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PHP 搜索脚本无法正确返回页面

转载 作者:行者123 更新时间:2023-11-29 09:16:06 25 4
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由于没有其他主题使用我使用的脚本,因此我需要认真的帮助。我目前正在处理一个工作项目,我的搜索功能正确返回第一页,但当我单击“下一步 10”时,它完全没有任何反应!请注意,我删除了一些代码,但留下了无用的注释,尽管这不会影响脚本的性能!这是我的代码,但我当然已经过滤了用户名和密码:<

<?php

// Get the search variable from URL

$var = @$_GET['q'] ;
$error = $_GET['error'];
$permnull = @$_GET['nullpermitted'] ;
$AdminorSuper = "";
$trimmed = trim($var); //trim whitespace from the stored variable


if(!$session->isSuperuser()){
$AdminorSuper = "Admin";
}

else
{
$AdminorSuper = "";
}
// rows to return
$limit=10;

if($error=="errdel")
{
echo "<p>Only Admins and Superusers may delete customers. </p>";
}
if($error=="errcret")
{
echo "<p>Only Admins and Superusers may create customers. </p>";
}
// 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("*FILTERED*","*FILTERED*","*FILTERED*"); //(host, username, password)


mysql_select_db("*FILTERED*") or die("Unable to select database");

// Build SQL Query
$query = "select * from customer where Surname OR TitleName OR PostCode like \"%$trimmed%\"
order by Surname";

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



if ($numrows == 0)
{
if($trimmed=="")
{
echo "No customers are in the database!";
}
else
{
echo "<h2>Results</h2>";
echo "<p>Sorry! No Results were found for: &quot;" . $trimmed . "&quot;.</p>";
}
}

// next determine if s has been passed to script, if not use 0
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
if($var=="")
{
echo "<p>Viewing all Results</p>";
}
else
{
echo "<p>Results for: &quot;" . $var . "&quot;</p>";
}
// begin to show results set
echo "";
$count = 1 + $s ;

// now you can display the results returned
echo '<table border="1">';
echo "<tr><td><b>Surname</b></td><td><b>Title/Name</b></td><td><b>Email</b></td><td><b>Telephone</b></td><td><b>Edit</b></td><td><b>Del</b></td></tr>\n";
while ($row= mysql_fetch_array($result)) {
$Surname = $row["Surname"];
$Title = $row["TitleName"];
$Email = $row["Email"];
$Telephone = $row["Telephone"];
$id = $row["id"];
echo '<tr><td>' .$Surname.'</td><td>'.$Title.'</td><td>'.$Email.'</td><td>'.$Telephone.'</td><td>' . '<a href="updateCustomerForm' . $AdminorSuper. '.php?id='.$id.'">[EDIT]</a></td>'.'<td>'. '<a href="deleteCustomer.php?id='.$id.'">[x]</a></td>'. '</tr>';
$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-$limit);
print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt;
Prev 10</a>&nbsp&nbsp;";
}

// 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
$pages++;
}
echo "</table>";
// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

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

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

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

?>

最佳答案

代码似乎缺少$s=$_GET['s'];

祝你好运:)

关于PHP 搜索脚本无法正确返回页面,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4173791/

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