Question

我正在尝试构建一个应用程序，我必须上传一个zip文件并提取其内容，然后将它们放入mysql数据库。我这样做了一半成功..这是将zip文件上传到名为的临时文件夹上传....现在我想访问每个单独的内容，并将其形成数据库并将真实内容移动到原始上传文件夹中说real_uploads ....我正在尝试，但我无法弄清楚如何继续.... 以下是我的代码........

<?php
if(isset($_FILES['zip'])){
    //defined an error erray
    $errors=array();
    $zip = new ZipArchive();
    //print_r($_FILES['zip']);
    //taking the extention of the last file (in lower case) of the zip file
    if(strtolower(end(explode('.',$_FILES['zip']['name'])))!=='zip'){
        $errors[]="this is not a zip file";
        //echo "file is not opened";
    }
    //if file bigger than 100mb
    if($_FILES['zip']['size']>104857600){
        $errors[]="there is a size limit of 100 mb";
    }
    if($zip->open($_FILES['zip']['tmp_name'])===FALSE){
        echo "file is opened";
        $errors[]="failed to open zip file";
    }
    if(empty($errors)){

        $path=$zip->extractTo('uploads');
        print_r($zip);
        $extracted_files=array();

        for($i=0;$i<$zip->numFiles;++$i){
            $zip_name=$zip->statIndex($i);
            $extracted_files[]=basename($zip_name['name']);

        $zip->close();
    }
}


?>
<html>
    <body>
        <form action="" method="POST" enctype="multipart/form-data">
            <div>
                <input type="file" name="zip"/>
                <input type="submit" value="Upload"/>
            </div>
        </form>
    </body>
</html>
<?php
if(isset($errors)){
    if(empty($errors)){
        echo 'your files '.implode("<br> ",$extracted_files).' been uploaded';

        }else{
            foreach($errors as $error){
                echo '<p>'.$error.'</p>';
            }
        }
}
?>

提取zip文件并访问zip文件的每个内容，然后将其移动到php中的新位置

0 个答案: