create a sql table with image - php

Question

I want to use php to create one table in sql which has images set content of the image and then I want fetch this table in html.
someone knows how this work? or someone has see the example on line?
please help. thanks a lot.

Answer 1

Here' is an article that will tell you all you want to know regarding storing images and the likes of in blob columns http://www.phpriot.com/articles/images-in-mysql

Answer 2

Ok, I understand that you want to upload an image file into a table in db, through your web app, using php, and then again you want to display that same image from the table in your web app(html page).
I will show the example for MySQL, using simple PHP 5.3
Create a sample table, say "photo_table" :
CREATE TABLE photo_table (photo_name_col VARCHAR(30),photo_img_col LONGBLOB);
Now for uploading the image file through your php page use the following in your html code:
<form name="photoForm" method="post" ENCTYPE="multipart/form-data" action="../insertImage.php" >
<input type="text" name="uploadFileName">
<!--commenting as I was not able to publish this note<input type="file" name="uploadFile" />-->
<!--Some other input fields if you want-->
<input type="submit" value="Submit"/>
</form>
Now, once the form "photoForm" is submitted, the form values are passed to "insertImage.php", where we will have the following code to upload the photo and its name into a table in our MySQL db:
<?php
$con=mysql_connect("ipAddressOfDB","username","password");
if(!$con){
die('Could not connect:'.mysql_error());
}
mysql_select_db("db_name",$con);
$image = $_FILES['uploadfile']['tmp_name'];
$imageName = $_POST['uploadFileName'];
$fp = fopen($image, 'r');
$content = fread($fp, filesize($image));
$content = addslashes($content);
fclose($fp);
$sql="insert into photo_table values('$imageName','$content')";
mysql_query($sql,$con);
?>
For displaying the photo from table into your html page, I use the following approach.
Fetch the image from db and copy it to a temporary location in the server.
<?php
$con=mysql_connect("ipAddressOfDB","username","password");
mysql_select_db("db_name",$con);
$sql="select * from photo_table where photo_name_col ='$uploadFileName'";
$result=mysql_query($sql,$con);
if($row=mysql_fetch_array($result)){
$fileName="generated/".$uploadFileName;
file_put_contents($fileName, $row['photo']);
}
echo "<img src='" . $fileName . "' />"; <!--this will display the image-->
?>