Remove whitespace after +44 and quotation marks around lat/long in JSON string with regex - regex

Question

In the example snippet below I have some JSON which needs to be edited (over 1400 entries). I need to achieve 2 things:
In this example line: "phone": "[phone]", I need to remove the whitespace between +44 and 2079693900 but for all records. Resulting in: "[phone]"
For latitude and longitude I need to get rid of the double quotes around the numbers, as the API I am using only accepts these values as floats.
Example: "latitude": "51.51736", needs to be: "latitude": 51.51736
I am most familiar with Ruby, and have done some parsing of JSON with this in the past, but I thought Regex would be the best tool to use for this kind of basic data cleaning task. I have referred to regex101.com and regular-expressions.info but I'm pretty stuck at this point. Thanks in advance!
[
{
"id": "101756",
"name": "1 Lombard Street
"email": "reception#1lombardstreet.com",
"website": "http://www.1lombardstreet.com",
"location": {
"latitude": "51.5129",
"longitude": "-0.089",
"address": {
"line1": "1 Lombard Street",
"line2": "",
"line3": "",
"postcode": "EC3V 9AA",
"city": "London",
"country": "UK"
}
}
},
{
"id": "105371",
"name": "108 Brasserie",
"phone": "[phone]",
"email": "enquiries#108marylebonelane.com",
"website": "http://www.108brasserie.com",
"location": {
"latitude": "51.51795",
"longitude": "-0.15079",
"address": {
"line1": "108 Marylebone Lane",
"line2": "",
"line3": "",
"postcode": "W1U 2QE",
"city": "London",
"country": "UK"
}
}
},
{
"id": "108701",
"name": "1901 Restaurant",
"phone": "[phone]",
"email": "london.restres#andaz.com",
"website": "http://www.andazdining.com",
"location": {
"latitude": "51.51736",
"longitude": "-0.08123",
"address": {
"line1": "Andaz Hotel",
"line2": "40 Liverpool Street",
"line3": "",
"postcode": "EC2M 7QN",
"city": "London",
"country": "UK"
}
}
},
{
"id": "102190",
"name": "2 Bridge Place",
"phone": "[phone]",
"email": "fb#dtlondonvictoria.com",
"website": "http://crimsonhotels.comdoubletreelondonvictoriadiningpre-theatre-dining",
"location": {
"latitude": "51.49396",
"longitude": "-0.14343",
"address": {
"line1": "2 Bridge Place",
"line2": "Victoria",
"line3": "",
"postcode": "SW1V 1QA",
"city": "London",
"country": "UK"
}
}
},
{
"id": "102063",
"name": "2 Veneti",
"phone": "[phone]",
"email": "2veneti#btconnect.com",
"website": "http://www.2veneti.com",
"location": {
"latitude": "51.5168",
"longitude": "-0.14673",
"address": {
"line1": "10 Wigmore Street",
"line2": "",
"line3": "",
"postcode": "W1U 2RD",
"city": "London",
"country": "UK"
}
}
},

Answer 1

You can use the following regex:
("phone":\s*"\+44)\s+|("(?:latitude|longitude)":\s*)"([^"]+)"
With the following replacement:
$1$2$3
The idea is to capture what we want and not capture what we do not, and then use backreferences to restore the substrings we want to keep.
Regex explanation:
The pattern contains 2 alternatives joined with | alternation operator:
("phone":\s*"\+44)\s+:
("phone":\s*"\+44) - the 1st capturing group matching literal "phone": + optional whitespace, then +44 literally
\s+ - 1 or more whitespaces that we'll remove
("(?:latitude|longitude)":\s*)"([^"]+)":
("(?:latitude|longitude)":\s*) - the second capturing group matching "latitude": or "longitude": and 0 or more whitespace characters
" - Literal " that we'll drop
([^"]+) - the third capturing group matching 1 or more characters other than " (we'll keep that)
" - again, a literal " that we'll drop.
See demo