如何在html页面中以表格格式输出，使用php脚本并从mysql数据库中提取？ [关闭]

问题描述：

I am trying to fetch the data from Mysql database and print it on new web page. I need to display the output in table format.
below is the PHP script, help me out to do so.
Thanks in advance.

<?php
    $userinput = $_POST['Sample_name'];
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "ProcessTrackingSystem";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_errno) {
        printf("Connect failed: %s
", $conn->connect_error);
        exit();
    }

    $result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails
                                   WHERE Sample_name = '$userinput'") or die(mysqli_error($conn));

    while ($row=mysqli_fetch_assoc($result))
    {
        printf("Sample_name:->");
        printf($row['Sample_name']);
        printf("<br>
");
        printf("<br>
");
        printf("SO_ID:->");
        printf($row['SO_ID']);
        printf("<br>
");
        printf("<br>
");
    }

    mysqli_free_result($result);
    $conn->close();
?>

I need output in tabular format.

答

You can create table as following way.

echo "<table border='1'>
    <tr>
        <th>Sample_name</th>
        <th>SO_ID</th>
    </tr>";

   while($row = mysqli_fetch_assoc($result))
   {
       echo "<tr>";
           echo "<td>" . $row['Sample_name'] . "</td>";
           echo "<td>" . $row['SO_ID'] . "</td>";
       echo "</tr>";
   }
   echo "</table>";

答

It will put in tabular format

<?php    
    $userinput = $_POST['Sample_name'];
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "ProcessTrackingSystem";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_errno) {
        printf("Connect failed: %s
", $conn->connect_error);
        exit();
    }
    $result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE Sample_name = '$userinput'") or die(mysqli_error($conn));
    echo "<table class='table' border='1' >";
    echo "<tr><th>Sample Name</th><th>SO ID</th></tr>";
    while ($row = mysqli_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td>{$row['Sample_name']}</td>";
        echo "<td>{$row['SO_ID']}</td>";
        echo "</tr>";
    }
    echo "</table>";
    mysqli_free_result($result);
    $conn->close();
    ?>

如何在html页面中以表格格式输出，使用php脚本并从mysql数据库中提取？ [关闭]

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