-1
請幫助我以下列表。我不知道我在哪裏犯了一個錯誤。在網址和條件查詢字符串
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.
一個完整的鏈接列表會出現在頁面的index.php
例如在:
1994 Nike
1995 O2
一旦鏈路(1994年耐克)點擊
,它會出現:1994 Nike Book
1994 Nike DVD
當你點擊最後一個步驟是 - 1994年耐克書:
1994 Nike Book John
1994 Nike Book Peter
我有以下的代碼,我不知道如何將它們連接在一起。
1.步驟
<?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 - $distributor</a></br>");
}
?>
2.步驟
<?php
if (isset($_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 - $item</a></br>");
}
}
?>
最後一步
<?php
if (isset($_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, $conn) or die(mysql_error());
echo TABLE LISTING;
}
?>
我試圖這樣的結構,但它不能如我所願。
<?php
if (isset($_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 - $distributor</a></br>");
}
}
?>
這是一個糟糕的結構嗎?
謝謝您的意見。
*「這是一個糟糕的結構?」* - 也許或可能不是,但你使用的api是。 –
...或者這真的是什麼問題呢?您的網址應該是,然後您可以通過$ _GET'訪問它們。今年是$ _GET ['year']'。 – chris85
這裏使用'codeigniter'的地方在哪裏? – Bira