storing image in mysql using php

Question

sugikrish 0 Junior Poster in Training

14 Years Ago

Hi frnds,

I want to store images in database and retrieve. I can successfully stored. But i dont know how retrieve it. If any one know pls help me. I paste my code here.

fimg.php

<body>
<form enctype="multipart/form-data" action="insert.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>
</body>

insert.php

<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "simg";

$link = mysql_connect($host, $username, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}


mysql_select_db ($database);  
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

      
      $tmpName  = $_FILES['image']['tmp_name'];  
       
     
      $fp      = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);
      



      $query = "INSERT INTO img";
      $query .= "(image) VALUES ('$data')";
      $results = mysql_query($query, $link);
      

      print "Thank you, your file has been uploaded.";
      
}
else {
   print "No image selected/uploaded";
}


mysql_close($link);

?>

seimg.php

<form action="seimg.php">
Enter image id:<input type="text" name="field" />
<input type="submit" name="submit" value="show" />

<?php

$username = "root";
$password = "";
$host = "localhost";
$database = "simg";

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

@mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$b =$_REQUEST['field'];

$query = mysql_query("SELECT * FROM img WHERE id='$b'");

$row = mysql_fetch_array($query);
$content = $row['image'];

?>

mysql php

Edited 14 Years Ago by nav33n because: Use [code] [/code] tags to wrap your code for easy readability.

7 Contributors
22 Replies
332 Views
2 Days Discussion Span
Latest Post 14 Years Ago Latest Post by sugikrish

shahramjaved 0 Newbie Poster

14 Years Ago

Please follow this complete tutorial for inserting and retrieving Image data from DB

http://www.phpriot.com/articles/images-in-mysql

vaultdweller123 32 Posting Pro

14 Years Ago

$query = mysql_query("SELECT * FROM img WHERE id='$b'");

$row = mysql_fetch_array($query);
echo "<img src='".$content = $row['image']."' />";

rajarajan2017

14 Years Ago

echo "<img src='". <?php echo $row['image']?>."' />";

do you stored the path of the image file in the database? if yes then you can use the above method.

Use $_POST or $_GET to the retrieve field.

vaultdweller123 32 Posting Pro

14 Years Ago

echo "<img src='". <?php echo $row['image']?>."' />";

lol of course its a parse error, i assume your inside a php tag coz you use echo, you cant nest another php inside php, instead you do it like this

echo "<img src='".$row['image']."' />";

REMEMBER

if you the code is inside the a php tag, use

echo "<img src='".row['image']."' />";

if not

<img src='<?=$row['image']?>' />

Edited 14 Years Ago by vaultdweller123 because: n/a

vaultdweller123 32 Posting Pro

14 Years Ago

theres nothing wrong in the echo, mind if i ask? where did you get $b =$_GET, and where actually the error is pointing?

Edited 14 Years Ago by vaultdweller123 because: n/a

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sugikrish 0 Junior Poster in Training · Answer 1 · 2010-06-29T12:46:18+00:00

hi frnd,

I use that link and do the same way. but i dont know how link one file to another in this program . successfully image downloaded, but not stored into database. can you help me?

sugikrish 0 Junior Poster in Training · Answer 2 · 2010-06-29T12:56:10+00:00

Hi..

1.
$query = mysql_query("SELECT * FROM img WHERE id='$b'");
2.

3.
$row = mysql_fetch_array($query);
4.
echo "<img src='".$content = $row."' />";

this coding working.But it not displayed image. It displayed some binary data. wat error in this ?

sugikrish 0 Junior Poster in Training · Answer 3 · 2010-06-29T12:59:12+00:00

Hi frnds,

I used $b =$_REQUEST; It shows error in this line."Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24" .

sugikrish 0 Junior Poster in Training · Answer 4 · 2010-06-29T13:40:51+00:00

I used this. But it shows an error."Parse error: parse error in C:\wamp\www\base\seimg.php on line 30"

line 30:echo "<img src='". <?php echo $row?>."' />";

sugikrish 0 Junior Poster in Training · Answer 5 · 2010-06-29T14:52:50+00:00

Hi frnds,

I did same wat u said. but it shows an error. I dont know. if u can pls say to me. I paste my code here.

seimg.php

<body>

<form action="seimg.php">
Enter image id:<input type="text" name="field" />
<input type="submit" name="submit" value="show" />

<?php

$username = "root";
$password = "";
$host = "localhost";
$database = "simg";

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

@mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$b =$_GET['field'];

$query = mysql_query("SELECT image FROM img WHERE id='$b'");

 $row = mysql_fetch_array($query);
 
 
 
echo "<img src='".$row['image']."' />";

   
      



?>  
   
  
</form>

it shows "Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24" this error.

sugikrish 0 Junior Poster in Training · Answer 6 · 2010-06-29T15:32:56+00:00

actually error pointing

$b =$_GET; this line only. i dont know wat error in this line. Any one know say to me.

rajarajan2017 · Answer 7 · 2010-06-29T15:43:07+00:00

You have to use $_GET if you mentioed the method as GET or otherwise use $_POST

sugikrish 0 Junior Poster in Training · Answer 8 · 2010-06-29T16:01:26+00:00

Hi frnd,

really thanks for reply. But again it shows an error. Can you help me for this?

rajarajan2017 · Answer 9 · 2010-06-29T17:05:31+00:00

What was the error?

sugikrish 0 Junior Poster in Training · Answer 10 · 2010-06-29T17:30:47+00:00

the same error:
"Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24"

vaultdweller123 32 Posting Pro · Answer 11 · 2010-06-29T17:49:17+00:00

you dont know what your doing bro... check you form if your using get or post

ayesha789 7 Posting Pro in Training · Answer 12 · 2010-06-29T17:58:35+00:00

Please use link method to save your pics. and just save reference of that link in your DB. It will improve efficiency

Jothe 0 Newbie Poster · Answer 13 · 2010-06-30T10:50:37+00:00

use the following to get the photo from where ur storing them. In this case i assume that the image is stored at the follder Photoz

<img src="photoz/<?php echo $imagename; ?>" width="200" height="300"alt="<?php echo $name; ?>"/>

rajarajan2017 · Answer 14 · 2010-06-30T12:26:58+00:00

I hope you missed the data somewhere, check your data with

Sample:

//check that $_POST['foo'] exists 
if (isset($_POST['foo'])) { 
    //$_POST['foo'] exists 
    $foo = $_POST['foo']; //we might perform input validation here 
} else { 
    //$_POST['foo'] does not exist 
    $foo = ''; //assign a default value 
} 

//use $foo 
echo $foo;

also check with empty($_POST). replace your variable instead of foo.

sugikrish 0 Junior Poster in Training · Answer 15 · 2010-06-30T12:42:53+00:00

Hi frnd,

I used this coding. but it didnt display the value. now i paste my code and output here. Otherwise if you can give full coding for retrieving image from database.

seimg.php

<body>

    <form action="seimg.php">
    Enter image id:<input type="text" name="foo" />
    <input type="submit" name="submit" value="show" />

    <?php

    $username = "root";
    $password = "";
    $host = "localhost";
    $database = "simg";

    @mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

    @mysql_select_db($database) or die("Can not select the database: ".mysql_error());



      //check that $_POST['foo'] exists

      if (isset($_POST['foo'])) {

      //$_POST['foo'] exists

      $foo = $_POST['foo']; //we might perform input validation here

      } else {

      //$_POST['foo'] does not exist

     $foo = ''; //assign a default value

      }


      //use $foo

      echo $foo;


    $query = mysql_query("SELECT * FROM img WHERE id='".$foo."'");
$row = mysql_fetch_array($query);
$content = $row['image'];

header('Content-type: image/jpg');
echo $content;


    ?>

       </form>


</body>
</html>

Output:

[url]http://localhost/base/seimg.php?foo=1&submit=show[/url]

shadiadiph 0 Junior Poster · Answer 16 · 2010-06-30T18:52:31+00:00

I strongly recommend against storing images in a database store the file name with extension ie picture.gif and save the actual picture to a directory on the server I would resize it first using php, then it is easy to call. Saved images in a database it takes up alot of space also writing the script to read them from a database is really hard.

sugikrish 0 Junior Poster in Training · Answer 17 · 2010-07-01T11:11:36+00:00

Hi frnd,

Actually i write program for storing image path in database and retrieve the path then display image. if you know the code pls send me.