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我构建了一个获取信息的表单,它可以很好地工作,但是我遇到了一个问题,URL最后没有更改。无法仅显示指向该搜索词的链接。PDO搜索表单断开的链接
说我搜索术语“谷歌搜索”
我的网址仍然是
http://localhost/search.php
当我与MySQL的工作我已经搜查一个术语它看起来有些像什么这样
后http://localhost/search.php?k=google+search
我为了做到这一点而改变了什么?
搜索引擎形式:
<form name="frmSearch" method="post" action="../search.php">
<input class="inp" name="var1" type="text" id="var1">
<input class="btn" type="submit" value="Search">
</form>
搜索引擎网页:
<?php
$nameofdb = 'xxxxxx';
$dbusername = 'xxxxxx';
$dbpassword = 'xxxxxx';
// Connect to MySQL via PDO
try {
$dbh = new PDO("mysql:dbname=$nameofdb;host=xxxxxx", $dbusername, $dbpassword);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}
$var1 = $_POST['var1'];
$query = "SELECT * FROM pages WHERE title LIKE :search OR keywords LIKE :search";
$stmt = $dbh->prepare($query);
$stmt->bindValue(':search', '%' . $var1 . '%', PDO::PARAM_INT);
$stmt->execute();
/* Fetch all of the remaining rows in the result set
print("Fetch all of the remaining rows in the result set:\n"); */
$result = $stmt->fetchAll();
foreach($result as $row) {
/* echo */ $row["title"];
/* echo */ $row["keywords"];
/* echo */ $row["photo"];
/* echo */ $row["link"];
echo "<a href=$row[link]> $row[photo] </a>";
}
if ($stmt->rowCount() > 0) {
$result = $stmt->fetchAll();
foreach($result as $row) {
echo $row["id"];
echo $row["title"];
}
} else {
echo 'There is nothing to show';
}
?>
有时我忘记了最简单的事情,谢谢! – Bianca 2014-12-19 08:23:01
我改变了方法=“GET”我认为它已经工作,但现在它显示我所有的结果与我搜索的任何术语?任何建议 – Bianca 2014-12-19 08:50:07
圣诞节即将来临^^!将'$ var1 = $ _POST ['var1'];'更改为'$ var1 = $ _GET ['var1'];' – Robert 2014-12-19 09:02:44