Question

我想在php中创建一个搜索功能，它获取输入的值并在数据库中搜索它。我有这段代码：

<center><form method="post" id="search" action="search.php?go">

        <p>

            <input type="text" name="name" id="name" />

            <input type="submit" id="search_btn" name="submit" value="Cerca utente" />
        </p>

    </form></center>

在php文件中我创建了这个：

public function searchUser($nomeUtente)
    {

        $sql = "SELECT uUsername FROM users WHERE='$nomeUtente' ";

        if( !$this->stmt = $this->mysqli->prepare($sql) )
            throw new Exception("MySQL Prepare statement failed: ".$this->mysqli->error);

        $this->stmt->execute();
        $this->stmt->store_result();
        $this->stmt->bind_result($nome);
        if( $this->stmt->num_rows == 0)
            return "Nessun nome corrispondente.";
        $nomi = array();
        $i = 0;
        while( $this->stmt->fetch() ){
        $nomi[$i]["nome"] = $nome;
           }


        return $nomi;

    }

如何在当前页面的表格中查看查询结果？我不想在另一个页面中查看结果，但是在搜索表单所在的页面中。

Answer 1

***HTML***

    <center><form method="post" id="search" action="search.php?go">

            <p>

                <input type="text" name="name" id="name" />

                <input type="submit" id="search_btn" name="submit" value="Cerca utente" />
            </p>

        </form></center>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"> </script>

    <script>
    $(document).ready(function(){

    $('#search_btn').click(function(){

    var Name=$('#name').val();
    $.Post('Search.php',{Name:Name},function(data)
{
alert(data);
});

    });


    });
    </script>

PHP（必须保存为search.php）

<?php
if($_isset($_POST["Name"]))
{
$Name=$_POST["Name"];


searchUser($Name);

public function searchUser($nomeUtente)
    {

        $sql = "SELECT uUsername FROM users WHERE='$nomeUtente' ";

        if( !$this->stmt = $this->mysqli->prepare($sql) )
            throw new Exception("MySQL Prepare statement failed: ".$this->mysqli->error);

        $this->stmt->execute();
        $this->stmt->store_result();
        $this->stmt->bind_result($nome);
        if( $this->stmt->num_rows == 0)
            return "Nessun nome corrispondente.";
        $nomi = array();
        $i = 0;
        while( $this->stmt->fetch() ){
        $nomi[$i]["nome"] = $nome;
           }


        return $nomi;

    }

}

?>

搜索功能php和mysql查看结果在当前页面

1 个答案: