所以我试图通过#texthere改变链接来获取任何内容,这样我就可以将它们导航到特定哈希的哈希页面。
根据我的代码,我只是不断定义'背部。有人可以看一看,指出我哪里出错了。
$json = array(
'userpost' => array()
);
$row = mysqli_fetch_assoc($check1);
$posts = array();
$posts['num'] = $num;
$posts['streamitem_id'] = $row['streamitem_id'];
$autoembed = new AutoEmbed();
$posts['streamitem_content'] = $autoembed->parse($row['streamitem_content']);
$regex = "/#(\w+)/";
$string=$row['streamitem_content'];
$string = preg_replace($regex, '<a href="hash.php?tag=$1">$1</a>', $string);
$posts['streamitem_content']=json_decode($string);
$posts['streamitem_creator'] = $row['streamitem_creator'];
$posts['streamitem_timestamp'] = $row['streamitem_timestamp'];
$posts['username'] = $row['username'];
$posts['id'] = $row['id'];
$posts['first'] = $row['first'];
$posts['middle'] = $row['middle'];
$posts['last'] = $row['last'];
$json['userpost'][] = $posts;
echo json_encode($json);
答案 0 :(得分:0)
好的,这就是我为解决这个问题所做的工作。不需要解码,这就是导致问题的原因。
$regex = "/#(\w+)/";
$posts['streamitem_content'] = $row['streamitem_content'];
$posts['streamitem_content'] = preg_replace($regex, "
<a href='hash.php?tag=$1'>$1</a>", $posts['streamitem_content'] );