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php - 如何在php搜索结果之后的新页面上显示每条记录

转载 作者:行者123 更新时间:2023-11-29 19:30:28 25 4
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我创建了一个简单的搜索功能,它可以根据用户输入的内容查找几个字段。然后,它以超链接的形式返回结果,仅显示学校名称。然后我试图允许用户在单独的页面上查看其余信息。我已经给每条记录一个唯一的 ID,但我正在努力弄清楚如何在所需的页面上调用它。这是我用于搜索结果页面 search.php 的代码

<?php
require_once 'connection.inc.php';

if(isset($_GET['keywords'])){
$keywords = $db->escape_string($_GET['keywords']);
$query = $db->query("
SELECT school, address_town
FROM School
WHERE school LIKE '%{$keywords}%'
OR address_town LIKE '%{$keywords}%'
");
?>
<div class ="result-count">
Found <?php echo $query->num_rows; ?> results
<br>
<?php
if ($query->num_rows == 0)
echo "Sory there were no results matching your search" ;{
}
?>
</div>
<?php

if($query->num_rows){
while($r = $query->fetch_object()){
?>

<div class ="result">
<a href='school.php?id=".$row['id']."'><?php echo $r-> school; ?></a>
</div>
<?php
}
}

这是页面的代码,我想在其中显示该特定记录的所有信息。此页面是 school.php

<?php
include ('connection.inc.php');
connect();

$sql = "SELECT * FROM School WHERE id = ".$_GET['id'];
$result = mysql_query($sql);
?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>School Details</title>
</head>
<body>
<?php
while ($record = mysql_fetch_object($result))
{
?>
<div class="post">
<p class="school"><? echo $record->school; ?></p>
<p class="address_1"><? echo $record->address_1; ?></p>
<p class="address_town"><? echo $record->address_town; ?></p>
<p class="postcode"><? echo $record->postcode; ?></p>
<p class="website"><? echo $record->website; ?></p>
<p class="phone"><? echo $record->phone; ?></p>
<p class="council"><? echo $record->council; ?></p>
<p class="pupils"><? echo $record->pupils; ?></p>
<p class="age_range"><? echo $record->age_range; ?></p>
<p class="gender"><? echo $record->gender; ?></p>
<p class="school_type"><? echo $record->school_type; ?></p>
<p class="ofsted_rating"><? echo $record->ofsted_rating; ?></p>
<p class="ofsted_report"><? echo $record->ofsted_report; ?></p>
<p class="gcse_percentage"><? echo $record->gcse_percentage; ?></p>
<p class="points_score"><? echo $record->points_score; ?></p>
<p class="pupil_teacher_ratio"><? echo $record->pupil_teacher_ratio; ?></p>
<p class="absence_percentage"><? echo $record->absence_percentage; ?></p>
<p class="language_not_english"><? echo $record->language_not_english; ?></p>
<p class="free_school_meals"><? echo $record->free_school_meals; ?></p>
<p class="happiness"><? echo $record->happiness; ?></p>
<p class="ofsted_parent_review"><? echo $record->ofsted_parent_review; ?></p>
<p class="events"><? echo $record->events; ?></p>
<p class="school_description"><? echo $record->school_description; ?></p>
<?php
}
mysql_free_result($result);
mysql_close();
?>
</body>
</html>

如果有人能提供帮助,那就太好了,我对此很陌生,并且会边走边学。

谢谢

最佳答案

您应该在 anchor 标记中添加值为 _blank 的目标属性。

例如:

<a target="_blank" href="http://your_url_here.html">Link</a>

就您而言:

<a target="_blank" href='add_your_base_url_here/school.php?id=".$row['id']."'><?php echo $r-> school; ?></a>

希望对您有所帮助,干杯!

关于php - 如何在php搜索结果之后的新页面上显示每条记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41812630/

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