Question

请帮我处理以下列表。我不知道我在哪里弄错了。

我的表：

year | distributor | item | nameautor
1994 | Nike | Book | John
1994 | Nike | Book | Peter
1994 | Nike | DVD | Jessie
1994 | Nike | DVD | Marc
1995 | O2 | Book | Heck
1995 | O2 | Book | Lars
etc.

完整的链接列表将显示在页面 index.php

上

示例：

1994 Nike
1995 O2

单击链接（1994 Nike）后，将显示：

1994 Nike Book
1994 Nike DVD

最后一步是你点击 - 1994 Nike Book：

1994 Nike Book John
1994 Nike Book Peter

我有以下代码，我不知道如何将它们链接在一起。

1。步骤

<?php
$query="(SELECT DISTINCT year, distributor FROM table)";
$back=mysql_query($query, $conn) or die(mysql_error());
while (list($year,$distributor) = mysql_fetch_row($back)){
    echo ("<a href=\"index.php?$year&$distributor\"><b></b>$year - $distributor</a></br>");
    }
?>

2。步骤

<?php
if (isset($_SERVER['QUERY_STRING']) && !empty($_SERVER['QUERY_STRING'])) {
    $year = $_SERVER['QUERY_STRING'];
    $distributor = $_SERVER['QUERY_STRING'];
    $namechoose = str_replace(array('%20', '&'), ' ' , $_SERVER['QUERY_STRING']);
    $query="(select distinct year, distributor, item FROM table WHERE CONCAT(year, ' ', distributor)='$namechoose')";
    $back=mysql_query($query, $conn) or die(mysql_error());
    echo ("<h3><center>you choose --- $namechoose ---</center></h3>");
    while (list($year,$distributor,$item) = mysql_fetch_row($back)){
        echo ("<a href=\"index.php?$year&$distributor&$item\"><b></b>$year - $distributor - $item</a></br>");   
        } 
} 
?>

最后一步

<?php
if (isset($_SERVER['QUERY_STRING']) && !empty($_SERVER['QUERY_STRING'])) {
    $year = $_SERVER['QUERY_STRING'];
    $distributor = $_SERVER['QUERY_STRING'];
    $item = $_SERVER['QUERY_STRING'];
    $detailedtable = str_replace(array('%20', '&'), ' ' , $_SERVER['QUERY_STRING']);
    $query="(select distinct year, distributor, item, nameautor FROM table WHERE CONCAT(year, ' ', distributor, ' ', item)='$detailedtable')";
        $back=mysql_query($query, $conn) or die(mysql_error());
    echo TABLE LISTING;
}
?>

我已经尝试过这种结构，但它并没有按照我的意愿运作。

<?php
if (isset($_SERVER['QUERY_STRING']) && !empty($_SERVER['QUERY_STRING'])) {
    if (!isset($_GET['item'])) {
            $year = $_SERVER['QUERY_STRING'];
            $distributor = $_SERVER['QUERY_STRING'];
            $namechoose = str_replace(array('%20', '&'), ' ' , $_SERVER['QUERY_STRING']);
            $query="(select distinct year, distributor, item FROM table WHERE CONCAT(year, ' ', distributor)='$namechoose')";
            $back=mysql_query($query, $conn) or die(mysql_error());
            echo ("<h3><center>you choose --- $namechoose ---</center></h3>");
            while (list($year,$distributor,$item) = mysql_fetch_row($back)){
                echo ("<a href=\"index.php?$year&$distributor&$item\"><b></b>$year - $distributor - $item</a></br>");   
                } 
     } else {
            $year = $_SERVER['QUERY_STRING'];
            $distributor = $_SERVER['QUERY_STRING'];
            $item = $_SERVER['QUERY_STRING'];
            $detailedtable = str_replace(array('%20', '&'), ' ' , $_SERVER['QUERY_STRING']);
            $query="(select distinct year, distributor, item, nameautor FROM table WHERE CONCAT(year, ' ', distributor, ' ', item)='$detailedtable')";
               $back=mysql_query($query, $conn) or die(mysql_error());
            echo TABLE LISTING;
         } 
    }
else {
    $query="(SELECT DISTINCT year, distributor FROM table)";
    $back=mysql_query($query, $conn) or die(mysql_error());
    while (list($year,$distributor) = mysql_fetch_row($back)){
        echo ("<a href=\"index.php?$year&$distributor\"><b></b>$year - $distributor</a></br>");
        }
}
?>

这是一个糟糕的结构吗？

感谢您的评论。

Answer 1

感谢您的帮助。

我已使用

修改了代码

echo ("<a href=\"index.php?year=$year&distributor=$distributor&item=$item\"><b></b>$year - $distributor - $item</a></br>");

现在它可以正常工作。

在url和条件中查询字符串

1 个答案: