0

我正在尝试使用过滤器,以查看文件是否存在,以及有多少:我正在尝试使用以下方式获取脚本:

$list = glob($filepath.$filenamePartial'*'.$ext);
var_dump($list);

解决了1:

$list   = glob($filepath.$FsisName.'_'.$count.'_'.'*'.$ext);
if (count($list) <= 0){
        $FFileUPL   = $filepath.$FsisName.'_'.$count.'_'.$FileDate.'.'.$ext;
        $event      =   move_uploaded_file($flTName, $FFileUPL);
    }else{
        echo 'this file Exist: '.$FsisName.'-'.$count.'-'.$FileDate.'.'.$ext.'<br>';
        $event  =   false;
    }

如何检索迭代器中的名称文件? 解决了2:

foreach($list as $fnames){
    $eFile['smg']       .= basename($fnames).'<br>';
}
4

1 回答 1

0

解决了:

$list   = glob($filepath.$FsisName.'_'.$count.'_'.'*'.$ext);
if (count($list) <= 0){
        $FFileUPL   = $filepath.$FsisName.'_'.$count.'_'.$FileDate.'.'.$ext;
        $event      =   move_uploaded_file($flTName, $FFileUPL);
    }else{
        echo 'this file Exist: '.$FsisName.'-'.$count.'-'.$FileDate.'.'.$ext.'<br>';
        $event  =   false;
    }
foreach($list as $fnames){
    $eFile['smg']       .= basename($fnames).'<br>';
}
于 2016-05-18T23:18:07.773 回答