提供数据以进行查询

时间：2017-01-22 09:55:20

标签： php mysql

我正在创建一个网页，找出并列出给定区域的医生。

我的两个php脚本如下所示。

<!DOCTYPE html>
    <html lang="en">
    <head>
    <meta charset="UTF-8">
    <title>Live Doctor 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;
            margin: 0 50%;
        }
        .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 term = $(this).val();
            var resultDropdown = $(this).siblings(".result");
            if(term.length){
                $.get("backend-search.php", {query: term}).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>

                Select Your City:
            <select name="city_nname">
                <option value="Udaipur">Udaipur</option>
                <option value="Jaipur">Jaipur</option>
                <option value="Jodhpur">Jodhpur</option>
                <option value="Sikar">Sikar</option>
                <option value="Surat">Surat</option>
                <option value="Ahmedabad">Ahmedabad</option>
                <option value="badoda">badoda</option>
                <option value="Pune">Pune</option>
                <option value="Bangalore">Bangalore</option>
            </select>


            <div class="search-box">
                <input type="text" autocomplete="off" placeholder="Search doctor..."  />
                <button type="submit" class="btn">Search</button>
            <div class="result"></div>
            </div>

    </body>
    </html>

以上代码是我的search_form.php。现在我想列出属于city_nname的医生名称。

这是我的php后端代码

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

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

// Escape user inputs for security
$query = mysqli_real_escape_string($link, $_REQUEST['query']);

if(isset($query)){
    // Attempt select query execution
    $sql = "SELECT * FROM doctors WHERE  doctor_name LIKE '" . $query . "%'";
    if($result = mysqli_query($link, $sql)){
        if(mysqli_num_rows($result) > 0){
            while($row = mysqli_fetch_array($result)){
                echo "<p>" . $row['doctor_name'] . "</p>";
            }
            // Close result set
            mysqli_free_result($result);
        } else{
            echo "<p>No matches found for <b>$query</b></p>";
        }
    } else{
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }
}

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

现在我想要一个带有city_nname和doctor_name（来自搜索框）的mysql查询，然后给出属于那个城市的医生列表。

2 个答案:

答案 0 :(得分：0)

SELECT * FROM doctors WHERE doctor_name LIKE 'doctor_name%' AND city_nname='city_nname'怎么样？我无法解决你在制定解决方案时遇到的麻烦。

答案 1 :(得分：0)

请添加评论行＆＃34; //添加此行＆＃34;和修改由＆＃34; //修改此行＆＃34;

的注释行

<强> search_form.php 

<!DOCTYPE html>
    <html lang="en">
    <head>
    <meta charset="UTF-8">
    <title>Live Doctor 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;
            margin: 0 50%;
        }
        .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 term = $(this).val();
            var city_nname = $("#city_nname").val();  // Add this line
            var resultDropdown = $(this).siblings(".result");
            if(term.length){
                $.get("backend-search.php", {query: term, city:city_nname}).done(function(data){     // Modify this line
                    // 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>

                Select Your City:
            <select name="city_nname" id="city_nname">  <!-- Modify this line -->
                <option value="Udaipur">Udaipur</option>
                <option value="Jaipur">Jaipur</option>
                <option value="Jodhpur">Jodhpur</option>
                <option value="Sikar">Sikar</option>
                <option value="Surat">Surat</option>
                <option value="Ahmedabad">Ahmedabad</option>
                <option value="badoda">badoda</option>
                <option value="Pune">Pune</option>
                <option value="Bangalore">Bangalore</option>
            </select>


            <div class="search-box">
                <input type="text" autocomplete="off" placeholder="Search doctor..."  />
                <button type="submit" class="btn">Search</button>
            <div class="result"></div>
            </div>

    </body>
    </html>

<强>后端-search.php中 

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

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

// Escape user inputs for security
$query = mysqli_real_escape_string($link, $_REQUEST['query']);
$city = mysqli_real_escape_string($link, $_REQUEST['city']); // Add this line

if(isset($query)){
    // Attempt select query execution
    $sql = "SELECT * FROM doctors WHERE  doctor_name LIKE '" . $query . "%' AND city_nname='".$city."'"; // modify the table column name "city_nname" as like yours
    if($result = mysqli_query($link, $sql)){
        if(mysqli_num_rows($result) > 0){
            while($row = mysqli_fetch_array($result)){
                echo "<p>" . $row['doctor_name'] . "</p>";
            }
            // Close result set
            mysqli_free_result($result);
        } else{
            echo "<p>No matches found for <b>$query</b></p>";
        }
    } else{
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }
}

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