Php 如何为自己的网站创建搜索引擎

我在互联网上得到了这种语法，并修改了一些部分。我想查找/搜索表中的记录。这个语法的结果是“SearchQueryNotFound”，我在这个语法中找不到问题。你能帮我找出错误吗？下面是语法

<?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>";

?>

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您提到您遇到此错误“搜索查询未找到”。您编写的代码清楚地表明url中没有设置参数“q”

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

如果必须这样做，您的url应该是

www.example.com?q=some_value

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@Rio Salonoy-由于未设置“p”的值，您的脚本在前3行之后无法继续。您正在告诉if（！isset（$\u GET['q']）。这意味着如果未设置该值，脚本将输出die（'Search Query not found'）。因此，首先确保在运行页面时，确保您的url已被删除。我希望你现在就明白了：）