如何使用php实现获取excel中的图片

要实现读取excel中的图片，可以通过phpspreadsheet来实现。phpspreadsheet是一个纯php编写的库，并引入了命名空间、PSR规范等。

使用composer安装phpspreadsheet

composer require phpoffice/phpspreadsheet

GitHub下载：

https://github.com/PHPOffice/PhpSpreadsheet

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excel图片如下图：

项目实例：

use PhpOffice\PhpSpreadsheet\Cell\Coordinate;use PhpOffice\PhpSpreadsheet\IOFactory;$imageFilePath = './uploads/imgs/'; //图片本地存储的路径if (!file_exists($imageFilePath)) { //如果目录不存在则递归创建 mkdir($imageFilePath, 0777, true);}try { $inputFileName = './files/1.xlsx'; //包含图片的Excel文件 $objRead = IOFactory::createReader('Xlsx'); $objSpreadsheet = $objRead->load($inputFileName); $objWorksheet = $objSpreadsheet->getSheet(0); $data = $objWorksheet->toArray(); foreach ($objWorksheet->getDrawingCollection() as $drawing) {  list($startColumn, $startRow) = Coordinate::coordinateFromString($drawing->getCoordinates());  $imageFileName = $drawing->getCoordinates() . mt_rand(1000, 9999);  switch ($drawing->getExtension()) {   case 'jpg':   case 'jpeg':    $imageFileName .= '.jpg';    $source = imagecreatefromjpeg($drawing->getPath());    imagejpeg($source, $imageFilePath . $imageFileName);    break;   case 'gif':    $imageFileName .= '.gif';    $source = imagecreatefromgif($drawing->getPath());    imagegif($source, $imageFilePath . $imageFileName);    break;   case 'png':    $imageFileName .= '.png';    $source = imagecreatefrompng($drawing->getPath());    imagepng($source, $imageFilePath, $imageFileName);    break;  }  $startColumn = ABC2decimal($startColumn);  $data[$startRow-1][$startColumn] = $imageFilePath . $imageFileName; } dump($data);die();} catch (\Exception $e) { throw $e;}public function ABC2decimal($abc){ $ten = 0; $len = strlen($abc); for($i=1;$i<=$len;$i++){  $char = substr($abc,0-$i,1);//反向获取单个字符  $int = ord($char);  $ten += ($int-65)*pow(26,$i-1); } return $ten;}

结果如图：

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