为jQuery post[复制]构造post数据

Question

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$.post("addNewItem.php", {itemid: '1', itemName: 'Foo'}, function(data, status) {
    }
);

但这并不是:

var postData = "";
var postInputs = document.querySelectorAll("[id^='itemInput']");
for (var i = 0; i < postInputs .length; i++) {
    var itemName = postInputs [i].name;
    var itemValue = postInputs [i].value ;
    postData += itemName + ": '" + itemValue  + "', ";
 }
 postData += "upload: 'Y'";

$.post("addNewItem.php", {
    postData
}, function(data, status) {
    }
);

null

null

Answer 1

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var postData = {
  itemid: '1',
  itemName: 'Foo'
};

$.post("addNewItem.php", postData, function(data, status) {

});