如何在json回调中删除“结果”：“成功”

Question

The result of the call to the api server is a json file, which begins with this string:

{
    "result": "success"
    ,   "data": {"total":16080,"pageCount":161,"result":[{"packWidth":250,"itemNo"

How do I remove the part that I do not care? that is, this

{
        "result": "success"
        ,   "data": {"total":16080,"pageCount":161,"result":

The complete result is:

{
    "result": "success"
    ,   "data": {"total":16080,"pageCount":161,"result":    [{"packWidth":250,"itemNo":"1203945","groupItemNo":"1203945","status":1,"categoryId":105096,"packType":"Color Box","barcode":"6922833439687","modelLabel":"Color","packQty":24,"packInclude":"USB Cable, User Manual, USB Charger, Earphone, 1pcs Li-Battery, LCD Protector, Leather Case, Plastic Case","clearance":false,"id":103928,"packWeight":"12.500","price":"181.2800","packLength":400,"description":"description test","unitWeight":"0.726","packHeight":300}]}}

I use the PHP language

I have to remove the initial part:

{
    "result": "success"
    ,   "data": {"total":16080,"pageCount":161,"result":

and the final:

}}

Answer 1

If you want to use part of a JSON to populate a CSV file, then parse the json using json_decode method and access the necessary information.

Try something like this:

var jsonObject = json_decode(myJson);
var interestingPart = jsonObject.data.result;

You can now access the data in an Object manner. Or if you want to get a json back from it, then use:

var interestingJson = json_encode(interestingPart);

Answer 2

Not tested, but it should work

preg_replace("/^.*?:(?=\w*\[) | (\w*}\w*}\w*$)/", "", $str); 

Edit: i wrote this before I knew of json_decode, you should really use the json functions like suggested in fazovskys answer.