i m 将该代码上载到一个网址:
var qreq = ".....myurl.php";
$.post(qreq, function(data){alert(data);});
在我的PHP档案中,有:
......
$prevtopic = $row["topic_id"];
echo $prevtopic;
警示已经发出,但只是空白。 由于某种原因,数据是空洞的,尽管有一米的对应数据来自购买力平价的档案。 i 还试图仅仅重复“hi”,但那就没有工作了。 什么是错了?
感谢。
您不能像这样提出交叉要求,而封杀你们的是。
您可为此使用JSONP,但不能像你那样获得正常的html/文本内容。 下面是一份简短的书面陈述,其中应更好地介绍关于JSONP或,但。 你们需要从PHP返回JSONP。
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