blocks|key|2157333|text|我想你应该重新考虑一下你在做什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2157334|为什么要将文件存储在mysql数据库中？我的意思是这有什么意义呢？|2157335|文件系统是用于存储文件的专家数据库。所以最好不要把你的文件放在一个普通的数据库里，而是放在文件系统里。|2157336|图像可以在目录(0102003.gif)中使用数字命名，并可以通过mysql记录中的文件名进行寻址。|2157337|如果你真的需要在数据库中存储文件，mysql不是一个合适的工具。|2157338|看看mongoDB吧。|2157339|PS:+mysql接口已弃用，请使用mysqli或PDO。|2157340|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|S|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|V|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

I guess you should re-think what you are doing.

Why do you store files in a mysql database? I mean what is the point in that? 

File system is an expert database for storing files. 
So better do not keep your files in a normal database but in the file system.

Images can be named with numbers in directories (0102003.gif) and can be addressed via the filenames in your mysql records. 

If you really need file storage in a database, mysql is not the tool for that. 

Have a look at mongoDB. 

PS: mysql interface is deprecated, please use mysqli or PDO.

blocks|key|5686642|text|要在php网站上存储图片，您只需将文件放在一个目录中，并将文件名保存在mySQL数据库中。祝你好运！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5686643|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

To store images on a php website, you just have to put the files in a directory and save the filenames on the mySQL database. Goodluck!

blocks|key|4269332|text|试试这段代码，它会对你有帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4269333|<?php
++++mysql_connect("localhost",+"root","")+or+die+("could+not+connect+to+the+server");
++++mysql_select_db("database1")+or+die+("that+database+could+not+be+found");

++++$file+=+$_FILES['image']['tmp_name'];

++++$image=+addslashes(file_get_contents($_FILES['image']['tmp_name']));
++++$image_name+=+addslashes($_FILES['image']['name']);
++++$image_size+=+getimagesize($_FILES['image']['tmp_name']);

++++mysql_query("INSERT+INTO+table1+(id,image)+VALUES+('1','{$image}')");++++++++
?>|code-block|syntax|javascript|4269334|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try This code it will help you. 

<pre><code>&lt;?php
 mysql_connect("localhost", "root","") or die ("could not connect to the server");
 mysql_select_db("database1") or die ("that database could not be found");

 $file = $_FILES['image']['tmp_name'];

 $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
 $image_name = addslashes($_FILES['image']['name']);
 $image_size = getimagesize($_FILES['image']['tmp_name']);

 mysql_query("INSERT INTO table1 (id,image) VALUES ('1','{$image}')"); 
?&gt;
</code></pre>

blocks|key|5686712|text|在表单声明中需要enctype=multipart/+form+-data。并通过$_FILES变量而不是$_POST变量访问文件。像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5686713|<form+action="testimage1.php"+method="post"+enctype="multipart/form-data">+<input+name="uploadimage"+type="file"+/>+</form>+<?php+$filename+=+$_FILES['uploadimage']['tmp_name'];+?>|code-block|syntax|javascript|5686714|5686715|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

You need enctype=multipart/form-data in your form declaration. And access the file through the $_FILES variable instead of the $_POST variable. Like:

<pre><code>&lt;form action="testimage1.php" method="post" enctype="multipart/form-data"&gt; &lt;input name="uploadimage" type="file" /&gt; &lt;/form&gt; &lt;?php $filename = $_FILES['uploadimage']['tmp_name']; ?&gt;
</code></pre>

<hr>

blocks|key|4269401|text|有两种不同的方法可以插入和检索图像：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4269402|4269403|+方法1:|blockquote|offset|length|style|BOLD|4269404|4269405|4269406|将整个图像以longblob类型存储在数据库中，并从数据库中检索它。|CODE|4269407|插入图片：|4269408|<?php
//Get+uploaded+file+data
$img_tmp_name++++=+$_FILES['image']['tmp_name'];
$img_name++++++++=+$_FILES['image']['name'];
$img_size++++++++=+$_FILES['image']['size'];
$img_type++++++++=+$_FILES['image']['type'];
$img_error+++++++=+$_FILES['image']['error'];
//Processed+uploaded+img+data
if($img_error+>+0)
{
++++die('Upload+error+or+No+img+uploaded');
}
$handle+=+fopen($img_tmp_name,+"r");
$content+=+fread($handle,+$img_size);
fclose($handle);
$encoded_content+=+addslashes($content);

$query="insert+into+user_info+(img_content,img_type)+values('".$encoded_content."','".$img_type."')";
$ex=mysqli_query($conn,$query);
if($ex){
++++echo+"image+uploaded";
}
else{
++++echo+"image+not+uploaded ".mysqli_error($conn);
}
?>|code-block|syntax|javascript|4269409|检索图片：|4269410|您可以通过两种方法获取图片，第一种是使用处理程序PHP脚本返回图片数据，第二种方法如下：|4269411|<?php
++++$query="select+*+from+user_info+WHERE+id='".$id."'";
++++$ex=mysqli_query($conn,$query);
++++$row=mysqli_fetch_array($ex)
?>++++++++++++++
<img+src="<?php+echo+"data:'".$row['img_type']."';base64,".base64_encode($row['img_content']);+?>"+/>|4269412|但如果图像大小较大，则不推荐使用方法1。|4269413|4269414|+方法2:|4269415|4269416|4269417|在此方法中，我们只将图像的路径存储在数据库中。我们将图像存储在一个目录中。|4269418|4269419|<?php
++++$target_dir+=+"image/";
++++$target_file+=+$target_dir+.+basename($_FILES["image"]["name"]);
++++move_uploaded_file($_FILES["image"]["tmp_name"],+$target_file);

++++$query="insert+into+user_info+(img_path)+values('".$target_file."')";
++++$ex=mysqli_query($conn,$query);
++++if($ex){
++++++++echo+"image+uploaded";
++++}
++++else{
++++++++echo+"image+not+uploaded ".mysqli_error($conn);
++++}
?>|4269420|4269421|<?php
++++$query="select+*+from+user_info+WHERE+id='".$id."'";
++++$ex=mysqli_query($conn,$query);
++++$row=mysqli_fetch_array($ex)
?>
<img+src="<?php+echo+$row['img_path'];+?>"+alt=""/>|4269422|entityMap^0|0|0|1|4|0|0|0|6|8|0|0|5|0|0|0|5|0|0|0|0|0|1|4|0|0|0|0|0|5|0|0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|1K|8|@$F|1L|G|1M|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1P|8|@$F|1Q|G|1R|H|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1S|8|@$F|1T|G|1U|H|I]]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1V|8|@]|9|@]|A|$T|U]]|$1|V|3|W|5|6|7|1W|8|@$F|1X|G|1Y|H|I]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|Z|3|10|5|S|7|20|8|@]|9|@]|A|$T|U]]|$1|11|3|12|5|6|7|21|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|22|8|@]|9|@]|A|$]]|$1|14|3|15|5|E|7|23|8|@$F|24|G|25|H|I]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|26|8|@]|9|@]|A|$]]|$1|17|3|-4|5|6|7|27|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|28|8|@]|9|@]|A|$]]|$1|1A|3|P|5|6|7|29|8|@$F|2A|G|2B|H|I]]|9|@]|A|$]]|$1|1B|3|1C|5|S|7|2C|8|@]|9|@]|A|$T|U]]|$1|1D|3|W|5|6|7|2D|8|@$F|2E|G|2F|H|I]]|9|@]|A|$]]|$1|1E|3|1F|5|S|7|2G|8|@]|9|@]|A|$T|U]]|$1|1G|3|-4|5|6|7|2H|8|@]|9|@]|A|$]]]|1H|$]]

There are 2 different ways by which you can insert and retrieve images:

<blockquote>
 Method 1:
</blockquote>

Store entire image in the database itself in <code>longblob</code> type and retrieve it from the database.

Insert image:

<pre><code>&lt;?php
//Get uploaded file data
$img_tmp_name = $_FILES['image']['tmp_name'];
$img_name = $_FILES['image']['name'];
$img_size = $_FILES['image']['size'];
$img_type = $_FILES['image']['type'];
$img_error = $_FILES['image']['error'];
//Processed uploaded img data
if($img_error &gt; 0)
{
 die('Upload error or No img uploaded');
}
$handle = fopen($img_tmp_name, "r");
$content = fread($handle, $img_size);
fclose($handle);
$encoded_content = addslashes($content);

$query="insert into user_info (img_content,img_type) values('".$encoded_content."','".$img_type."')";
$ex=mysqli_query($conn,$query);
if($ex){
 echo "image uploaded";
}
else{
 echo "image not uploaded&lt;br/&gt;".mysqli_error($conn);
}
?&gt;
</code></pre>

Retrieve image:

You can retrieve an image by 2 methods, first by using a handler PHP script to return the image data and the second one as follows:

<pre><code>&lt;?php
 $query="select * from user_info WHERE id='".$id."'";
 $ex=mysqli_query($conn,$query);
 $row=mysqli_fetch_array($ex)
?&gt; 
&lt;img src="&lt;?php echo "data:'".$row['img_type']."';base64,".base64_encode($row['img_content']); ?&gt;" /&gt;
</code></pre>

But method 1 is not a recommended way if the image size will be large.

<blockquote>
 Method 2:
</blockquote>

In this method, we only store the path of the image in the database. We store the images in a directory.

Insert image:

<pre><code>&lt;?php
 $target_dir = "image/";
 $target_file = $target_dir . basename($_FILES["image"]["name"]);
 move_uploaded_file($_FILES["image"]["tmp_name"], $target_file);

 $query="insert into user_info (img_path) values('".$target_file."')";
 $ex=mysqli_query($conn,$query);
 if($ex){
 echo "image uploaded";
 }
 else{
 echo "image not uploaded&lt;br/&gt;".mysqli_error($conn);
 }
?&gt;
</code></pre>

Retrieve image:

<pre><code>&lt;?php
 $query="select * from user_info WHERE id='".$id."'";
 $ex=mysqli_query($conn,$query);
 $row=mysqli_fetch_array($ex)
?&gt;
&lt;img src="&lt;?php echo $row['img_path']; ?&gt;" alt=""/&gt;
</code></pre>

im trying to upload an image for a member profile and store it in a database using php then retrieve it but its does not work with me 

this is what im trying for inserting the image :

<pre><code>&lt;html&gt;
&lt;head&gt;
&lt;title&gt; Upload an image &lt;/title&gt;
&lt;/head&gt;
&lt;body&gt;
&lt;form action="index.php" method="POST" enctype="multipart/from/data"&gt;
 File:
 &lt;input type="file" name="image" &gt; &lt;input type="submit" value="upload"&gt;
&lt;/form&gt;
&lt;/body&gt;
&lt;/html&gt;
&lt;?php

 mysql_connect("localhost", "root","") or die ("could not connect to the server");
 mysql_select_db("project") or die ("that database could not be found");

 $file = $_FILES['image']['tmp_name'];

 if (!isset($file))
 echo "please select file";
 else
 {

 $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
 $image_name = addslashes($_FILES['image']['name']);
 $image_size = getimagesize($_FILES['image']['tmp_name']);

 if($image_size == FALSE)
 echo "that not image ";

 else
 {

 if (!$insert= mysql_query("INSERT INTO user_info (image) VALUES ('$image')"))
 echo "Problem";

 else
 {

 echo "image: &lt;p /&gt; your image &lt;p /&gt;&lt;img src='view.php?id="id"'&gt;";
 }
 }


 }
</code></pre>

and this for retrieving the image

<pre><code>&lt;?php

 // do some validation here to ensure id is safe
 mysql_connect("localhost", "root","") or die ("could not connect to the server");
 mysql_select_db("project") or die ("that database could not be found");
 $id = addslashes($_REQUEST['id']);

 $image = mysql_query("SELECT * FROM user_info WHERE id='$id'");
 $image = mysql_fetch_assoc($image);
 $image = $image['image'];

 header("Content-type: image/jpeg");
 echo $image;
?&gt;
</code></pre>

how insert and retrieve images to and from database using php

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试上传会员资料的图片，并使用php将其存储在数据库中，然后检索它，但它对我不起作用这就是我尝试插入图像的方法：<html><head><title> Upload an image </title></head><body><form action="index.php" method="POST" encty...

问如何使用php在数据库中插入和检索图像
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