Php 如何将值从数据库显示到字段中 改变形象 >

Php 如何将值从数据库显示到字段中 改变形象 >,php,html,mysqli,Php,Html,Mysqli,编辑数据的代码 <form action="" method="post" id="validation-form" class="form-horizontal" enctype='multipart/form-data' > <div class="control-group">


编辑数据的代码
                                        <form action="" method="post" id="validation-form" class="form-horizontal" enctype='multipart/form-data' >

                                            <div class="control-group">
                                                <label class="control-label" for="file">Change Image</label>
                                                <div class="controls">
                                                    <input type="file"  name="file" id="imgae" value="<?php echo $image;?>">
                                                </div>                                                
                                            </div>

                                            <div class="control-group">
                                                <label class="control-label" for="name">Change Name</label>
                                                <div class="controls">
                                                    <input type="text" class="input-large" name="name" id="name" value="<?php echo $name;?>">
                                                </div>
                                            </div>

                                            <div class="control-group">
                                                <label class="control-label" for="message">Change Description</label>
                                                <div class="controls">
                                                    <textarea class="span4" name="message" id="message" rows="4"><?php echo $message;?></textarea>
                                                </div>
                                            </div>

                                            <div class="controls">
                                               <input type="hidden" name="id" value=<?php echo $_GET['id'];?>>
                                                <input type='submit' class="btn btn-danger btn" name='but_upload' value="Submit" >
                                                <!--                              <button type="submit" class="btn btn-danger btn">Validate</button>&nbsp;&nbsp;-->
                                                <a href="elements.php" class="btn"> back </a>&nbsp;&nbsp;&nbsp;
$time_stamp=time();
$image=$_文件['file']['name'];
$image=$time\u stamp.'\u'$形象;
$target_dir=“img/upload/”;
$target\u file=$target\u dir。基本名称($_文件[“文件”][“名称]);
$imageFileType=strtolower(路径信息($target_文件,路径信息_扩展名));
$extensions_arr=数组(“jpg”、“jpeg”、“png”、“gif”);
$location='img/upload/'。$image;
if(在数组中($imageFileType,$extensions\u arr)){
移动上传的文件($\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\;
}
$name=$_POST['name'];
$message=$_POST['message'];
$query=“插入图像(图像、位置、名称、消息)值('$image'、'$location'、'$name'、'$message')”;
mysqli_query($con,$query)或die(mysqli_error($con));
echo“document.location.href='elements.php?msg=Updated';
更新数据的代码

    $time_stamp = time();
        $image = $_FILES['file']['name'];
        $image = $time_stamp . '_' . $image;
        $target_dir = "img/upload/";
        $target_file = $target_dir . basename($_FILES["file"]["name"]);
        $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
        $extensions_arr = array("jpg","jpeg","png","gif");
        $location = 'img/upload/'.$image;
        if( in_array($imageFileType,$extensions_arr) ){
            move_uploaded_file($_FILES['file']['tmp_name'],'img/upload/'.$image);
        }


     $name = $_POST['name'];
    $message = $_POST['message'];
     $query = "insert into images(image,location,name,message) values('$image', '$location', '$name', '$message')";
     mysqli_query($con,$query) or die(mysqli_error($con));

    echo "<script>document.location.href='elements.php?msg=Updated'</script>";

     <!-- -->
$name=$\u POST['name'];
$message=$_POST['message'];
$query=“更新图像集图像='$image',位置='$location',名称='$name',消息='$message',其中id='$id';
mysqli_查询($con,$query);
echo“document.location.href='elements.php?msg=Updated';
}
包括“footer.php”;?>

这是用于向数据库中添加和更新数据的代码。使用此代码,我可以向数据库中添加或编辑数据。我无法成功地将数据编辑到数据库中,但问题是在我尝试更新数据时,数据没有显示在插入字段中,因此请在更新数据时帮助在插入字段中显示数据。

希望将您的问题更新为。虽然此代码片段可以解决问题,确实有助于提高你的文章质量。请记住,您将在将来回答读者的问题,而这些人可能不知道您的代码建议的原因。先生,此解释是可以的,或者需要更多的解释。通过使用上述代码,wwe可以从数据库中获取数据
     $name = $_POST['name'];
    $message = $_POST['message'];
     $query = "UPDATE images SET image = '$image', location = '$location', name = '$name', message = '$message' where id = '$id'";
     mysqli_query($con,$query);


    echo "<script>document.location.href='elements.php?msg=Updated' </script>"  ;

}

 include 'footer.php'; ?>
     $name = $_POST['name'];
    $message = $_POST['message'];
     $query = "UPDATE images SET image = '$image', location = '$location', name = '$name', message = '$message' where id = '$id'";
     mysqli_query($con,$query);


    echo "<script>document.location.href='elements.php?msg=Updated' </script>"  ;

}

 include 'footer.php'; ?>
<?php


$id = $_GET['id'];
$sql = "SELECT * FROM users WHERE id ='1'";
$result =mysqli_query($con,$sql);
$row = mysqli_fetch_assoc($result);
  ?>
By using this code on the top of the page then we can show the values in the fields and able to update the existing data