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我有以下的JSON字符串。嵌套的JSON对象长度
{
"data": [
{
"city_id": "1",
"city_name_eng": "Multan",
"0": {
"category_id": "1",
"city_id": "1",
"category_name_eng": "Mango",
"0": {
"product_id": "1",
"category_id": "1",
"product_name_eng": "Mango1"
},
"1": {
"product_id": "2",
"category_id": "1",
"product_name_eng": "Mango2"
},
"2": {
"product_id": "3",
"category_id": "1",
"product_name_eng": "Mango3"
},
"3": {
"product_id": "4",
"category_id": "1",
"product_name_eng": "Mango1"
}
},
"1": {
"category_id": "2",
"city_id": "1",
"category_name_eng": "Shoes"
},
"2": {
"category_id": "3",
"city_id": "1",
"category_name_eng": "Bank"
}
},
{
"city_id": "2",
"city_name_eng": "Lahore",
"3": {
"category_id": "4",
"city_id": "2",
"category_name_eng": "Food"
},
"4": {
"category_id": "5",
"city_id": "2",
"category_name_eng": "Computer"
},
"5": {
"category_id": "6",
"city_id": "2",
"category_name_eng": "Mobile"
}
}
]
}
我想拥有JSONObject的长度。基本上,具有"city_id": "1"
的第一个对象具有4个嵌套的JSONObjects和2个值,即city_id和city_name_eng。 当我使用objJSON.length()
时,它给了我6的长度。但相反,我想拥有4个嵌套JSONObjects,我如何在运行时获得它? (无论嵌套的jsonobjects是3或4或任何数字)
如何区分嵌套的JSONObjects和键值?
为什么不能使用JSONArray而不是将JSONObjects命名为0,1,2等?这将使它更方便处理.. –
如何做到这一点?我有一个来自PHP网站的JSON响应。 在那里我用'json_encode(array('data'=> $ data));'' – Zeeshan