Remove Whitespace Inside Json Object Key Value Pair

remove whitespace inside json object key value pair

Assign the trimmed value to the specific user[key]:

var user = { first_name: "CSS", last_name: " H ", age: 41, website: "java2s.com" };
for (var key in user) {    user[key] = user[key].toString().trim()    console.log(key+"-->"+user[key]);}

Trim white spaces in both Object key and value recursively

You can clean up the property names and attributes using Object.keys to get an array of the keys, then Array.prototype.reduce to iterate over the keys and create a new object with trimmed keys and values. The function needs to be recursive so that it also trims nested Objects and Arrays.

Note that it only deals with plain Arrays and Objects, if you want to deal with other types of object, the call to reduce needs to be more sophisticated to determine the type of object (e.g. a suitably clever version of new obj.constructor()).

function trimObj(obj) {
  if (!Array.isArray(obj) && typeof obj != 'object') return obj;
  return Object.keys(obj).reduce(function(acc, key) {
    acc[key.trim()] = typeof obj[key] == 'string'? obj[key].trim() : trimObj(obj[key]);
    return acc;
  }, Array.isArray(obj)? []:{});
}

Regex to remove spaces from Key but not Value

You can probably try the following expression as a quick fix granted the JSON you need to fix is in the format you have shown (it might fail to work with other JSONs):

\s(?=[^"]*":\s*")

See the regex demo

Details

\s - a whitespace
(?=[^"]*":\s*") - a positive lookahead that, immediately to the right of the current position, requires
- [^"]* - zero or more chars other than "
- ": - ": string
- \s* - 0+ whitespaces
- " - a double quote.

Remove spaces in Json keys

You can use JSON.stringify(), JSON.parse(), String.prototype.replace() with RegExp /\s(?=\w+":)/g to match space character followed by one or more word characters followed by " followed by :

var arr = [{    "Id": "ALFKI",    "Contact Name": "Maria Anders",    "Contact Title": "Sales Representative",    "City": "Berlin",    "Slider": 10}, {    "Id": "ANATR",    "Contact Name": "Ana Trujillo",    "Contact Title": "Owner",    "City": "México D.F.",    "Slider": 5}];
arr = JSON.parse(JSON.stringify(arr).replace(/\s(?=\w+":)/g, ""));
console.log(arr);

Remove leading and trailing whitespaces in JSON keys

In addition to @BillKarwin's suggestion to trim whitespace before you enter it into the database, you can also update all the values in the database to remove the spurious whitespace:

UPDATE product_json_table
SET description = REGEXP_REPLACE(description, '\\s|\\r|\\n','');

Then your original query will work:

SELECT id, product, description 
FROM product_json_table
WHERE JSON_EXTRACT(description, '$.wheels') > 2;

Output:

id  product         description
1   truck_space     {"wheels":4,"seats":3,"fuel":"diesel","mileage":8}

Demo on dbfiddle

Update

You can also perform the whitespace replacement on the fly although this will be much less efficient than permanently removing it with the UPDATE query above:

SELECT id, product, REGEXP_REPLACE(description, '\\s|\\r|\\n','') AS description 
FROM product_json_table
WHERE JSON_EXTRACT(REGEXP_REPLACE(description, '\\s|\\r|\\n',''), '$.wheels') > 2

Output:

id  product         description
1   truck_space     {"wheels":4,"seats":3,"fuel":"diesel","mileage":8}

Demo on dbfiddle

Python - remove spaces between key and values from dictionary object

All you need to care about is the length of a string, so use len() to compare results.

If you care about generated JSON sizes, you can trivially tell json.dumps() to not use any spaces:

string_data = json.dumps(data, separators=(',', ':'))

Demo:

>>> import json
>>> sample = '''{"metadata":{"info":"important info"},"timestamp":"2018-04-06T12:19:38.611Z","content":{"id":"1","name":"name test","objects":[{"id":"1","url":"http://example.com","properties":[{"id":"1","value":"1"}]}]}}'''
>>> data = json.loads(sample)
>>> len(sample)
205
>>> len(json.dumps(data))
224
>>> len(json.dumps(data, separators=(',', ':')))
205

With separators specified, the output length matches the original input. The separators option is documented as:

If specified, separators should be an (item_separator, key_separator) tuple. The default is (', ', ': ') if indent is None and (',', ': ') otherwise. To get the most compact JSON representation, you should specify (',', ':') to eliminate whitespace.

(Bold emphasis mine).

sys.getsizeof() is the wrong tool to make comparisons here. The function gives you the memory footprint of an object (without recursing), and this memory footprint is a implementation detail of your current Python implementation and operating system specific data type sizes.

The dict object uses 240 bytes on your OS, but this is a over-allocated hash table of references. On a different OS with different size pointers, the memory size would be different, and either way, you didn't include the size of the referenced string objects. For string objects, the memory footprint depends heavily on the highest Unicode codepoint; the string 'a' has a different footprint from '', even both have length 1, because the latter contains a non-BMP codepoint so 4 bytes are needed to store just that one character, plus Python object overhead.

Next, all you saved was 19 spaces. Compression would have saved you much more, and spaces don't matter much when compressing:

>>> import zlib
>>> len(zlib.compress(json.dumps(data).encode('utf8'), 9))
155
>>> len(zlib.compress(json.dumps(data, separators=(',', ':')).encode('utf8'), 9))
154

Compression saved 50 bytes there, while using the compact JSON separators saved another single byte.

Remove Whitespace Inside Json Object Key Value Pair