跳轉到推特鏈接頁面上的位置

我有一個頁面，我已經編程了聖經的順序訪問，每個詩句都包含一個鳴叫按鈕。跳轉到推特鏈接頁面上的位置

當我推送詩句時，鏈接重定向到詩句頁面，但從頂部開始。

我想要的是，當tweet鏈接被點擊時，它會進入經文頁面，然後跳轉到該頁面上的經文位置。

這可能嗎？你能幫我嗎？

這是我當前的代碼：

<html> 
<title>The Bible (Sequential)</title> 
<font face="arial"> 
<?php 
define ('HOSTNAME', 'host'); 
define ('USERNAME', 'user'); 
define ('PASSWORD', 'pass'); 
define ('DATABASE_NAME', 'db'); 

$db = mysql_connect(HOSTNAME, USERNAME, PASSWORD) or die ('I cannot connect to 
MySQL.'); 

mysql_select_db(DATABASE_NAME); 

$query = "SELECT id,book,cap,verse,line FROM bible_kjv ORDER BY id LIMIT 0, 30 "; 

$result = mysql_query($query); 
while ($data_fetch=mysql_fetch_array($result)) { 

echo "<p>", ($data_fetch['book']), " ", ($data_fetch['cap']), " ",  
($data_fetch['verse'])," </p>"; 
echo "<p>", ($data_fetch['line']), "</p>"; 

echo "<a href='twitter.com/share' class='twitter-share-button' data-url='bible-query- 
seq-1.php?id=$data_fetch[id]' 
data-text='The Bible (Sequential): http://mysite/bible-query-seq/bible-query-seq-1.php? 
id=$data_fetch[id]'>Tweet</a>"; 

echo "<hr>"; 

} 

mysql_free_result($result); 
mysql_close(); 
?> 

<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s) 
[0],p=/^http:/.test(d.location)?'http':'https';if(!d.getElementById(id)) 
{js=d.createElement 

(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore 
(js,fjs);}}(document, 'script', 'twitter-wjs');</script> 

<center><a href="bible-query-seq-2.php">Page 2</a></center> 
</font> 
</html>

來源

2013-07-07 adeoba

您應該標記頁面的各個部分，在每一節中都包含一個''。在生成鏈接時將#verseNUM添加到網址，請查看http://www.w3.org/TR/html401/struct/links.html#h-12.2.1 – Orestes

謝謝奧雷斯特斯和帕維爾，這是我做的;我改變了以下內容：

echo "<p>", ($data_fetch['book']), " ", ($data_fetch['cap']), " ",  
($data_fetch['verse'])," </p>"; 
echo "<p>", ($data_fetch['line']), "</p>"; 

echo "<a href='twitter.com/share' class='twitter-share-button' data-url='bible-query- 
seq-1.php?id=$data_fetch[id]' 
data-text='The Bible (Sequential): http://mysite/bible-query-seq/bible-query-seq-1.php? 
id=$data_fetch[id]'>Tweet</a>";

要：

echo "<p>", "<a name='$data_fetch[id]'>", ($data_fetch['book']), " ",  
($data_fetch['cap']), " ", ($data_fetch['verse'])," </p>"; 
echo "<p>", ($data_fetch['line']), "</p>"; 

echo "<a href='twitter.com/share' class='twitter-share-button' data-url='bible-query- 
seq-1.php?id=$data_fetch[id]' 
data-text='The Bible (Sequential): http://mysite/bible-query- 
seq/bible-query-seq-1.php?id=$data_fetch[id]#$data_fetch[id]'>Tweet</a>";

它現在的作品！

來源

2013-07-08 04:56:47 adeoba

完全。您需要您想要滾動到的HTML元素的ID。假設你有一個ID爲'verseToScrollTo'的div。在網址末尾追加#verseToScrollTo將使頁面在頁面加載時自動滾動到該位置。

來源

2013-07-07 19:43:11

跳轉到推特鏈接頁面上的位置

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