php - 使用输入文件类型编辑图像

Question

我做了一个简单的编辑来编辑mysql中的数据，一切正常，除了当我想编辑输入文件类型图像时它不起作用，它不会给出错误消息它只是不编辑任何内容当我删除输入文件类型图像时它可以工作。 通过编辑图像，我的意思是输入新图像将替换旧图像。

这是我的代码:

<?php

require("db.php");

$id     = $_REQUEST['theId'];
$result = mysql_query("SELECT * FROM table WHERE id  = '$id'");
$test   = mysql_fetch_array($result);

$name   = $test['Name'] ;
$email  = $test['Email'] ;                  
$image  = $test['Image'] ;

if (isset($_POST['submit']))
{   
    $name_save  = $_POST['name'];
    $email_save = $_POST['email'];

    if (isset($_FILES['image']['tmp_name']))
    {
        $file       = $_FILES['image']['tmp_name'];
        $image      = addslashes(file_get_contents($_FILES['image']['tmp_name']));
        $image_name = addslashes($_FILES['image']['name']);

        move_uploaded_file($_FILES["image"]["tmp_name"],"photos/" . $_FILES["image"]["name"]);
        $image_save ="photos/" . $_FILES["image"]["name"];

        mysql_query("UPDATE table SET Name ='$name_save', Email  ='$email_save',Image ='$image_save' WHERE id = '$id'") or die(mysql_error()); 

        header("Location: index.php");
    }
}
?>



<form method="post">
    <table>
        <tr>
            <td>name:</td>
            <td>
                <input type="text" name="name" value="<?php echo $name ?>"/>
            </td>
        </tr>
        <tr>
            <td>email</td>
            <td>
                <input type="text" name="email" value="<?php echo $email ?>"/>
            </td>
        </tr>
        <tr>
            <td>image</td>
            <td>
                <input type="file" name="image" value="<?php echo $image ?>"/>
            </td>
        </tr>
        <tr>
            <td>&nbsp;</td>
            <td>
                <input type="submit" name="submit" value="submit" />
            </td>
        </tr>
    </table>

Answer 1

表单中缺少 enctype="multipart/form-data"，并且您的表单中没有 type="file"。

给出以下代码并尝试。

<?php
require("db.php");
$id =$_REQUEST['theId'];

$result = mysql_query("SELECT * FROM table WHERE id  = '$id'");
$test = mysql_fetch_array($result);

$name=$test['Name'] ;
$email= $test['Email'] ;                    
$image=$test['Image'] ;

if(isset($_POST['submit'])){    
$name_save = $_POST['name'];
$email_save = $_POST['email'];
$image_save=$image //Added if image is not chose from the form post

if (isset($_FILES['image']['tmp_name'])) {
$file=$_FILES['image']['tmp_name'];
$image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name= addslashes($_FILES['image']['name']);
move_uploaded_file($_FILES["image"]["tmp_name"],"photos/" . $_FILES["image"]["name"]);
$image_save ="photos/" . $_FILES["image"]["name"];
}
mysql_query("UPDATE table SET Name ='$name_save', Email  ='$email_save',Image ='$image_save' WHERE id = '$id'")
or die(mysql_error()); 
header("Location: index.php");      }
?>



<form method="post" enctype="multipart/form-data">
<table>
<tr>
<td>name:</td>
<td><input type="text" name="name" value="<?php echo $name ?>"/></td>
</tr>
<tr>
<td>email</td>
<td><input type="text" name="email" value="<?php echo $email ?>"/></td>
</tr>

<tr>
<td>image</td>
<td><input type="file" name="image" /></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input type="submit" name="submit" value="submit" /></td>
</tr>
</table>

此外，如果更新时没有选择图像，则应通过sql获取之前的图像值并更新。