我应该如何在csv导出中用“-”替换任何空值 - PHP

如何解决我应该如何在csv导出中用“-”替换任何空值 - PHP

public function exportCsv()
    {
        $this->download_send_headers("subscriber_list_export_" . date("Y-m-d") . ".csv");
        $model = $this->getModel('clientsdatas');
        $array = $model->getCsvArray();
        echo $this->array2csv($array);  
        die();
    }

    function array2csv(array &$array)
    {
        if (count($array) == 0) {
            return null;
        }
        ob_start();
        $df = fopen("PHP://output",'w');
        fputcsv($df,array_keys(reset($array)));
        foreach ($array as $row) {
            $row['id'] = '="' . $row['id'] . '"';
            fputcsv($df,$row);
        }

        fclose($df);
        return ob_get_clean();
    }

    function download_send_headers($filename) {
        // disable caching
        $Now = gmdate("D,d M Y H:i:s");
        header("Expires: Tue,03 Jul 2001 06:00:00 GMT");
        header("Cache-Control: max-age=0,no-cache,must-revalidate,proxy-revalidate");
        header("Last-Modified: {$Now} GMT");
    
        // force download  
        header("Content-Type: application/force-download");
        header("Content-Type: application/octet-stream");
        header("Content-Type: application/download");
    
        // disposition / encoding on response body
        header("Content-disposition: attachment;filename={$filename}");
        header("Content-transfer-encoding: binary");
    }
}

解决方法

如果您认为问题出在 PHP 或 CSV 上，那么您可能没有注意数据源，因为可以根据需要查询数据，而无需操作数组：

CASE WHEN column IS NULL THEN "-" ELSE column END AS column