用图像创建一个SQL表

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

这是一篇文章，将告诉您所有有关在blob列中存储图像等内容的知识， 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-->
?>