Question

我想创建一个表单，允许用户键入不同字段的搜索，并让它从数据库中选择正确的值并显示它们。出于某些原因，我无法弄清楚如何使用Mysqli查询定义搜索页面。

如何根据搜索表单和mysqli查询创建一个显示表中记录的search.php页面？

这是我迄今为止所做的尝试：

PHP代码：

$DBhost = "localhost";
$DBuser = "root";
$DBpass = "";
$DBname = "search";

$DBcon = new MySQLi($DBhost,$DBuser,$DBpass,$DBname);

if ($DBcon->connect_errno) {
    die("ERROR : -> ".$DBcon->connect_error);
}

$output = '';
if(isset($_POST['search'])) {
    $query = mysqli_query($DBcon,"SELECT * FROM data WHERE name LIKE '%" .  $_POST['name'] . "%'
          OR address LIKE '%" . $_POST['address'] . "%'
          OR city LIKE '%" . $_POST['city'] . "%'
          OR state LIKE '%" . $_POST['state'] . "%'") or die ("Could not search");
}

HTML代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>MySQL table search</title>
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js"></script>
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js"></script>
    <link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" rel="stylesheet" type="text/css"/>
    <style>
        BODY, TD {
            font-family:Arial, Helvetica, sans-serif;
            font-size:12px;
        }
    </style>
</head>

<body>
    <form action ="index.php" method = "post">
        <table>
            <tr>
                <td>Name:</td>
                <td><input type="text" name="name" /></td>
            </tr>
            <tr>
                <td>Address:</td>
                <td><input type="text" name="address" /></td>
            </tr>
            <tr>
                <td>City:</td>
                <td><input type="text" name="city" /></td>
            </tr>
            <tr>
                <td>State:</td>
                <td><input type="text" name="state" /></td>
            </tr>
            <tr>
                <td></td>
                <td><input name="search" type="submit" value="Search"/></td>
            </tr>
        </table>
    </form> 
    <?php echo ("$output");?>
</body>
</html>

Answer 1

根据我对你的问题的理解，你喜欢在sql中搜索表并回显输出。所以我在这里使用mysqli和sql LIKE运算符来按用户搜索模式。

注意：在我使用LIKE tblname %$vars%的sql查询中，您也可以使用这种方式%$vars%但两者都有不同的含义。要了解更多信息，请访问此LINK

输出图像如下：

 <?php

$link = new mysqli ('localhost','root','admin','demo');
if($link->connect_error){
    die("failed to connect".$link->connect_error);
}

?>
<!DOCTYPE html>
<html>
<head>

</head>
<body>
    <form action ="" method = "post">
        <table>
            <tr>
                <td>Name:</td>
                <td><input type="text" name="name" /></td>
            </tr>
            <tr>
                <td>Address:</td>
                <td><input type="text" name="address" /></td>
            </tr>
            <tr>
                <td>City:</td>
                <td><input type="text" name="city" /></td>
            </tr>
            <tr>
                <td>State:</td>
                <td><input type="text" name="state" /></td>
            </tr>
            <tr>
                <td></td>
                <td><input name="search" type="submit" value="Search"/></td>
            </tr>
        </table>
    </form> 

    <?php 
        if(isset($_POST['search'])){
            $output = '';
            $name = $_POST['name'];
            $address = $_POST['address'];
            $city = $_POST['city'];
            $state = $_POST['state'];

            $sql = "SELECT name,address,city,state FROM sql_like_operator WHERE name LIKE '%$name%' OR address LIKE '%$address%' OR  city LIKE '%$city%' OR state LIKE '%$state%'";

            $res= $link->query($sql);

            if($res->num_rows > 0){
                while($row = $res->fetch_assoc()){
                    $output .= $row['name']."<br>";
                    $output .= $row['address']."<br>";
                    $output .= $row['city']."<br>";
                    $output .= $row['state']."<br>";
                }
            }
            else
            {
                echo "0 results";
            }
        }
    ?>

    <h5><?php echo @$output;?></h5>
</body>
</html>

PHP Mysqli：多键搜索表单

1 个答案: