The variable "headerKey" cannot be assigned a value of type 'string' because its type is 'never'

Can someone please help me understand why I am encountering an error with this specific line of code:

rowData[headerKey] = value;

The error message I am getting is:

Type 'string' is not assignable to type 'never'.ts(2322) const headerKey: never

Below is the block of code in question:

const exportData = (
    table.getSortedRowModel().rows.length > 0
      ? table.getSortedRowModel().rows // Use sorted data if available
      : table.getCoreRowModel().rows
  ).map((row) => {
    const rowData: Partial<TableDataGeneralTemp> = {};
    row.getVisibleCells().forEach((cell) => {
      if (cell.column.columnDef.header !== "IsSelected") {
        const value = cell.getValue() as string;
        const headerKey = cell.column.columnDef.header as keyof TableDataGeneralTemp; 

        rowData[headerKey] = value;
      }
    });
    return rowData;
  });

Here are the respective interfaces for reference:

interface TableData1 {
  application_date: string;
  application_id: string;
  applicant_name: string;
  institution?: string;
  applicant_email?: string;
  status: string;
  payment_status: string;
}

// Other interface definitions...

interface CommonTableData {
  isSelected: boolean;
}

type TableDataGeneralTemp =
  | TableData1
  | TableData2
  | TableData3
  | DocStatusData
  | InvoiceData
  | ReceiptData;

// More type definitions...

I attempted to resolve the error by including this check:

if (headerKey && headerKey in rowData) {
    rowData[headerKey as keyof TableDataGeneralTemp] = value;
}

Unfortunately, this correction did not successfully address the issue. Previously, the function was error-free until additional code was integrated into the component. Now, I am struggling to troubleshoot and eliminate this type error.

Answer №1

Instead of:

if (headerKey && rowData.hasOwnProperty(headerKey)) {
    rowData[headerKey as keyof TableDataGeneralTemp] = value;
}

Consider using:

(rowData as any)[headerKey] = value;

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