在JSON對象上使用jQuery的find（）與brnwdrng的問題類似，我正在尋找一種搜索類似JSON的對象的方法。假設我的對象的結構是這樣的：TestObj = {
"Categories": [{
"Products": [{
"id": "a01",
"name": "Pine",
"description": "Short description of pine."
},
{
"id": "a02",
"name": "Birch",
"description": "Short description of birch."
},
{
"id": "a03",
"name": "Poplar",
"description": "Short description of poplar."
}],
"id": "A",
"title": "Cheap",
"description": "Short description of category A."
},
{
"Product": [{
"id": "b01",
"name": "Maple",
"description": "Short description of maple."
},
{
"id": "b02",
"name": "Oak",
"description": "Short description of oak."
},
{
"id": "b03",
"name": "Bamboo",
"description": "Short description of bamboo."
}],
"id": "B",
"title": "Moderate",
"description": "Short description of category B."
}]};我想得到一個id =“A”的對象。我嘗試過各種各樣的東西，比如：$(TestObj.find(":id='A'"))但似乎沒什么用。任何人都可以想到一種基于某些標準檢索項目而不使用“每個”的方法嗎？
3 回答
拉莫斯之舞
TA貢獻1820條經驗 獲得超10個贊
這適用于[{“id”：“data”}，{“id”：“data”}]
function getObjects(obj, key, val) {
var newObj = false;
$.each(obj, function()
{
var testObject = this;
$.each(testObject, function(k,v)
{
//alert(k);
if(val == v && k == key)
{
newObj = testObject;
}
});
});
return newObj;}- 3 回答
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