Php 在表中显示SQL搜索表单结果

我试图通过“提交”表单从“SQL”数据库中提取信息。我认为连接是建立的，因为它的代码与条目的代码相同，并且工作正常。我可以将结果显示为项目块，但不能将其集成到表中

目前，代码只是返回一个空白页

我找到了很多几年前的例子，但后来情况发生了变化

下面是PHP页面上的代码，它应该在表中显示结果

在来这里寻求帮助之前，我尝试了不同的方法将结果集成到表中，但都返回相同的结果。我对这方面还不太熟悉

 <?php

$servername = "localhost";
$username = "root";
$password = "Rmvs03ff";
$dbname = "EmployeeListing";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if (isset($_POST['submit-search'])) {
    $search = mysqli_real_escape_string($conn, $_POST['search']);
    $sql = "SELECT * FROM BasicEmployee WHERE EmployeeID LIKE '%$search%' OR FirstName LIKE '%$search%' OR LastName LIKE '%$search%' OR DoB LIKE '%$search%'";
    $result = mysqli_query($conn, $sql);
    $queryResult = mysqli_num_rows($result);

    echo "<br/>There are " .$queryResult. " matches found";

    echo "<table>
            <tr>
            <th>Employee I.D.</th>
            <th>Sex</th>
            <th>First Name</th>
            <th>Last Name</th>
            <th>Date of Birth</th>
            </tr>";

    if ($queryResult > 0) {
        while ($row = mysqli_fetch_assoc($result)) {
            echo    "<tr>"
            echo    "<td><strong>
                    ".$row['EmployeeID']."
                    </strong></td>"
            echo    "<td>
                    ".$row['Prefix']."
                    </td>"
            echo    "<td><p>
                    ".$row['FirstName']."
                    </p></td>"
            echo    "<td><p>
                    ".$row['LastName']."
                    </p></td>"
            echo    "<td><p>
                    ".$row['DoB']."
                    </p></td>";
            echo "</table>";
        }

    } else {
        echo "<br/>No results found";
    }

}

echo "<br/>Approved";

$conn->close();

?>
   </body>
   </html>

---

我有一个html的开头，当然是头和身体，我不想用这个来阻塞空间

您需要回显整个表一次：

<?php

$servername = "localhost";
$username = "root";
$password = "Rmvs03ff";
$dbname = "EmployeeListing";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if (isset($_POST['submit-search'])) {
    $search = mysqli_real_escape_string($conn, $_POST['search']);
    $sql = "SELECT * FROM BasicEmployee WHERE EmployeeID LIKE '%$search%' OR FirstName LIKE '%$search%' OR LastName LIKE '%$search%' OR DoB LIKE '%$search%'";
    $result = mysqli_query($conn, $sql);
    $queryResult = mysqli_num_rows($result);

    echo "<br/>There are " .$queryResult. " matches found";

    $table = "<table>
            <tr>
            <th>Employee I.D.</th>
            <th>Sex</th>
            <th>First Name</th>
            <th>Last Name</th>
            <th>Date of Birth</th>
            </tr>";

    if ($queryResult > 0) {
        while ($row = mysqli_fetch_assoc($result)) {
            $table .= "<tr>";
            $table .= "<td><strong>
                    ".$row['EmployeeID']."
                    </strong></td>";
            $table .= "<td>
                    ".$row['Prefix']."
                    </td>";
            $table .= "<td><p>
                    ".$row['FirstName']."
                    </p></td>";
            $table .= "<td><p>
                    ".$row['LastName']."
                    </p></td>";
            $table .= "<td><p>
                    ".$row['DoB']."
                    </p></td>";
            $table .= "</table>";
        }

    } else {
        echo "<br/>No results found";
    }

}
echo $table;


echo "<br/>Approved";

$conn->close();

?>
   </body>
   </html>

谢谢你的评论，我已经编辑了我的答案，解决了这个问题，谢谢！如果有多个结果，我将查找如何正确显示！这不是对上面“引用-这个错误在PHP中是什么意思？”的回答，因为我得到的是一个空白页，而不是一个错误。。。。现在日志似乎没有记录任何东西。为什么我在每个echo语句的末尾没有看到分号？当没有行时，您是否看到表html被破坏了？