php - 编辑表中的链接未连接到mysql

Question

我在通过 html 表编辑 mysql 数据库时遇到问题。

当我运行一个简单的 while 循环时，它可以工作，我可以提取数据并更新数据库:

工作代码

<?php


    while ($row = mysqli_fetch_array($res))
        echo "$row[id]. $row[Key_Role] .$row[Incumbant] .$row[Attrition_Risk] .$row[Ready_Now] .$row[lowerYears] .$row[higherYears]<a href='edit.php?edit=$row[id]'>edit</a> <br />";

?>

但是，当我通过表运行 php 时，编辑链接不再从 mysql 中提取数据，并且我无法更新信息。我已将其放在索引下方并编辑 php 文件。

索引.php

<?php

        include_once('db.php');


        if(ISSET($_POST['Key_Role']))

        {

        $Key_Role = $_POST['Key_Role'];
        $Incumbant = $_POST['Incumbant'];
        $Attrition_Risk = $_POST['Attrition_Risk'];
        $Ready_Now = $_POST['Ready_Now'];
        $lowerYears = $_POST['1-2_Years'];
        $higherYears = $_POST['3-5_Years'];

        $sql = "INSERT INTO tmdata VALUES('','$Key_Role','$Incumbant','$Attrition_Risk','$Ready_Now','$lowerYears','$higherYears')";
        $res = mysqli_query($conn, $sql);

    echo "<meta http-equiv='refresh' content='0;url=index.php'>"; 
        if ($res)
                echo "<meta http-equiv='refresh' content='0;url=index.php'>";
        else 
                echo "Failed";
        } else {

            echo "please enter a key Role";
        }


        $res = mysqli_query($conn, "SELECT * FROM tmdata");



    ?>
    <style>

    <?php include 'style.css' ?>

    </style>

    <H1 class="Title">Talent Management System</H1>

    <form action="." method="post">

    Key Role:<input type="text" name="Key_Role">
    Incumbant:<input type="text" name="Incumbant">
    Attrition_Risk:<input type="text" name="Attrition_Risk">
    Ready_Now:<input type="text" name="Ready_Now">
    1-2_Years:<input type="text" name="1-2_Years">
    3-5_Years:<input type="text" name="3-5_Years">


    <input type ="submit" value="Enter">


    </form>



    <h1> List Of Key Roles</h1>

    <?php
    /*

        while ($row = mysqli_fetch_array($res))
            echo "$row[id]. $row[Key_Role] .$row[Incumbant] .$row[Attrition_Risk] .$row[Ready_Now] .$row[lowerYears] .$row[higherYears]<a href='edit.php?edit=$row[id]'>edit</a> <br />";
    */
    ?>


    <table>
            <tr>
                <th>id</th>
                <th>Key_Role</th>
                <th>Incumbant</th>
                <th>Attrition_Risk</th>
                <th>Ready_Now</th>
                <th>1-2_Years</th>
                <th>3-5_Years</th>  
                <th>Edit</th>  
            </tr>
            <?php while ($row = mysqli_fetch_array($res)):;?>
            <tr>
                <td><?php echo $row['id'];?></td>
                <td><?php echo $row['Key_Role'];?></td>
                <td><?php echo $row['Incumbant'];?></td>
                <td><?php echo $row['Attrition_Risk'];?></td>
                <td><?php echo $row['Ready_Now'];?></td>
                <td><?php echo $row['lowerYears'];?></td>
                <td><?php echo $row['higherYears'];?></td>
                <td><a href='edit.php?edit=$row["id"]'>edit</a></td>
            </tr>
            <?php endwhile;?>
        </table>

编辑.php

<?php

    include_once('db.php');

    if (isset($_GET["edit"]))
    {

        $id = $_GET["edit"];
        $res = mysqli_query($conn, "SELECT * FROM tmdata WHERE id='".$id."'");
        $row = mysqli_fetch_array($res);


    }

    if( isset($_POST['newKey_Role']) || isset($_POST['newIncumbant']) || isset($_POST['newAttrition_Risk']) || isset($_POST['newReady_Now']) || isset($_POST['newLowerYears']) || isset($_POST['newHigherYears']) )
    {

        $newKey_Role = $_POST['newKey_Role'];
        $newIncumbant = $_POST['newIncumbant'];
        $newAttrition_Risk = $_POST['newAttrition_Risk'];
        $newReady_Now = $_POST['newReady_Now'];
        $newLowerYears = $_POST['newLowerYears'];
        $newHigherYears = $_POST['newHigherYears'];
        $id = $_POST['id'];

        $sql = "UPDATE tmdata SET Key_Role='$newKey_Role', Incumbant='$newIncumbant', Attrition_Risk='$newAttrition_Risk', Ready_Now='$newReady_Now', lowerYears='$newLowerYears', higherYears='$newHigherYears' WHERE id='".$id."'";
        $res = mysqli_query($conn, $sql) or die(mysqli_error($conn));
        echo "<meta http-equiv='refresh' content='0;url=index.php'>";

    }


?>


<form action="edit.php" method="post">

Key Role:<input type="text" name="newKey_Role" value ="<?php echo $row[1]; ?>"></input> <br />
Incumbant:<input type="text" name="newIncumbant" value ="<?php echo $row[2]; ?>"></input> <br />
Attrition_Risk:<input type="text" name="newAttrition_Risk" value ="<?php echo $row[3]; ?>"></input> <br />
Ready_Now:<input type="text" name="newReady_Now" value ="<?php echo $row[4]; ?>"></input> <br />
1-2 Years:<input type="text" name="newLowerYears" value ="<?php echo $row[5]; ?>"></input> <br />
3-5 Years:<input type="text" name="newHigherYears" value ="<?php echo $row[6]; ?>"></input> <br />
<input type="hidden" name="id" value ="<?php echo $row[0]; ?>">

<input type ="submit" value="Update">


</form>

<style>

<?php include 'style.css' ?>

</style>

看起来好像 ID 没有拉到编辑 php.ini 中。我也尝试过手动输入表中的id:

<td><a href='edit.php?edit=$row[184]'>edit</a></td>

但这也不起作用。

任何帮助将不胜感激。

提前谢谢您。

Answer 1

我发现你需要将链接放在 php.ini 中。

<td><?php echo "<a href='edit.php?edit=$row[id]'>edit</a>"?></td>

最终成功了。