Question

用户是否可以输入让我们说“红色汽车”的信息。它会在数据库和php / html中找到图像，在搜索结果中会显示图像吗？

我的代码＆＃34;搜索＆＃34;

<?php
    $query = $_GET['query']; 


    $min_length = 3;


    if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then

        $query = htmlspecialchars($query); 
        // changes characters used in html to their equivalents, for example: < to &gt;

        $query = mysql_real_escape_string($query);
        // makes sure nobody uses SQL injection

        $raw_results = mysql_query("SELECT * FROM articles
            WHERE (`title` LIKE '%".$query."%') OR (`text` LIKE '%".$query."%')") or die(mysql_error());

        // * means that it selects all fields, you can also write: `id`, `title`, `text`
        // articles is the name of our table

        // '%$query%' is what I'm looking for, % means anything, for example if $query is Hello
        // it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
        // or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'

       if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

    while($results = mysql_fetch_array($raw_results)){
    // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop



    echo "<div class='successmate'><h2></h2></a>

    </div><br><br><br>";

echo "<div class='search69'><a href='../pages/{$results['page_name']}'><h2>{$results['title']}</h2></a><p>{$results['text']}</‌p>";



    }


}
    else{ // if there is no matching rows do following
    echo ("<br><br><div class='search1'><h2>No results</h2></br></br>");
}

}

else{ // if query length is less than minimum
echo ("<br><br><div class='search1'><h2>Minnimum Length Is</h2><h2>".$min_length);
}

?>

我希望找到关键字的图片：

http://puu.sh/cCHv1/9f58d770f3.jpg

这些是DB中图像的名称：

http://puu.sh/cCHwa/a82d2cc7fe.png

谢谢！

Answer 1

如果没有测试，我会说你的搜索功能已关闭，显示结果只是简单的事情，只需将下面的代码放在你想要显示图像结果的地方（这只会用于显示图像，但是让它与其他结果一起工作是一个简单的返工）

<?php
while($data = $result->fetch_assoc()){
    echo '<img src="'.$data['image_path'].'" alt="" title="" />';
}
?>

然后你可以根据需要简单地将图像包装在容器中，另一种方法是将结果存储到'数组中并使用foreach循环显示它

<?php
while($data = $result->fetch_assoc()){
   $imageArray[] = $data['image_path'];
}

/*Code below you can place where ever you want, no need to place it directly after the 
 above query execute*/

foreach($imageArray as $imgPATH){
   echo '<img src="'.$imgPATH.'" alt="" title="" />';
}
?>

Answer 2

我理解这个问题的方式，我会这样做

编辑数据库表，以便在包含“ser_pic1.jpg”的图像之后得到“image_name”

SELECT * FROM {your_database} WHERE image_name =“Red car”

并发布如

link here

显示从数据库到结果页面的图像

2 个答案: