如何使搜索引擎到您自己的網站

Question

我在互聯網上得到這種語法並修改了一些部分。我想在表格中找到/搜索記錄。此語法最終爲「搜索查詢未找到」我無法在此語法中找到問題。你能幫我找一些錯誤嗎？這是語法。

<?php // Get the search variable from URL 
if(!isset($_GET['q'])) 
die('Search Query not found'); 
$var = $_GET['q']; 
$trimmed = trim($var); //trim whitespace from the stored variable 

// rows to return 
$limit=10; 

// check for an empty string and display a message. 
if ($trimmed == ""){ 
echo "<p>Please enter a search…</p>"; 
exit; 
} 

// check for a search parameter 
if (!isset($var)){ 
echo "<p>We dont seem to have a search parameter!</p>"; 
exit; 
} 

//connect to your database ** EDIT REQUIRED HERE ** 
mysql_connect("localhost","user1","test123"); 

//specify database ** EDIT REQUIRED HERE ** 
mysql_select_db("inventory") or die("Unable to select database"); 
// Build SQL Query 
$query = ("SELECT * FROM client WHERE account_name LIKE \"%$trimmed%\" or maintenance_type like \"%$trimmed%\" or ma_status like \"%$trimmed%\" ma_contract_start like \"%$trimmed%\" ma_contract_end like \"%$trimmed%\" ma_reference_no like \"%$trimmed%\" ORDER BY account_name DESC"); 

// EDIT HERE and specify your table and field names for the SQL query 

$numresults=mysql_query($query); 
$numrows=mysql_num_rows($numresults); 

// If we have no results, offer a google search as an alternative — this is optional 

if ($numrows == 0) 
{ 
echo "<h4>Results</h4>"; 
echo "<p>Sorry, your search: $trimmed returned zero results</p>"; 

// google 
echo "<p><a href=\"http://www.google.com/search?q=" 
. $trimmed . "\" target=\"_blank\" title=\"Look up 
" . $trimmed . " on Google\">Click here</a> to try the 
search on google</p>"; 
} 

// next determine if s has been passed to script, if not use ZERO (0) to Limit the output 
if (empty($s)) { 
$s=0; 
} 

// get results 
$query = " limit $s,$limit"; 
$result = mysql_query($query) or die("Couldn’t execute query"); 

// display what the person searched for 
echo "<p>You searched for: $var </p>"; 

// begin to show results set 
echo "Results: <br/>"; 
$count = 1 + $s ; 

// now you can display the results returned 
while ($row= mysql_fetch_array($result)) { 
$name = $row['account_name']; 
$mtype = $row['maintenance_type']; 
$status = $row['ma_status']; 
$start = $row['ma_contract_start']; 
$end = $row['ma_contract_end']; 
$reference = $row['reference_no']; 


echo "$count.> $name $mtype $status $start $end $reference <br/>" ; 
$count++ ; 
} 

$currPage = (($s/$limit) + 1); 

//break before paging 
echo "<br />"; 

// next we need to do the links to other results 
if ($s>=1) { 

// bypass PREV link if s is 0 
$prevs=($s-$limit); 
print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< 
Prev 10</a> "; 
} 

// calculate number of pages needing links 
$pages=intval($numrows/$limit); 

// $pages now contains int of pages needed unless there is a remainder from division 

if ($numrows%$limit) { 
// has remainder so add one page 
$pages++; 
} 

// check to see if last page 
if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { 

// not last page so give NEXT link 
$news=$s+$limit; 

echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; 
} 

$a = $s + ($limit) ; 
if ($a > $numrows) { $a = $numrows ; } 
$b = $s + 1 ; 
echo "<p>Showing results $b to $a of $numrows</p>"; 

?> 

Answer 1

你提到你收到此錯誤「搜索查詢未找到」。你寫的代碼清楚地告訴參數'q'沒有在url中設置。

if(!isset($_GET['q'])) 
die('Search Query not found'); 

如果有工作，您的網址應該是

www.example.com?q=some_value