用户是否可以输入“红色汽车”,它会在数据库和 php/html 中找到该图像,并在搜索结果中显示该图像?
我的“搜索”代码
<?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 >
$query = mysql_real_escape_string($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);
}
?>
我想通过关键字查找的图片:
http://puu.sh/cCHv1/9f58d770f3.jpg
这些是数据库中图像的名称:
http://puu.sh/cCHwa/a82d2cc7fe.png
谢谢!
最佳答案
未经测试,我想说您已经关闭了搜索功能,显示结果将是一件简单的事情,只需将下面的代码放在您想要显示图像结果的位置(这仅用于显示图像,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,如果您愿意,您可以简单地将图像包装在任何容器中,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="" />';
}
?>
关于php - 将数据库中的图像显示到结果页面,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26729687/