更新但不删除我在seprate文件中尝试的文件夹中的图像，但不起作用

时间：2018-05-24 13:49:03

标签： php mysqli

`这是我的新index.php文件

<?php
include('db.php');
if(isset($_POST['submit']))
{
    $name = $_POST['name'];

    $image = $_FILES['image']['name'];
    $query = "INSERT INTO `img_table`(`name`, `imgtitle`) 
               VALUES ('$name','$image')";
    header("location:fetch.php");
    if(mysqli_query($con,$query))
    {
        move_uploaded_file($_FILES['image']['tmp_name'],"image/$image");
    }


}

?>
<html>
<head><title>hi</title></head>

<body>
    <form method="POST" enctype="multipart/form-data" >
    <table border=1px;>
        <tr>
        <th>name</th>
        <td><input type="text" name="name" /></td>
        </tr>

        <tr>
        <th>img_title</th>
        <td><input type="file" name="imgtitle"/>
        </tr>

        <tr>
        <input type="submit" name="submit" value="add"/>
        </tr>

    </table>
    </form>
</body>

`这是我的db.php页面

<?php
$con = mysqli_connect("localhost" ,"root", "","crud");
if($con){
    echo "connected successfully";
}

?>

this is my index.php page 

<?php
include('db.php');
if(isset($_POST['submit']))
{
    $name = $_POST['name'];
    $sname = $_POST['surname'];
    $city = $_POST['city'];
    $state = $_POST['state'];
    $image = $_FILES['image']['name'];
    $query = "INSERT INTO `table`(`name`, `surname`, `city`, `state`, `image`) 
               VALUES ('$name','$sname','$city','$state','$image')";
    header("location:fetch.php");
    if(mysqli_query($con,$query))
    {
        move_uploaded_file($_FILES['image']['tmp_name'],"image/$image");
    }


}

?>



<html>
    <head>
    <title> crud operation </title>
    </head>

    <body>
        <table border=1px; solid;>
        <form method = "POST" action = "" enctype = "multipart/form-data">
            <label> name </label>
            <input type = "text" name = "name"/><br>

            <label> Surname </label>
            <input type = "text" name = "surname"/><br>

            <label> city </label>
            <input type = "text" name = "city"/><br>

            <label> state </label>
            <input type = "text" name = "state"/><br>

            <label> image </image>
            <input type= "file" name = "image"/>

            <tr>
            <td><input type = "submit" name= "submit" value="upload"/></td>
            </tr>

        </form>
        </table>
    </body>

</html>

这是我的fetch.php 

<?php 
include('db.php');
?>

<html>
<table align = "center" border = 1px; >
    <tr><th>id</th>
        <th>name</th>
        <th>surname</th>
        <th>city</th>
        <th>state</th>
        <th>image</th>
        <th colspan = "2">Action</th>
    </tr>
    <?php
        $query ="SELECT * FROM `table`";

        $fire = mysqli_query($con,$query)or die("could not fetch". mysqli_error($con));

        while($row = mysqli_fetch_array($fire))
        {   
    ?>
        <tr><td><?php echo $row['id'];?></td>
            <td><?php echo $row['name'];?></td>
            <td><?php echo $row['surname'];?></td>
            <td><?php echo $row['city'];?></td>
            <td><?php echo $row['state'];?></td>

            <td><img = src = "image/<?php echo $row['image'];?>" style="width:100px; height:100px;" > </td>
            <td><a href ='update.php? id=<?php echo $row['id']?> &name=<?php echo $row['name']?> &sname=<?php echo $row['surname']?>&city=<?php echo $row['city']?>
                                      &state=<?php echo $row['state']?>
                                      &image=<?php echo $row['image']?>
                                     '/> EDIT </a></td>

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

        </tr>


    <?php } ?>
        </table>
        </html>

这是我的update.php页面问题，图片已成功更新，但未从文件夹中删除旧图片。

 <?php
        include('db.php');
        if(isset($_POST['UPDATE']))
        {   
        $id = $_REQUEST['id'];
        $name = $_REQUEST['name'];
        $surname = $_REQUEST['sname'];
        $city = $_REQUEST['city'];
        $state = $_REQUEST['state'];
        $image = $_FILES['image']['name'];

        $query = "UPDATE `table` SET `name`='$name' ,`surname`='$surname' ,
                 `city`='$city',`state`='$state',`image`='$image'  WHERE id = '$id' ";
            $fire = mysqli_query($con,$query);

            move_uploaded_file($_FILES['image']['tmp_name'],"image/$image");
            unlink('image/$image');
            header('location:fetch.php');

        }
        ?>

         <html>
            <head>
            <title> crud operation </title>
            </head>

            <body>
                <table  border=1px:solid;>
                <form  method = "POST" action = "#" enctype = "multipart/form-data">
                    <label> name </label>
                    <input type = "text" name = "name" value = <?php echo "$_REQUEST[name]";?> /> <br>

                    <label> Surname </label>
                    <input type = "text" name = "surname" value = <?php echo "$_REQUEST[sname]";?> /><br>

                    <label> city </label>
                    <input type = "text" name = "city" value= <?php echo "$_REQUEST[city]";?> /><br>

                    <label> state </label>
                    <input type = "text" name = "state" value= <?php echo "$_REQUEST[state]";?> /><br>

                    <label> image </image>
                    <input type= "file" name = "image" value= <?php echo "$_REQUEST[image]";?> /><br>

                    <tr>
                    <td><input type = "submit" name= "UPDATE" value="UPDATE"/></td>
                    </tr>

                </form>
                </table>
            </body>

        </html>

我想从表中更新任何图像时从文件夹中删除旧图像。但是这段代码只更新数据库中的文件以及在文件夹中保存新图像，但不会从文件夹中删除旧图像，因为我是初学者，在这里我知道专家可以让我理解我的代码有什么问题并提出一些更好的代码。请提前帮助和感谢。

1 个答案:

答案 0 :(得分：0)

检查下面的代码，我已经更新了一些代码，特别是我新增了隐藏字段旧图像，然后获取该值并删除图像的一个条件是否在编辑时上传。 另请更改查询中的表，因为我们不能使用表名table，因为它是保留关键字。

<?php
    include('db.php');
    if(isset($_POST['UPDATE']))
    {   
    $id = $_REQUEST['id'];
    $name = $_REQUEST['name'];
    $surname = $_REQUEST['sname'];
    $city = $_REQUEST['city'];
    $state = $_REQUEST['state'];

    print_r($_FILES);
    //check new image is upload
    if(isset($_FILES['image']['name']) && $_FILES['image']['name'] != '' ){

        $image = $_FILES['image']['name'];
        move_uploaded_file($_FILES['image']['tmp_name'],"image/$image");
        unlink('image/'.$_POST['old_image']);
    }
    else{
        $image = $_POST['old_image'];
    }

    $query = "UPDATE `user` SET `name`='$name' ,`surname`='$surname' ,
             `city`='$city',`state`='$state',`image`='$image'  WHERE id = '$id' ";
        $fire = mysqli_query($con,$query);

        header('location:fetch.php');

    }
    ?>

     <html>
        <head>
        <title> crud operation </title>
        </head>

        <body>
            <table  border=1px:solid;>
            <form  method = "POST" action = "#" enctype = "multipart/form-data">
                <label> name </label>
                <input type = "text" name = "name" value = <?php echo "$_REQUEST[name]";?> /> <br>

                <label> Surname </label>
                <input type = "text" name = "surname" value = <?php echo "$_REQUEST[sname]";?> /><br>

                <label> city </label>
                <input type = "text" name = "city" value= <?php echo "$_REQUEST[city]";?> /><br>

                <label> state </label>
                <input type = "text" name = "state" value= <?php echo "$_REQUEST[state]";?> /><br>

                <label> image </image>
                <input type= "hidden" name = "old_image" value= <?php echo "$_REQUEST[image]";?> /><br>        
                <input type= "file" name = "image"  /><br>

                <tr>
                <td><input type = "submit" name= "UPDATE" value="UPDATE"/></td>
                </tr>

            </form>
            </table>
        </body>

    </html>
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