CODEWITHSUNDEEP
phpmysqlsql
Otachan
Otachan

Reputation: 89

how to display an image from the database without using a directory

I looked up some tutorials from youtube, like 5 of them but they're all not working for me for some reason, I'm only getting broken image icon.

I've tried using a folder to get the images like img src="images/$row['picture'];

But what I want is to display it straight from the database.

If possible please send me a code for it, so I will know if it's my browser or my coding.

I copied these codes but it's not even working

https://vikasmahajan.wordpress.com/2010/07/07/inserting-and-displaying-images-in-mysql-using-php/

It doesn't show the image:

Upvotes: 1

Views: 768

Answers (1)

tinonetic
tinonetic

Reputation: 8036

It is generally discouraged to approach it that way. <-(Good to know)

...it can still be done though and your use case may call for it...

With full credit to @Andomar:

First you create a MySQL table to store images, like for example:

create table testblob (
    image_id        tinyint(3)  not null default '0',
    image_type      varchar(25) not null default '',
    image           blob        not null,
    image_size      varchar(25) not null default '',
    image_ctgy      varchar(25) not null default '',
    image_name      varchar(50) not null default ''
);

Then you can write an image to the database like:

$imgData = file_get_contents($filename);
$size = getimagesize($filename);
mysql_connect("localhost", "$username", "$password");
mysql_select_db ("$dbname");
$sql = sprintf("INSERT INTO testblob
    (image_type, image, image_size, image_name)
    VALUES
    ('%s', '%s', '%d', '%s')",
    mysql_real_escape_string($size['mime']),
    mysql_real_escape_string($imgData),
    $size[3],
    mysql_real_escape_string($_FILES['userfile']['name'])
    );
mysql_query($sql);

You can display an image from the database in a web page with:

$link = mysql_connect("localhost", "username", "password");
mysql_select_db("testblob");
$sql = "SELECT image FROM testblob WHERE image_id=0";
$result = mysql_query("$sql");
header("Content-type: image/jpeg");
echo mysql_result($result, 0);
mysql_close($link);

Upvotes: 1

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