Question

我创建了一个每行都有一个编辑按钮的表

ID Name Age Class  EDIT
1  sam  12  12th   edit

用于编辑按钮的代码是

echo "<td><a href=\"admin_edit_members.php?id=".$row['id']."\">Edit</a></td>";

admin_edit_members.php页面上使用的代码是：

<form class="form-horizontal" role="form" action="<?php $_PHP_SELF ?>" enctype="multipart/form-data" method="post">
    <div class="form-group">
        <label class="col-lg-3 control-label">Name:</label>
        <div class="col-lg-8">
            <input class="form-control" value="" type="text" name="name">
        </div>
    </div>

    <div class="form-group">
        <label class="col-lg-3 control-label">Age:</label>
        <div class="col-lg-8">
            <input class="form-control" value="" type="text" name="age">
        </div>
    </div>

    <div class="form-group">
        <label class="col-lg-3 control-label">Class:</label>
        <div class="col-lg-8">
            <input class="form-control" value="" type="text" name="class">
        </div>
    </div>

    <div class="form-group">
        <label class="col-md-3 control-label"></label>
        <div class="col-md-8">
            <input class="btn btn-primary" value="Save Changes" type="submit" name="submit">
            <span></span>
        </div>  
    </div>
</form>

<?php
    $con=mysqli_connect("localhost","root","","db");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    // escape variables for security
    $id = $_GET['id'];
    $name = mysqli_real_escape_string($con, $_POST['name']);
    $age = mysqli_real_escape_string($con, $_POST['age']);
    $class = mysqli_real_escape_string($con, $_POST['class']);

    $sql = "UPDATE student SET name='".$name."',age='".$age."',class='".$class."' WHERE id ='".$id."'";


    if (!mysqli_query($con,$sql)) {
        die('Error: ' . mysqli_error($con));
    }
    header("Location: student_list.php");
    exit;
    mysqli_close($con);
?>

我的问题是，当我编辑表单时，我需要编辑所有字段或在数据库中输入的空白值。有没有办法让用户只能编辑他想要的特定字段

Answer 1

运行选择查询

$query="SELECT * FROM student";
$result=  mysqli_query($con, $query) or die(mysqli_error());
//get the value from database
while ($row=  mysqli_fetch_array($result))
{
    $name_data=$row['name'];
    $name_age=$row['age'];
}


if(isset($_POST['submit']))
{
    $name=$_POST['name'];
    $age=$_POST['age'];
    if(empty($name))
    {
        //if the value is empty its going to set it equal to the database value
        $name=$name_data;
    }
    else 
        $name=$name;

    if(empty($age))
    {
        $age=$name_age;
    }
    else 
        $age=$age;
}

现在运行更新查询

Answer 2

请尝试以下方法： -

<?php
    $sql = "select * from student where id = $_GET['id']";
    $res = mysqli_query($con,$sql);
    $row =  mysqli_fetch_array($res);
?>
<form class="form-horizontal" role="form" action="<?php $_PHP_SELF ?>" enctype="multipart/form-data" method="post">
    <div class="form-group">
        <label class="col-lg-3 control-label">Name:</label>
        <div class="col-lg-8">
            <input class="form-control" value="<?php echo $row["name"]; ?>" type="text" name="name" >
        </div>
    </div>
//likewise in all inputbox.
</form>

编辑表格的特定字段

2 个答案: