Question

我正在尝试让搜索引擎结果页面显示一条消息，说“如果没有输入密钥，则不会输入关键字，请再试一次”。但我一直忽略了下面脚本的问题，如果你没有输入密钥，脚本不允许你按回车，但我只有那里，因为我没有关于如何进行搜索的线索引擎结果页面显示一条消息“没有搜索关键词请再试一次”任何人？

搜索栏：

<form action="/search.php" method="GET"> 
<input class="term" type="text" id="term" name="term" required />  
<input type="submit" class='submit'  id="submit" value="search" disabled />
</form>
<script type="text/javascript"> 
document.getElementById('term').oninput = function() {
    document.getElementById('submit').disabled = !this.value.trim();
}
</script>

显示搜索结果：

<?php
    $db = mysqli_connect('localhost','root', '', 'searchengine');

    if(!$db) {
        die('sorry we are having some problbems');
    }

    $sql = mysqli_query(
        $db,
        sprintf(
            "SELECT * FROM searchengine WHERE name LIKE '%s' LIMIT 0,20",
            '%'. mysqli_real_escape_string($db,$_GET['term']) .'%'
        )
    );

    while($ser = mysqli_fetch_array($sql)) {
        echo "<a href='$ser[pageurl]'>$ser[img]</a>";
    }


    mysqli_close($db);
?>

Answer 1

<?php
$db = mysqli_connect('localhost','root', '', 'searchengine');

if(!$db) {
    die('sorry we are having some problbems');
}

// SET GETTER AS A VARIABLE
$searchTerm = mysqli_real_escape_string($db,$_GET['term']);

if ( empty($searchTerm))
{
echo("no key words searched please try again");
}
else
{
$sql = mysqli_query(
    $db,
    sprintf(
        "SELECT * FROM searchengine WHERE name LIKE '%s' LIMIT 0,20",
        '%'. $searchTerm .'%'
    )
);

while($ser = mysqli_fetch_array($sql)) {
    echo "<a href='$ser[pageurl]'>$ser[img]</a>";
}
}

mysqli_close($db);
?>

我认为，基于你的问题，这就是你想要的。

我希望有所帮助！

Answer 2

您可以检查该术语是否为空，如果是，则退出并显示消息

<?php
if(empty($_GET['term']){
    exit("no key words entered please try again");
}
$db = mysqli_connect('localhost','root', '', 'searchengine');
//rest of your code

Answer 3

    <div id="result" style="display:none">
        <!-- Here is for result -->
    </div>

    <script type="text/javascript">
        $.sendComment = function(){
            var deger = $("form#search").serialize();
            $.ajax({
                url: "yourpost.php",
                type: "POST",
                data: result,
                dataType: "json",
                success: function(answer){
                    if(answer.error){
                        $("#result").html(answer.error).show();
                    }else{
                        $("#result").html(answer.ok).show();

                    }
                }
            });
        }
</script>

yourpost.php

     <?php
        require "connect.php";
        if(@$_SERVER["HTTP_REFERER"]==""){
            $array["error"]='No post';
        }else{
            if($_POST){
                @$name = strip_tags(mysql_real_escape_string($_POST['name']));
            if(isempty($name)){
               $array["error"] = "no post";
            }else{
               //SQL query..
               $array["ok"] = "Search is success";
            }
        }
        echo json_encode($array);
        ?>

简单的mysqli搜索引擎？

3 个答案: