PHP 搜索建议答案

【问题标题】：PHP search suggestionsPHP 搜索建议
【发布时间】：2011-05-15 10:07:58
【问题描述】：

我有一个 PHP 脚本，它与 jQuery 一起提供搜索建议。它从 MySQL 数据库中提取结果。但是，对于用户键入的字母，我只希望一次显示 5 个结果，但似乎所有结果都出现了。为什么会这样？

我的代码是：

<p id="searchresults"><?php

$db=new mysqli('localhost','username','password','database');

if(isset($_POST['queryString'])){
$queryString=$db->real_escape_string($_POST['queryString']);
            if(strlen($queryString)>0){
                $query = $db->query("SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%'");
                if($query){
                    while ($result = $query ->fetch_object()){
                        echo '<a href="/search/'.$result->name.'/1/">';                     
                        $name=$result->name;            
                        echo ''.$name.'';
                    }
                }
            }
        }
?></p>

我希望你能理解我要描述的内容。

【问题讨论】：

real_escape_string 真的是 SQL 注入安全的吗？
@Uwe: real_escape_string 提供 SQL 注入安全性。所以，是的。（虽然 PDO 更好。）

标签： php jquery mysql html

【解决方案1】：

更改"SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%'"

致"SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%' LIMIT 5"

如果您想将其限制为 5 个结果。

【讨论】：

【解决方案2】：

您需要在页面中添加分页代码：

有示例代码：

<?php 
 // Connects to your Database 
 mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()); 
 mysql_select_db("address") or die(mysql_error()); 
 //This checks to see if there is a page number. If not, it will set it to page 1 
 if (!(isset($pagenum)))  
 { 
 $pagenum = 1; 
 } 
 //Here we count the number of results 
 //Edit $data to be your query 
 $data = mysql_query("SELECT * FROM topsites") or die(mysql_error()); 
 $rows = mysql_num_rows($data); 
 //This is the number of results displayed per page 
 $page_rows = 4; 
 //This tells us the page number of our last page 
 $last = ceil($rows/$page_rows); 
 //this makes sure the page number isn't below one, or more than our maximum pages 
 if ($pagenum < 1) 
 { 
 $pagenum = 1;
 } 
 elseif ($pagenum > $last) 
 { 
 $pagenum = $last; 
 } 
 //This sets the range to display in our query 
 $max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; 
 //This is your query again, the same one... the only difference is we add $max into it 
 $data_p = mysql_query("SELECT * FROM topsites $max") or die(mysql_error()); 
 //This is where you display your query results
 while($info = mysql_fetch_array( $data_p )) 
 { 
 Print $info['Name']; 
 echo "<br>";
 }
 echo "<p>";
 // This shows the user what page they are on, and the total number of pages
 echo " --Page $pagenum of $last-- <p>";
 // First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page.
 if ($pagenum == 1) 
 {
 } 
 else 
 {
 echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=1'> <<-First</a> ";
 echo " ";
 $previous = $pagenum-1;
 echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$previous'> <-Previous</a> ";
 } 
 //just a space
 echo " -- ";
 //This does the same as above, only checking if we are on the last page, and then generating the Next and Last links    
 if ($pagenum == $last)
 { 
 }
 else {
 $next = $pagenum+1;
 echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next -></a> ";
 echo " ";
 echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last ->></a> ";
 }
 ?>

来源：www.twitter.com/ZishanAdThandar

【讨论】：