php 使用输入文件类型编辑图像
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editing image by using input file type
提问by Mj Jam
I made a simple editing for to edit data in mysql, everything works fine except when I want to edit an input file type image it doesn't work, it doesn't give an error message it just doesn't edit anything and when I remove the input file type image it works. and by editing an image I mean entering a new image the will replace the old image.
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，它没有给出错误消息它只是不编辑任何东西，当我删除它工作的输入文件类型图像。并通过编辑图像我的意思是输入一个新图像将替换旧图像。
here is my code:
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> </td>
<td>
<input type="submit" name="submit" value="submit" />
</td>
</tr>
</table>
回答by Saranya Sadhasivam
In form enctype="multipart/form-data" is missing and in your form there is no type="file".
在表单 enctype="multipart/form-data" 中丢失，并且在您的表单中没有 type="file"。
Give the below code and try.
给出下面的代码并尝试。
<?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> </td>
<td><input type="submit" name="submit" value="submit" /></td>
</tr>
</table>
Moreover you should get the previous image value through sql and update if image is not chose while updating.
此外，如果更新时未选择图像，您应该通过 sql 获取先前的图像值并更新。
回答by Parfait
<tr>
<td>image</td>
<td><input type="file" name="image" ></td>
</tr>
回答by zkanoca
You have to use input:type=fileelement instead of input:type=textin order to handle image file using $_FILES. Or you cannot get the image file. So your if statement returns false and nothing happens.
您必须使用input:type=fileelement 而不是input:type=text为了使用$_FILES. 或者您无法获取图像文件。所以你的 if 语句返回 false 并且什么也没有发生。
<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>image preview</td>
<td><img src="photos/<?php echo $image ?>" /></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="submit" value="submit" /></td>
</tr>
</table>
</form>
