As a newbie, I'm looking for the best approach to achieve the below:
Here is the Array I get from my DB query that contains a left join on the "class" table
[
{"legnumber":1,
"classcode" : "J"},
{"legnumber":1,
"classcode" : "Y"},
{"legnumber":2,
"classcode" : "J"}
]
And I would like to get something like this:
{
"legs": [
{
"legnumber" : 1,
"classes" : [
{"classcode" : "J"},
{"classcode" : "Y"}
]
},
{
"legnumber" : 2,
"classes" : [
{"classcode" : "J"}
]
}
]
}
Thanks a lot for your suggestions. I'm using Sequelize in this project but I'm writing raw queries as I find it more convenient for my DB model.
Regards, Nico
You can group your array items based on legnumber using array#reduce and then get all the values to create your result using Object.values().
const arr = [ {"legnumber":1, "classcode" : "J"}, {"legnumber":1, "classcode" : "Y"}, {"legnumber":2, "classcode" : "J"} ],
output = arr.reduce((r, {legnumber, classcode}) => {
r[legnumber] ??= {legnumber, classes: []};
r[legnumber].classes.push({classcode});
return r;
},{}),
result = {legs: Object.values(output)};
console.log(result);Hassan's answer is the more concise way to handle this, but here is a more verbose option to help understand what's happening:
const queryResults = [
{ legnumber: 1, classcode: 'J' },
{ legnumber: 1, classcode: 'Y' },
{ legnumber: 2, classcode: 'J' },
]
// create an object to store the transformed results
const transformedResults = {
legs: [],
}
// loop through each item in the queryResult array
for (const result of queryResults) {
// try to find an existing leg tha matches the current leg number
let leg = transformedResults.legs.find((leg) => leg.legnumber === result.legnumber)
// if it doesn't exist then create it and add it to the transformed results
if (!leg) {
leg = {
legnumber: result.legnumber,
classes: [],
}
transformedResults.legs.push(leg)
}
// push the classcode
leg.classes.push({ classcode: result.classcode })
}
console.log(transformedResults)