如何在PHP中导出mysql表数据 [英] How to export mysql table data in PHP

问题描述

这是我导出到excel的代码，但它显示数据列表而不是以excel导出

this is my code to export to excel, but it display the data list and not to export in excel

function ExportToExcel($requests){		
	//print_r($requests);//exit;
		extract($requests);		
		 
		$query = "SELECT * FROM ".$tname." P WHERE 1";
		$cond_search = '';
		$Querycond = '';
		if(isset($dwnld) && ($dwnld == "dwnld_all")){
			$Querycond    .= " AND  1 ";
		}
		
		if(isset($dwnld) && ($dwnld == "new_submission")){
			$Querycond    .= " AND apply_on >= CURDATE()";
		}
		
		if(isset($dwnld) && ($dwnld == "all_submission")){
			$Querycond    .= " AND  3";
		}

		$query .=  $Querycond;
	
		$query1 = $query . " ORDER BY P.apply_on desc";
		
		$setRec = mysql_query($query1);
		  
		$developer_records = array();
		while( $rows = mysql_fetch_assoc($setRec) ) {
		$developer_records[] = $rows;
		} 	
		$filename = "export_".date('Ymd') . ".xls";
		header("Content-type: application/octet-stream");
		header("Content-Type: application/vnd.ms-excel; name='excel'");
		header("Content-Disposition: attachment; filename='".$filename."'");
		header('Content-Type: application/vnd.openxmlformats-officedocument.spreadsheetml.sheet');
		header('Cache-Control: max-age=0');

		$show_coloumn = false;
		if(!empty($developer_records)) {
		foreach($developer_records as $record) {
		if(!$show_coloumn) {
		// display field/column names in first row
		echo implode("\t", array_keys($record)) . "\n";
		$show_coloumn = true;
		}
		echo implode("\t", array_values($record)) . "\n";
		}
		}
		
		exit;		
		return true;
	}

我的尝试：



请帮帮我，如何解决它，我已经在ajax调用php文件然后将php文件转到另一个php文件

What I have tried:

please help me, how to solve it,i have did in ajax call to php file then that php file to another php file

推荐答案

请求）{
// print_r（

requests){ //print_r(

requests）; //退出;
extract（

requests);//exit; extract(

requests）;

requests);

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