formData = new FormData();
formData.append('xxx', xxxxxx);
formData.append('type', $("#1").val());
...
$.ajax({
url: '/test',
data: formData,
processData: false,
contentType: false,
type: 'POST',
success: function(data){
alert(data);
}
});