php - 从推文链接跳转到页面上的位置

标签 php mysql

我有一个页面,我在其中编程了圣经的顺序访问,每节经文都包含一个推文按钮。

当我推文一首诗时,链接会重定向到诗句页面,但从顶部开始。

我想要的是,当点击推文链接时,它会转到诗句页面,然后跳转到该页面上的诗句位置。

这可能吗?如果可以的话你能帮我吗?

这是我当前的代码:

<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>

最佳答案

谢谢奥瑞斯特斯和帕维尔,这就是我所做的;我更改了以下内容:

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>";

致:

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>";

现在可以了!

关于php - 从推文链接跳转到页面上的位置,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17515577/

相关文章:

php - View Helper 中的 Zend Framework 2 服务

php - Paypal PHP Rest API用于信用卡支付,如何处理错误以显示给用户

php - 从一个数据库中选择一个数组,并将其与另一个数据库中的数组进行比较

php - 使用 foreach from 变量将多行插入数据库

php - 正则表达式从第一个大写匹配到查询字符串的句子结尾

php - 使用 Facebook 的 PHP SDK v4 创建用户拥有的 Open Graph 对象

php - 数据表中的总和是多少? PHP

java - MySQL Java根据两个字符串列删除行

mysql - 算在where语句中吗?

php - 在 mysql 中使用条件选择 SUM()