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我有一个表包含几个字段（id, firstname, surname, username, search_count）

我已经构建了一个小型搜索引擎，可以搜索我的表以查找存在于firstname或中的任何匹配项，surname并且我得到的结果没有问题。

现在，我要做的是search_count每次匹配时将字段增加 1！

例如，假设我们有下表users：

id | firstname | surname  | username | search_count 
1  | John      | Mike     | un1      | 0
2  | John      | Jeff     | un2      | 0
3  | Dale      | John     | un3      | 0
4  | Mike      | Gorge    | un4      | 0

假设我们正在搜索Jeff关键字

因此，查询将返回 1 条记录

我想要做的是增加search_count匹配1记录

所以结果将类似于：

id | firstname | surname  | username | search_count 
2  | John      | Jeff     | un2      | `1`

如果我们进行新的搜索（例如John），结果应该是这样的：

id | firstname | surname  | username | search_count 
1  | John      | Mike     | un1      | 1
2  | John      | Jeff     | un2      | 2
3  | Dale      | John     | un3      | 1

我尝试了几种方法，但没有运气..所以我感谢任何提示和帮助

这是我的代码...

<?php
// open the HTML page
include 'html_open.php';
// require the db connection
require '/inc/db.inc.php';
// require the error messages
require '/inc/echo.inc.php';



    if (isset($_GET['keyword'])) {
        $keyword = $_GET['keyword'];
        if (!empty($keyword)) {
            // build our search query
            $search_query = "SELECT * FROM `users` WHERE `firstname` = '".mysql_real_escape_string($keyword)."' OR `surname` = '".mysql_real_escape_string($keyword)."' ORDER BY `search_count` DESC";
            // run the search query
            $search_query_run = mysql_query($search_query);
            // search results
            $search_results_num = mysql_num_rows($search_query_run);
            // check query return results
            if ($search_results_num>0) {        
                echo 'Search engine returns <strong>[ '.$search_results_num.' ]</strong> result(s) for <strong>[ '.$keyword.' ]</strong>:<br>';
                // retrieving the information found
                echo '<ol>';
                while ($search_result_information = mysql_fetch_assoc($search_query_run)) {
                    //$current_search_count = ;
                    echo '<li>'.$search_result_information['username'].'.  This user has been searched: '.$search_result_information['search_count'].' times before.</li>';
                }
                echo '</ol><hr>';
                include 'search_form.php';
            } else {
                echo '<hr>Search engine returns no result for <strong>[ '.$keyword.' ]</strong>, please try another keyword.<hr>'; // hint: no result found
                include 'search_form.php';
            }               
        } else {
            echo $err20_002; // hint: must insert input
            include 'search_form.php';
        }
    } else {
            echo $err20_001; // hint: form has not been submitted
            include 'search_form.php';
    }
// close the HTML page  
include 'html_close.php';   
?>

PS我是PHP / MySQL的新手，这是我的第一个代码:)

4

2 回答 2

0 
...
while ($search_result_information = mysql_fetch_assoc($search_query_run)) {
    # add this following line
    mysql_query('UPDATE `users` SET search_count=search_count+1 WHERE id='.$search_result_information['id']);
    # Edit. Change result to show new number
    $search_result_information['search_count']++; # this adds 1 to the value (does not affect stored data in database)
    echo '<li>'.$search_result_information['username'].'.  This user has been searched: '.$search_result_information['search_count'].' times before.</li>';
}
...
于 2013-10-20T13:12:06.417 回答
0

在您的情况下，您需要获取找到的值并执行更新语句，将 +1 添加到search_count列

$search_query = "SELECT id, firstname, surname, username, search_count FROM `users` WHERE `firstname` = '".mysql_real_escape_string($keyword)."' OR `surname` = '".mysql_real_escape_string($keyword)."' ORDER BY `search_count` DESC";
$search_query_run = mysql_query($search_query);
        // search results
        $search_results_num = mysql_num_rows($search_query_run);
        // check query return results
        if ($search_results_num>0) {        
            echo 'Search engine returns <strong>[ '.$search_results_num.' ]</strong> result(s) for <strong>[ '.$keyword.' ]</strong>:<br>';
            // retrieving the information found
            echo '<ol>';
            while ($search_result_information = mysql_fetch_assoc($search_query_run)) {
                //$current_search_count = ;
                $update_search = "UPDATE users SET search_count = search_count + 1 WHERE id = {$search_result_information['id']}"; // so every `id` will increment its search_count with 1. You will need to select the rows once again, to take this count, or to manually increment in PHP
                mysql_query($update_search);
                echo '<li>'.$search_result_information['username'].'.  This user has been searched: '.$search_result_information['search_count'].' times before.</li>';

PS：强烈不推荐使用 mysql_* lib。从官方文档http://www.php.net/manual/en/function.mysql-query.php中的红框可以看出，您应该选择当前实际支持的 API 之一

于 2013-10-20T13:12:08.377 回答