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VisualizaçõesUncaught TypeError despite undefined check
I am getting this error.
I am trying to filter by time of day, and I'm receiving the inputs from two HTML time fields.
export function timeBetweenFilterFn(rows, id, filterValues) {
const startVals = filterValues[0] ? filterValues[0].split(':') : undefined;
const endVals = filterValues[1] ? filterValues[1].split(':') : undefined;
const [startHour, startMinutes] = startVals ?
startVals.map( (val) => {
return parseInt(val);
})
: undefined;
const [endHour, endMinutes] = endVals ?
endVals.map( (val) => {
return parseInt(val);
})
: undefined;
if (startVals || endVals) {
return rows.filter( (row) => {
const [cellHour, cellMinutes] = row.values[id].split(':').map( (val) => {
return parseInt(val);
})
if (cellHour === startHour){
return cellMinutes >= startMinutes;
}
else if(cellHour === endHour){
return cellMinutes <= endMinutes;
}
else{
return cellHour >= startHour && cellHour <= endHour;
}
})
}
return rows;
}
when either the start field or end field is not given, I get the uncaught type error when I try to map the array to the new values despite the ternary check.
1 Respostas
Responde à pergunta0
If you use undefined as the fallback value, you will get an error cause you are trying to get variables [startHour, startMinutes] and [endHour, endMinutes] from undefined, which is not iterable.
You can't try to get variables from undefined.
As you can see in this example made in the browser console, i get the same error, because undefined is not iterable.
Instead of undefined, try using [] as default value in order to avoid errors.
In this case the variables a and b will simply be undefined.