[英]Import Excel file to MySQL using PHP Yii2
I just want to import data from excel (xls, xlsx) to mySql db in Yii2 using PHPExcel. 我只想使用PHPExcel将数据从excel(xls,xlsx)导入到Yii2中的mySql数据库。 This is my code in Controller:
这是我在Controller中的代码:
$modelFile ->file = $firstName. '_' .$middleName. '_' .date('Y-m-d'). '_' .$file ->getBaseName(). "." .$file ->getExtension();
$objPHPExcel = new \PHPExcel();
$inputFiles = fopen("../file/".$modelFile ->file, "r");
try {
$inputFileType = \PHPExcel_IOFactory::identify($inputFiles);
$objReader = \PHPExcel_IOFactory::createReader($inputFileType);
$objPHPExcel = $objReader ->load($inputFiles);
} catch (Exception $ex) {
die('Error');
}
$sheet = $objPHPExcel ->getSheet(0);
$highestRow = $sheet ->getHighestRow();
$highestColumn = $sheet ->getHighestColumn();
//$row is start 2 because first row assigned for heading.
for ($row = 2; $row <= $highestRow; ++$row) {
$rowData = $sheet ->rangeToArray('A'.$row. ':' .$highestColumn.$row, NULL, TRUE, FALSE);
//save to branch table.
$modelHeader = new FakturOut();
$modelDetail = new FakturOutDetail();
$modelHeader ->name = $rowData[0][0];
$modelHeader ->age = $rowData[0][1];
$modelHeader ->address = $rowData[0][2];
$modelHeader ->academic_id = $rowData[0][3];
$modelHeader ->mother_name = $rowData[0][4];
$modelHeader ->father_Name = $rowData[0][5];
$modelHeader ->gender = $rowData[0][6];
$modelHeader ->height = $rowData[0][7];
$modelHeader ->weight= $rowData[0][8];
$modelHeader ->save();
}
And then the browser return an error notification like pathinfo() expects parameter 1 to be string, resource given . 然后浏览器返回错误通知,如pathinfo()期望参数1为字符串,资源已给定 。 Please help to solve this error.
请帮助解决此错误。
in your code you have used 在您使用的代码中
$inputFileType =\PHPExcel_IOFactory::identify($inputFiles);
which is used to identify the valid excel file and it expects the parameter as filename
. 用来识别有效的Excel文件,期望参数为
filename
。
I can see in your code, you have passed $inputFiles
as a parameter to identify
method, which is not a file name but resource handler. 我可以在您的代码中看到,您已将
$inputFiles
作为参数来identify
方法,这不是文件名而是资源处理程序。 and identify
method expects it to be string (file name). 并
identify
方法期望它是字符串(文件名)。
This is the reason you are getting error. 这就是您出错的原因。
Note : fopen() returns a file pointer resource on success, or FALSE on error. 注意:fopen()成功返回文件指针资源,错误返回FALSE。
$inputFiles = fopen("../file/" . $modelFile->file, "r");
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