PHPSpreadsheet Excel Report Generation Error

Question

I am generating excel file in PHP using PHPSpreadsheet library. When I directly call the url 'http://10.150.18.22/ppc/dicreport.php?report_date=2022-10-07&report_type=Daily_Report' excel file gets generated successfully. But when same url is called through AJAX call, following output is generated on browser console.

Ajax Call:

<script>    
  $("#generateReportForm").on("submit", function (e) {
    e.preventDefault();
    var report_date=$('#report_date').val();
    var report_type=$('#report_type').val();
    console.log(report_date);
    console.log(report_type);
    $.ajax({
      url: "dicreport.php?report_date="+report_date+"&report_type="+report_type,
      type: "GET",
      data: $("#generateReportForm").serialize(),
      success: function (response) { 
           console.log(response);          
      },
    });
  });
</script>

dicreport.php

<?php
  $REPORT_DATE=$_GET['report_date'];
  $REPORT_TYPE=$_GET['report_type'];
  
  require 'vendor/autoload.php';

  use PhpOffice\PhpSpreadsheet\Spreadsheet;
  use PhpOffice\PhpSpreadsheet\IOFactory;

  $spreadsheet = new Spreadsheet();
  $sheet = $spreadsheet->getActiveSheet();


  $spreadsheet->getDefaultStyle()
            ->getFont()
            ->setName('Arial')
            ->setSize(10);
            

  $file=$REPORT_TYPE."_".$REPORT_DATE.".xlsx";

  header('Content-disposition: attachment; filename='.$file);
  header('Content-type: application/vnd.openxmlformats-officedocument.spreadsheetml.sheet');

  $writer = IOFactory::createWriter($spreadsheet,'Xlsx');
  $writer->save('php://output');
?>

Headers have been sent to the browser window before the ajax's return of header(s) — Ken Lee, Oct 08 '22 at 06:29
See whether this [SO Post](https://stackoverflow.com/questions/27701981/phpexcel-download-using-ajax-call) helps — Ken Lee, Oct 08 '22 at 06:47
The same issue is solved for PHPExcel. Check this: https://stackoverflow.com/questions/27701981/phpexcel-download-using-ajax-call . — D. Dimopoulos, Oct 08 '22 at 07:20

score 0 · Answer 1 · answered Mar 04 '23 at 03:30

Try this.

check your AJAX code and output $REPORT_DATE=$_GET['report_date']; $REPORT_TYPE=$_GET['report_type'];

to ensure they are both displaying their value as expected. do this by printing its value via your PHP script.

Make sure you are not outputting any value in the spreadsheet code such as echo .... or print ...

PHPSpreadsheet Excel Report Generation Error

1 Answers1