I have a NodeJS app and I have two async blocks that I am calling but the condition is that if the first block returns a value, I don't want to fire the second block. This is how my code looks:
//BLOCK 1
await canFunction1HandleID(id)
.then( () => {
const result = await functionOne(id);
if(!result){
//throw error
}
return result;
})
//BLOCK 2
await canFunction2HandleID(id)
.then( () => {
const result = await functionTwo(id);
if(!result){
//throw error
}
return result;
})
If the first block returns a value, I just want to return that value and not have the second block execute. But right now both blocks are executed, one after another. Any idea why and how can I achieve that I am trying to do here?
TIA.
Just put the first value into a variable?
const result1 = await canFunction1HandleID(id)
.then(() => {
const result = await functionOne(id);
if (!result) {
//throw error
}
return result;
});
if (!result1) {
const result2 = await canFunction2HandleID(id)
.then(() => {
const result = await functionTwo(id);
if (!result) {
//throw error
}
return result;
});
};
If canFunction1HandleID and canFunction2HandleID are functions that return Promise<bool>, what about the following?
//BLOCK 1
if (await canFunction1HandleID(id)) {
const result = await functionOne(id);
if (!result) {
//throw error
}
return result;
}
//BLOCK 2
if (await canFunction2HandleID(id)) {
const result = await functionTwo(id);
if (!result) {
//throw error
}
return result;
}
If the first block returns a value, I just want to return that value and not have the second block execute.
Suppose what you expect is: if functionOne has valid return, use it; otherwise try functionTwo and throw error if no valid return neither.
The code to implement the logic can be simply like:
const result = await functionOne(id).catch(()=>{}) || await functionTwo(id).catch(()=>{});
if (!result) {
// throw error
}