Question

搜索页：

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>PHP Live MySQL Database Search</title>
<style type="text/css">
    body{
        font-family: Arail, sans-serif;
    }
    /* Formatting search box */
    .search-box{
        width: 300px;
        position: relative;
        display: inline-block;
        font-size: 14px;
    }
    .search-box input[type="text"]{
        height: 32px;
        padding: 5px 10px;
        border: 1px solid #CCCCCC;
        font-size: 14px;
    }
    .result{
        position: absolute;        
        z-index: 999;
        top: 100%;
        left: 0;
    }
    .search-box input[type="text"], .result{
        width: 100%;
        box-sizing: border-box;
    }
    /* Formatting result items */
    .result p{
        margin: 0;
        padding: 7px 10px;
        border: 1px solid #CCCCCC;
        border-top: none;
        cursor: pointer;
    }
    .result p:hover{
        background: #f2f2f2;
    }
</style>
<script src="https://code.jquery.com/jquery-1.12.4.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    $('.search-box input[type="text"]').on("keyup input", function(){
        /* Get input value on change */
        var inputVal = $(this).val();
        var resultDropdown = $(this).siblings(".result");
        if(inputVal.length){
            $.get("backend-search.php", {term: inputVal}).done(function(data){
                // Display the returned data in browser
                resultDropdown.html(data);
            });
        } else{
            resultDropdown.empty();
        }
    });

    // Set search input value on click of result item
    $(document).on("click", ".result p", function(){
        $(this).parents(".search-box").find('input[type="text"]').val($(this).text());
        $(this).parent(".result").empty();
    });
});
</script>
</head>
<body>
    <div class="search-box">
        <input type="text" autocomplete="off" id="id" name="id" placeholder="Search Products or codes..." />
        <div class="result"></div>
    </div>
</body>
</html>

后端搜索页面：

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "username", " ", "root");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

if(isset($_POST['term'])){
    // Prepare a select statement
    $sql = "SELECT * FROM ProductCodes WHERE item_name LIKE ?";
    //$sql = "SELECT * FROM ProductCodes WHERE ('item_name' LIKE ?) OR ('id' LIKE ?)";


    if($stmt = mysqli_prepare($link, $sql)){
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "s", $param_term);

        // Set parameters
        $param_term = $_REQUEST['term'] . '%';

        // Attempt to execute the prepared statement
        if(mysqli_stmt_execute($stmt)){
            $result = mysqli_stmt_get_result($stmt);

            // Check number of rows in the result set
            if(mysqli_num_rows($result) > 0){
                // Fetch result rows as an associative array
                while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
                    echo "<p>BA" . $row["id"] . " " . $row["item_name"] . "</p>";
                }
            } else{
                echo "<p>No matches found</p>";
            }
        } else{
            echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
        }
    }
     //var_dump(mysqli_error($db_conx));
    // Close statement
    mysqli_stmt_close($stmt);
}

// close connection
mysqli_close($link);
?>

请放轻松-新手警报！我已经用包含两个列的表（ProductCodes）设置了以上内容-item_name和id。我需要进行搜索以检查两个列，但是我尝试替换

$sql = "SELECT * FROM ProductCodes WHERE item_name LIKE ?";

与

"$sql = "SELECT * FROM ProductCodes WHERE ('item_name' LIKE ?) OR ('id' LIKE ?)";

我遇到的另一个问题是，我只需要将表单的值设置为id字段，但看不到如何分配它。感谢您的帮助，我已经尝试了各种组合，但我只是绕圈而行。

Answer 1

您可以像这样绑定参数：

$stmt->bind_param('sssd', $code, $language, $official, $percent);

从这里获取：http://php.net/manual/ro/mysqli-stmt.bind-param.php

在您的情况下，这将是：

$stmt->bind_param('si', $yourStringValue, $yourIntValue);

，您将需要修复查询，以免对like使用id，该"SELECT * FROM ProductCodes WHERE item_name LIKE ? or id = ?"可能是数字：

from GameFrame import GameFrame
global game

def click(x, y):
    global game
    print("%s %s" %(x,y))
    game.changeColor(x, y)

if __name__ == '__main__':
    global game
    game = GameFrame("title", click) 
    pass

Answer 2

好吧凯特，我想我理解您的问题。您想使用一个输入来搜索此查询的两个字段。

让我们从适应查询开始，然后：

$sql = 'SELECT id, item_name FROM ProductCodes WHERE item_name LIKE ? OR id = ?';

然后我们将参数绑定到它，因为有两个询问标记（？），我们将需要两个参数。

$stmt->bind_param('ss', $searchName, $searchId);

$searchName = '%' . $_REQUEST['term'] . '%';
$searchId = $_REQUEST['term'];

或使用与您相同的功能。

mysqli_stmt_bind_param($stmt, "ss", $searchName, $searchId);

$searchName = '%' . $_REQUEST['term'] . '%';
$searchId = $_REQUEST['term'];

对不起，一团糟！修正了考虑参考。

Answer 3

您可以在查询中使用两个参数，如下所示：

// prepare and bind
$stmt = $conn->prepare("SELECT * FROM ProductCodes WHERE id = ? OR item_name LIKE ? ");
$stmt->bind_param("sss", $ProductCodes,$ID);
$ProductCodes = '%' . $_REQUEST['term'] . '%';
$ID = '%' . $_REQUEST['id'] . '%';

为数据库中的即时搜索分配值

3 个答案: