Question

这是index.php。当我给出一个输入时，它会获取特定的名称和年份。没关系。但是当我提交表单时，没有任何输入它给出了电影的所有名称和年份，但我不想要这样，用户无法显示保存在数据库中的所有数据。我给了priventdefault（）方法，但它不起作用。我怎么解决这个问题？

<!DOCTYPE html>
<html>
    <head>
        <title>ajax</title>
        <script
            src="https://code.jquery.com/jquery-2.2.4.min.js"
            integrity="sha256-BbhdlvQf/xTY9gja0Dq3HiwQF8LaCRTXxZKRutelT44="
            crossorigin="anonymous"></script>
        <script type="text/javascript">



            $(function() // this function excited if the jquery is ready i mean after jquery successfully loaded
            {
                function loaddata()
                {

                    var moviename= $("#moviename").val(); // read moviename value and assign;

                    $.ajax({
                        type: "GET",
                        url: "query.php",
                        data: {
                            name:moviename // there is no variable name so you have to assign the moveiname to name vairable ;
                        },
                        success: function (data) {
                           $("#result").html(data);

                        }

                    });

                }


                $("#submit").click(function(event) // Click Event Listener.
                {
                    event.preventDefault();
                    loaddata()
                });
            });



        </script>
    </head>
    <body>


        <p>Enter movie name </p>
        <form action="" method="POST">
            <input type="text" name="moviename" id="moviename" placeholder="Enter Movie Name" required autocomplete="off">
            <input type="submit" name="submit" id="submit" value="Search"/>
            <!-- if you want ot use jquery you have to use event listener. like $("#submit").click(function(event){}); code from line 31 to 35 -->
        </form>
        <br>

        <div id="result">

        </div>

    </body>
</html

///这是query.php

<?php

include 'dbcon.php';

$name =isset($_GET['name'])?$_GET['name']:'';


$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name like '%$name%'");

while($row = mysqli_fetch_array($query))
{
    echo "<p>".$row['name']."</p>";
    echo "<p>".$row['year']."</p>";
}

?>

Answer 1

仅当$ name不为空时才执行查询。

if(!empty($name)) { 
  $query = mysqli_query( $conn,"SELECT * FROM movie WHERE name like '%$name%'");
  while($row = mysqli_fetch_array($query)){
    echo "<p>".$row['name']."</p>";
    echo "<p>".$row['year']."</p>";
  }
}

Answer 2

$name =isset($_GET['name'])?$_GET['name']:'';

if(!empty($name)){
$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name like '%$name%'");

while($row = mysqli_fetch_array($query))
{
    echo "<p>".$row['name']."</p>";
    echo "<p>".$row['year']."</p>";
}
}

在javascript中

     var moviename= $("#moviename").val(); // read moviename value and assign;
    if(moviename){
                        $.ajax({
                            type: "GET",
                            url: "query.php",
                            data: {
                                name:moviename // there is no variable name so you have to assign the moveiname to name vairable ;
                            },
                            success: function (data) {
                               $("#result").html(data);

                            }

                        });
}

Answer 3

在query.php中，如果为空，请始终将空字符串分配给行中的$name＆＃39; name＆＃39;值传递给query.php

$name =isset($_GET['name'])?$_GET['name']:'';.

因此，查询将返回db中的所有结果，因为$name是一个空字符串，并与db中的所有数据匹配。

您可以验证$ name是否为空，而不是运行查询。

当我不提供任何输入时，为什么表单是获取所有数据？

3 个答案: