如何使用 PHP、mysqli 和 html 表单创建搜索
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How to create a search using PHP, mysqli and a html form
提问by Rebekah
I want to create a form which allows the user to type in a search and have it pick up the right values from a database and display them, for some reason I can't get my query to work it just displays "could not search"
我想创建一个表单，它允许用户输入搜索并让它从数据库中选择正确的值并显示它们，由于某种原因我无法让我的查询工作它只显示“无法搜索”
Here is my php code
这是我的 php 代码
<?php
include "connect.php";
$output = '';
if(isset($_POST['search'])) {
$search = $_POST['search'];
$search = preg_replace("#[^0-9a-z]i#","", $search);
$query = mysqli_query("SELECT * FROM house WHERE town LIKE '%$search%'") or die ("Could not search");
$count = mysqli_num_rows($query);
if($count == 0){
$output = "There was no search results!";
}else{
while ($row = mysqli_fetch_array($query)) {
$town = $row ['town'];
$street = $row ['street'];
$bedrooms = $row ['bedrooms'];
$bathroom = $row ['bathrooms'];
$output .='<div> '.$town.''.$street.''.$bedrooms.''.$bathrooms.'</div>';
}
}
}
?>Here is my form
这是我的表格
<form action ="home.php" method = "post">
<input name="search" type="text" size="30" placeholder="Belfast"/>
<input type="submit" value="Search"/>
</form>
<?php print ("$output");?>采纳答案by Funk Forty Niner
You're not connecting to your DB in your query:
您没有在查询中连接到数据库：
$query = mysqli_query("SELECT
^ missing connection variable
there is no connection variable (unknown what you are using to connect with)
没有连接变量（不知道你用什么来连接）
$query = mysqli_query($connection, "SELECT ...
^^^^^^^^^^^^
From the manual http://php.net/manual/en/mysqli.query.php
从手册http://php.net/manual/en/mysqli.query.php
Object oriented style
mixed mysqli::query ( string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
面向对象的风格
mixed mysqli::query ( string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
Procedural style
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
程序风格
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
figuring you are using mysqli_to connect with. If you're using mysql_or PDO to connect with, that won't work. Those different MySQL APIs do not intermix with each other.
确定您正在使用mysqli_连接。如果您正在使用mysql_或 PDO 进行连接，那将不起作用。这些不同的 MySQL API 不会相互混合。
Plus, instead of or die ("Could not search")do or die(mysqli_error($connection))to catch any errors, if any.
此外，而不是or die ("Could not search")做or die(mysqli_error($connection))赶上任何错误，如果有的话。
Add error reportingto the top of your file(s) which will help find errors.
将错误报告添加到文件顶部，这将有助于查找错误。
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote:Error reporting should only be done in staging, and never production.
旁注：错误报告应该只在登台时进行，而不要在生产中进行。
Example mysqliconnection:
示例mysqli连接：
$connection = mysqli_connect("myhost","myuser","mypassw","mybd")
or die("Error " . mysqli_error($connection));
For more information, visit:
欲了解更多信息，请访问：
回答by reyven
$query = mysqli_query("SELECT * FROM house WHERE town LIKE '%$search%'") or die ("Could not search");
$result = mysqli_query($connection,$query);
$count = mysqli_num_rows($result);
The $connectionis the variable declared in your connect.php to connect to your database .
该$connection是你connect.php声明的变量，以连接到数据库。
You should have out the $resultbefore the $count.
你应该$result在$count.
回答by butter bee
Try to remove the space between if($count==0)it will start working!!
尝试删除if($count==0)它之间的空间将开始工作！
if($count==0){
$output = "There was no search results!";}
