Question

我想知道是否有人可以帮助我尝试将实时搜索结合到我的网站中，我已经找到了一个教程并且外观方面的内容现在正在运行，有人可以告诉我如何编辑以下PHP来拉来自一个表的信息。

表格 - 电影要为每个结果回显的字段 - 电影，描述，图像

目前它成功地从2个表中提取信息，一个显示搜索内容，另一个提供类别分隔符的信息，我需要的是删除类别方面并从单个表中提取信息。

道歉我的PHP知识非常有限，希望这能最好地描述问题。

<p id="searchresults">
<?php
// PHP5 Implementation - uses MySQLi.
// mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase');
$db = new mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase');

if(!$db) {
    // Show error if we cannot connect.
    echo 'ERROR: Could not connect to the database.';
} else {

    // Is there a posted query string?
    if(isset($_POST['queryString'])) {
        $queryString = $db->real_escape_string($_POST['queryString']);

        // Is the string length greater than 0?
        if(strlen($queryString) >0) {
            $query = $db->query("SELECT * FROM search s INNER JOIN categories c ON s.cat_id = c.cid WHERE name LIKE '%" . $queryString . "%' ORDER BY cat_id LIMIT 8");

            if($query) {
                // While there are results loop through them - fetching an Object.

                // Store the category id
                $catid = 0;
                while ($result = $query ->fetch_object()) {
                    if($result->cat_id != $catid) { // check if the category changed
                        echo '<span class="category">'.$result->cat_name.'</span>';
                        $catid = $result->cat_id;
                    }
                    echo '<a href="'.$result->url.'">';
                    echo '<img src="search_images/'.$result->img.'" alt="" />';

                    $name = $result->name;
                    if(strlen($name) > 35) { 
                        $name = substr($name, 0, 35) . "...";
                    }                       
                    echo '<span class="searchheading">'.$name.'</span>';

                    $description = $result->desc;
                    if(strlen($description) > 80) { 
                        $description = substr($description, 0, 80) . "...";
                    }

                    echo '<span>'.$description.'</span></a>';
                }
                echo '<span class="seperator"><a href="http://www.marcofolio.net/sitemap.html" title="Sitemap">Nothing interesting here? Try the sitemap.</a></span><br class="break" />';
            } else {
                echo 'ERROR: There was a problem with the query.';
            }
        } else {
            // Dont do anything.
        } // There is a queryString.
    } else {
        echo 'There should be no direct access to this script!';
    }
}
?>
</p>

Answer 1

我想：

<p id="searchresults">
<?php
// PHP5 Implementation - uses MySQLi.
// mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase');
$db = new mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase');

if(!$db) {
    // Show error if we cannot connect.
    echo 'ERROR: Could not connect to the database.';
} else {

    // Is there a posted query string?
    if(isset($_POST['queryString'])) {
        $queryString = $db->real_escape_string($_POST['queryString']);

        // Is the string length greater than 0?
        if(strlen($queryString) >0) {
            $query = $db->query("SELECT * FROM movies  WHERE Movie LIKE '%" . $queryString . "%' ORDER BY Movie LIMIT 8");

            if($query) {
                // While there are results loop through them - fetching an Object.


                while ($result = $query ->fetch_object()) {
                    echo '<a href="'.$result->url.'">';
                    echo '<img src="search_images/'.$result->img.'" alt="" />';

                    $name = $result->movie;
                    if(strlen($name) > 35) { 
                        $name = substr($name, 0, 35) . "...";
                    }                       
                    echo '<span class="searchheading">'.$name.'</span>';

                    $description = $result->description;
                    if(strlen($description) > 80) { 
                        $description = substr($description, 0, 80) . "...";
                    }

                    echo '<span>'.$description.'</span></a>';
                }
                echo '<span class="seperator"><a href="http://www.marcofolio.net/sitemap.html" title="Sitemap">Nothing interesting here? Try the sitemap.</a></span><br class="break" />';
            } else {
                echo 'ERROR: There was a problem with the query.';
            }
        } else {
            // Dont do anything.
        } // There is a queryString.
    } else {
        echo 'There should be no direct access to this script!';
    }
}
?>
</p>

Answer 2

替换

SELECT * FROM search s INNER JOIN categories c ON s.cat_id = c.cid WHERE name LIKE '%" . $queryString . "%' ORDER BY cat_id LIMIT 8

与

SELECT * FROM movies WHERE movie LIKE '%" . $queryString . "%' ORDER BY movie LIMIT 8

将是一个良好的开端。

你可以从那里继续吗？

除此之外 - 你写过电影＆＃39;对于上面描述中的字段名称 - 我使用了小写字母＆＃39; M＆＃39;所以你可能需要改变它。

[edit]回复评论＃1：

让我们开始简单 - 用{/ 1>替换if ($query) { ... }

if ($query) {
    while ($result = $query ->fetch_object()) { // this line loops through all the results
        echo '<img src="search_images/'.$result->image.' />'; // check capital I on 'image'
        echo '<strong>'.$result->movie.'</strong>';
        echo $result->description.'<br />';
    }
}

基本上，您只需将$result->myFieldName放在while循环中即可获得所需的数据。希望这应该足够让你去=]

MySQL实时查询

2 个答案: