使用PHP Mysqli从下拉列表中选择多个选项时，如何显示数据库中的数据

Question

脚本页面运行良好。 当我在下一个仪表板页面中选择多个选项时，不会显示任何记录。 请解决此问题。 我认为所选值无法在仪表板页面中识别

script.php的

<?php include("connection.php") ?>
<form id="script" name="script" action="dashboard.php" method="post">
    <strong>Choose Script Name : </strong><select name="script[]" id="select3" multiple=multiple style="margin: 20px;width:300px;">   
        <?php
        $result = $conn->query("select script_name from script_details ORDER BY script_name");
        while ($row = $result->fetch_assoc()) {
            unset($script_name);
            $script_name = $row['script_name'];
            echo '<option value="' . $id . '">' . $script_name . '</option>'; // Generated From database
        }
        ?>
    </select>
    <input type="submit" name="submit" id="button" value="View Dashboard" />
</form>

Dashboard.php

<table border="1">
    <tr align="center">
        <th>Number </th>      <th>Script Name</th>    <th> Date</th> 
    </tr> 
    <?php
    include("connection.php");
    $select = $_POST['script'];
    $selects = "SELECT * FROM script_details where script_name='$select'";
    $result = $conn->query($selects);
    echo "<table>";
    while ($row = $result->fetch_assoc()) {
        echo "<tr><td>" . $row["id"] . "</td><td>" . $row["script_name"] . "</td></tr>" . "</td><td>" . $row["date"] . "</td></tr>";
    }
    echo "</table>";
[This is script page Image. Selecting option from script_details database. Field name : script_name.][1]?>

这是“仪表板”页面。 选择script2，script3选项时。 不显示所选项目的记录。

Answer 1

我将通过以下方式进行处理：

$scriptsArr = $_POST['script'];
$scriptsStr = implode(',', $scriptsArr);

$selects = "SELECT * FROM script_details where script_name IN ($scriptsStr)";

我将其拆分为几个变量，以便您可以了解过程。 希望我能帮上忙！

希望您的理解一点都不安全，建议您阅读更多有关准备好的语句的信息： http : //php.net/manual/en/mysqli.quickstart.prepared-statements.php

Answer 2

首先，您的代码易受sql攻击

在Scrip中，您没有在<select>标记中定义选项的值。 首先定义值，为此您需要从数据库中获取

script.php的

<?php include("connection.php") ?>
<form id="script" name="script" action="dashboard.php" method="post">
    <strong>Choose Script Name : </strong>
    <select name="script[]" id="select3" multiple=multiple style="margin: 20px;width:300px;">   
        <?php
        $result = $conn->query("select id, script_name from script_details ORDER BY script_name");
        while ($row = $result->fetch_assoc()) {
            unset($script_name);
            $script_name = $row['script_name'];
            $id = $row['id'];
            echo '<option value="' . $id . '">' . $script_name . '</option>'; // Generated From database
        }
        ?>
    </select>
    <input type="submit" name="submit" id="button" value="View Dashboard" />
</form>

在仪表板中进行适当的标记

Dashboard.php

<table border="1">
    <tr align="center">
        <th>Number </th>      <th>Script Name</th>    <th> Date</th> 
    </tr> 
    <?php
    include("connection.php");
    $select = $_POST['script'];
    $ids = "'" . implode("','", $select) . "'";
    $selects = "SELECT * FROM script_details WHERE id IN ($ids)";
    $result = $conn->query($selects);
    while ($row = $result->fetch_assoc()) {
        echo "<tr>"
                . "<td>" . $row["id"] . "</td>"
                . "<td>" . $row["script_name"] . "</td>"
                . "<td>" . $row["date"] . "</td>"
            . "</tr>";
    }
    ?>
</table>