0.2.0 • Published 1 year ago
json-string-values-parser v0.2.0
json-string-values-parser
Recursively traverse JSON properties and call transformation function on the value.
Example
import { jsonParseStringValues } from "json-string-values-parser";
const source = {
a: "1",
b: "false",
c: "string",
d: "null",
e: "undefined",
};
const result = jsonParseStringValues(source);
console.log(result);
/*
{
a: 1,
b: false,
c: "string",
d: null,
e: "undefined",
}
*/
Install
npm install json-string-values-parser
API
Function
import { jsonParseStringValues } from "json-string-values-parser";
jsonParseStringValues(source, options); // options is optional
Class instantiation
import { JsonStringValuesParser } from "json-string-values-parser";
const parser = new JsonStringValuesParser(options); // options is optional
parser.parse(source);
Options
All options are optional.
options.nullValues: string[] | false
String values converted to null
.
- If set to
false
, conversion won't happen. - If omitted,
JSON.parse
will be applied ("null"
will becomenull
). - If passed array, only values from the array will be converted to
null
.
options.falseValues: string[] | false
String values converted to false
.
- If set to
false
, conversion won't happen. - If omitted,
JSON.parse
will be applied ("false"
will becomefalse
). - If passed array, only values from the array will be converted to
false
.
options.trueValues: string[] | false
String values converted to true
.
- If set to
false
, conversion won't happen. - If omitted,
JSON.parse
will be applied("true"
will becometrue
). - If passed array, only values from the array will be converted to
true
.
options.parseNumbers: boolean
default:
true
Try to convert string to number using JSON.parse
method.
options.arrayDelimiter
Array delimiter. If set, string will be splitted into array by given pattern(s). Array values will be parsed again.
jsonParseStringValues("1, 2, 3", { arrayDelimiter: ", " });
// returns [1, 2, 3]
License
Licensed under the MIT license. Copyright (c) 2023-present Vladislav Alexeev