Question

我有一个网站，我提供随机内容（文字，图片等）。

我创建了一个链接，显示网址中编码内容的mysql记录ID（例如mysite.com/country-query.php?id=23）。

我的问题是，当页面加载时，网址没问题，但页面只是空白。

你能帮忙吗？

以下是代码：

mysite.com/country-query.php：

<?php

    define ('HOSTNAME', 'hostname');
    define ('USERNAME', 'username');
    define ('PASSWORD', 'password');
    define ('DATABASE_NAME', 'database_name');

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

    mysql_select_db(DATABASE_NAME);

    $query = "SELECT country_id, iso2, short_name, long_name, iso3, numcode, un_member, calling_code, cctld FROM table_name ORDER by rand() LIMIT 1";

    $result = mysql_query($query);

    while ($row = mysql_fetch_array($result))
    {
        echo "<p>", "<b>", "Short Name: ", "</b>", ($row['short_name']), "</p>";
        echo "<p>", "<b>", "Long Name: ", "</b>", ($row['long_name']), " ", "<b>","Calling Code: ", "</b>", ($row['calling_code']), " ", "<b>", "ccTLD: ", "</b>",($row['cctld'])," </p>";
        echo "<a href='action.php?id=$row[country_id]'>NEXT COUNTRY</a>";
    }

    mysql_free_result($result);
    mysql_close();

?>

'action.php'是显示网址的位置，的网页内容应该显示。以下是我的代码：

<?php

    $id = mysql_escape_string($_GET['country_id']);

    define ('HOSTNAME', 'hostname');
    define ('USERNAME', 'username');
    define ('PASSWORD', 'password');
    define ('DATABASE_NAME', 'database_name');

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

    mysql_select_db(DATABASE_NAME);

    $query = "SELECT country_id, iso2, short_name, long_name, iso3, numcode, un_member, calling_code, cctld FROM table_name WHERE country_id = '$id' ";

    $result = mysql_query($query);

    while ($row = mysql_fetch_array($result))
    {
        echo "<p>", "<b>", "Short Name: ", "</b>", ($row['short_name']), "</p>";
        echo "<p>", "<b>", "Long Name: ", "</b>", ($row['long_name']), " ", "<b>", "Calling Code: ", "</b>", ($row['calling_code']), " ", "<b>", "ccTLD: ", "</b>", ($row['cctld'])," </p>";
    }

    mysql_free_result($result);
    mysql_close();

?>

实际上，加载mysite.com/country-query.php会首先加载一个随机国家但该页面上的链接显示该国家/地区的ID 已经加载了，所以我想在这里实现的是首先加载mysite.com/country-query.php的登录页面，然后加载显示在该页面上的链接的内容。 PHP

感谢。

PS：第一个代码的第一行有一个开始的php标记，如果没有显示的话！

Answer 1

更改action.php

$id = mysql_escape_string($_GET['country_id']);

到

$id = mysql_escape_string($_GET['id']);

country-query.php

中的

OR

echo "<a href='action.php?id=$row[country_id]'>NEXT COUNTRY</a>";

到

echo "<a href='action.php?country_id=$row[country_id]'>NEXT COUNTRY</a>";

您似乎对参数名称感到困惑

显示URL的内容

1 个答案: