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我想从递归AJAX获取文件名,但到现在为止它不工作,在我的代码是这样AJAX,如何返回文件名?
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>GET NAME</title>
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<script src="/assets/mediaelements/build/jquery.js"></script>
<script src="/assets/mediaelements/build/mediaelement-and-player.min.js"></script>
<link href="/assets/mediaelements/build/mediaelementplayer.min.css" rel="stylesheet" />
<script type="text/javascript" src="jquery.min.js"></script>
<script>
(function rekurse(){
setTimeout(function()
{
/* ---------------------------------- */
$.ajax({
type: "POST",
cache: false,
url: 'shift.php',
data: {offi: 'E:/DataText/OFFICE/BI/FAR1'},
success: function(data){
alert(data);
rekurse();
},
error: function(){
alert(data);
rekurse(); // recurse, if you'd like.
}
});
/* ---------------------------------- */
}, 1000);
})();
</script>
</head>
<body onload="rekurse();return false;">
</body>
<html><body
和shift.php,它就像
<?PHP
//
CLEARSTATCACHE();
//
$grps = '_';
$offi = $_POST['offi'];
$temp = $offi.'/'.$grps.'*.*';
$arrs = GLOB($temp );
$coun = COUNT($arrs);
//
IF($coun<1):
ECHO json_encode("ok");
ENDIF;
//
$text = $offi."/call.htm";
$hand = FOPEN($text, 'w');
$text = FWRITE($hand,"\r\n");
$hand = FCLOSE($hand);
//
$file = $arrs[0];
$hand = FOPEN($file, 'r');
$temp = FREAD($hand, FILESIZE($file));
$hand = FCLOSE($hand);
//
$arrs = EXPLODE(',',$temp);
IF(COUNT($arrs)>0):
ECHO json_encode($temp);
ELSE:
ECHO json_encode("ok");
ENDIF;
//
?>
什么IM错过了这个和抱歉,我的英语
谢谢
方面 ·班邦·