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我嘗試編寫一個函數,返回一個jQuery圖像滑塊的圖像數組。PHP讀取目錄和返回數組與相關圖像鏈接的路徑
現在我在獲得正確的邏輯代碼方面陷入困境。我在想整個結構是錯的嗎? 我可以得到1個文件的正確路徑,但相關文件的數組似乎不起作用。
##info past to the function:
$id = '11101';
$dir = "myimgdir/";
$thumbs = 'TN';
$medium = 'M';
##the files in the dir are named like this:
11101x1xTN //thumbnail
11101x2xTN
11101x3xTN
11101x1xM //some img but in medium size
11101x2xM
11101x3xM
所以:
$dir = "$dir$id/";
$get_opts = array($thumbs, $medium);
// array to hold return value
$retval = array();
//full dir
$fulldir = "{$_SERVER['DOCUMENT_ROOT']}/$dir";
$d = @dir($fulldir) or die("getImages: Failed opening directory $dir for reading");
while(false !== ($entry = $d->read())) {
$fullpath = escapeshellarg("$fulldir$entry");
$file_extension = end(explode('.', $fullpath));
$file_name = basename($entry, $file_extension);
####here I get stuck
foreach ($get_opts as $get_opt) {
if (strpos($file_name, $get_opt)) { //so 11101x1xTN gontains TN
$retval[] = array(
'TN' => "/$dir$entry",
'TNsize' => getimagesize("$fulldir$entry"),
####how to return the medium path as well (11101x1xM)
'M' => "/$dir",
);
}
}
}
$d->close();
return $retval;
爲什麼'$ fullpath = escapeshellarg(「$ fulldir $ entry」);'?只是好奇 –