Display image with description

Jene200

Hi,

I read the following article and i was trying to use it on my webpage.

http://www.phpbuilder.com/columns/florian19991014.php.

As per the script i created a table binary_data.

CREATE TABLE binary_data (
id INT(4) NOT NULL AUTO_INCREMENT PRIMARY KEY,
description CHAR(50),
bin_data LONGBLOB,
filename CHAR(50),
filesize CHAR(50),
filetype CHAR(50)
);

In the next step i copied the store.php

<?php

// store.php3 - by Florian Dittmer <dittmer@gmx.net>
// Example php script to demonstrate the storing of binary files into
// an sql database. More information can be found at http://www.phpbuilder.com/
?>

<html>
<head><title>Store binary data into SQL Database</title></head>
<body>

<?php
// code that will be executed if the form has been submitted:

if ($submit) {

// connect to the database
// (you may have to adjust the hostname,username or password)

MYSQL_CONNECT("localhost","root","password");
mysql_select_db("binary_data");

$data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));

$result=MYSQL_QUERY("INSERT INTO binary_data (description,bin_data,filename,filesize,filetype) ".
    "VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");

$id= mysql_insert_id();
print "<p>This file has the following Database ID: <b>$id</b>";

MYSQL_CLOSE();

} else {

// else show the form to submit new data:

<form method="post" action="<?php echo $PHP_SELF; ?>" enctype="multipart/form-data">
File Description:<br>
<input type="text" name="form_description"  size="40">
<input type="hidden" name="MAX_FILE_SIZE" value="1000000">
<br>File to upload/store in database:<br>
<input type="file" name="form_data"  size="40">
<p><input type="submit" name="submit" value="submit">
</form>

<?php

}

</body>
</html>

It displays a page with a form for uploding images and description.

Next to get image i copied the getdat3.php.

<?php

// getdata.php3 - by Florian Dittmer <dittmer@gmx.net>
// Example php script to demonstrate the direct passing of binary data
// to the user. More infos at http://www.phpbuilder.com
// Syntax: getdata.php3?id=<id>

if($id) {

// you may have to modify login information for your database server:
@MYSQL_CONNECT("localhost","root","password");

@mysql_select_db("binary_data");

$query = "select bin_data,filetype from binary_data where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");

Header( "Content-type: $type");
echo $data;

};
?>

and i called the following from a html page and it worked fine for me.

But i need to display the images with description.I tried all the ways i know .But didn't worked.I am not good in php,just started learning it..

Please help me to display the image with description.

If anybody have any sample script like this,please send to me

Thanks

Jene

bastien

Since the image is coming from the DB its needs the image header sent before the broswer can properly display the image...the trick here is to call that page that show's the image as the source for the image

try this:

<?php

// show_desc.php3 

if($id) {

// you may have to modify login information for your database server:
@MYSQL_CONNECT("localhost","root","password");

@mysql_select_db("binary_data");

$query = "select description from binary_data where id=$id";
$result = @MYSQL_QUERY($query);

$description = @MYSQL_RESULT($result,0,"description  ");

echo $description ;
echo "<br>";
echo "<img src=\"getdat3.php?id=$id\">";


};
?>

Jene200

Hi Bastien,

Thank you so much.But i tried the code and i got a blank page.

echo "<img src=\"getdat3.php?id=$id\">";

I changed $id to 1 and tried,but i got a blank page again.

I have a database with image and image description.I wated to display both in a table.I mean all the images with description i need to display.If i have 10 images in my database i need to display all images with description.Is there any simple way to do that..?

If i am using the getdat3.php, i need to call the code everytime i need to display image.If i have 10 images in my database, i need to call them like

img src="getdata.php?id=1">
img src="getdata.php?id=2">
.
.
.
img src="getdata.php?id=10">

I need to display each image with its own description.

Please help me to solve my problem.

I am not good in PHP.I wasted three days to figure it out the problem.Then i posted my question here.

Thanks in advance.

Jene.

bastien

okay, that means most likely the page got nothing from the db. Time to start debugging. Begin with checking the sql statement to make sure that its correct, echoing out the statement means you can copy it and paste it into the db gui of your choice to see if the statement works. The $_REQUEST['id'] change is to accomodate the changes in PHP which don't allow variables to be created from form elements.

try this:

<?php 

// show_desc.php3 

if($_REQUEST['id']) { 
  $id = $_REQUEST['id'];

// you may have to modify login information for your database server: 
@MYSQL_CONNECT("localhost","root","password"); 

@mysql_select_db("binary_data"); 

$query = "select description from binary_data where id=$id"; 

echo $query."<br>";

$result = @MYSQL_QUERY($query); 

$description = @MYSQL_RESULT($result,0,"description  "); 

echo $description ; 
echo "<br>"; 
echo "<img src=\"getdat3.php?id=$id\">"; 


}; 
?>

Jene200

Thank you for you reply.

I tried it and still getting a blank page.

As you said i checked the select statement in databse using phpmyadmin.

select description from binary_data where id=$id

When i tried like this it got the following error

#1054 - Unknown column '$id' in 'where clause'

Then i tried

select description from binary_data where id=1

It showed me the description.

Is this means the problem with where clause,thats why i am getting blank page..?

Thanks,
Jene.

bastien

The @ symbol suppresses the errors and lets the code keep running. Remove it to get error messages.

It also looks like the value of $id is not being passed. I assumed from the code that this value is being passed from somewhere else(ie another page).

Jene200

Thank you bastien.

As you said the "value of $id is not being passed", i tried the following code and it showed me both description and photo.

<?php

// show_desc.php3

if($id=id) {

// you may have to modify login information for your database server:
@MYSQL_CONNECT("localhost","user","pass");

@mysql_select_db("dbname");

$query = "select description from binary_data where id=id";
$result = @($query);

$description = @MYSQL_RESULT($result,0,"description");

echo $description;
echo "<br>";

echo "<img src=\"getdat3.php?id=id\">";

};
?>

It also worked when i gave $id=$id instead of giving id.

But i have one problem. It displays only the first recod from the database.How can i display all the records(both image and description together)..?

I tried the following select statment in database and it showed me all descriptions which i stored in the database.

SELECT description
FROM binary_data
WHERE id = id

Thanks,
Jene.

bastien

build a loop to move thru the data

<?php

// show_desc.php3

if($id=id) {

// you may have to modify login information for your database server:
@MYSQL_CONNECT("localhost","user","pass");

@mysql_select_db("dbname");

$query = "select description from binary_data ";
$result = @MYSQL_QUERY($query);

while ($rows = MYSQL_FETCH_ARRAY($result))
{
echo $rows['description'];
echo "<br>";

echo "<img src=\"getdat3.php?id=".$rows['id']."\">";


};
?> 

to make it look nicer, play with putting table tags in the code

Jene200

Thank you so much bastien .

But now i have another problem.It displays both description and Image space,but the image is not showing properly,i mean it showing read X there.I guess its not getting it correct path. (if the path is not correct the image shows one red X mark like that way it displays).

I clicked on the pictures property and it showed me like that

Address
(URl) : http://www.site.com/getdat3.php?id=

So i think the image is not getting its id value.

I am confused what to do.Please help me.

Thanks
Jene.

bastien

can you post the structure of your tables...and how you are passing the initial id (if any)

Jene200

I am posting my table structure below.

Field Type Attributes Null Default Extra

id int(4) - No - auto_increment

description varchar(255) - Yes NULL

bin_data longblob BINARY Yes NULL

filename varchar (50) - Yes NULL

filesize varchar (50) - Yes NULL

filetype varchar (50) - Yes NULL

('-" means empty)

I am directly using this code to get both description and image.

 
<?php 

// show_desc.php3 

if($id=id) { 

// you may have to modify login information for your database server: 
@MYSQL_CONNECT("localhost","user","pass"); 

@mysql_select_db("dbname"); 

$query = "select description from binary_data "; 
$result = @MYSQL_QUERY($query); 

while ($rows = MYSQL_FETCH_ARRAY($result)) 
{ 
echo $rows['description']; 
echo "<br>"; 

echo "<img src=\"getdat3.php?id=".$rows['id']."\">"; 


}; 
?>

do we need some way to get the id value before executing the above code...?

Thanks ,
Jene.

bastien

here are the two working pages (tested my machine (xp, apache2, php4.3.9 mysql 4.01.x)

<?php 

// show_desc.php
require("conn.php");

  // you may have to modify login information for your database server

  $query = "select description, id from binary_data ";
  $result = connect($query);

  while ($rows = MYSQL_FETCH_ARRAY($result)) 
  { 
  echo $rows['description']; 
  echo "<br><br>";

  echo "<img src=\"show_image.php?id=".$rows['id']."\">\n";


}; 
?>

<?php
//show_image.php

require("conn.php");
//check to see if the id is passed
if(isset($_GET['id'])) {
  $id=$_GET['id'];


  $query = "select bin_data, filetype from binary_data where id=$id";
  //echo $query;
  $result = connect($query);
  $row = mysql_fetch_array($result);
  {
   $data = $row['bin_data'];
   $type = $row['filetype'];
  }
  if ($type=="pjpeg") $type = "jpeg";
  Header( "Content-type: $type");

  echo $data;

}
?>

hth

Jene200

Thank you so much bastien. Its working fine now.

I got an error here

$result = connect($query);

and changed that to

$result = @($query);

and it worked.

Thanks a lot for your help.

Jene.

Jene200

Hi,

I am trying to write a code to update the records in binary_data table. I tried so many ways but its not working.
I don't really know how to do that.
I think, first we need to display all the records and we need to select the one which we need to edit/update the record.

This is the code which i was trying to update the records in binary_data table.

update_record.php

<?php 

$id=$_GET['id'];

@MYSQL_CONNECT("localhost","user","pass");
@mysql_select_db("dbname");

$query=" SELECT * FROM binary_data WHERE id='id'";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();

$i=0;
while ($i < $num) {
$description=mysql_result($result,$i,"description");
$image=mysql_result($result,$i,"bin_data");
$name=mysql_result($result,$i,"name");

++$i;
}
?>

<form method="post" action="update.php">
<input type="hidden" value="<? echo $id; ?>" name="id">
Description: 
<input type="text" value="<? echo $description;?>" name="description"><br>
Image: <input type="file" name="data" value="<? echo $bin_data; ?>"><br>
Filetype: <input type="hidden" name="name" value="<? echo $filename; ?>"><br>
Filesize: <input type="hidden" name="size" value="<? echo $filesize; ?>"><br>
Filesize: <input type="hidden" name="type" value="<? echo $filetype; ?>"><br>
Name: <input type="text" name="name" value="<? echo $name; ?>"><br>
<input type="Submit" value="Update">
</form>

update.php

<?php 


$id=$_GET['id'];
$description=$_POST['description'];
$bin_data=$_POST['bindata'];
$filename=$_POST['filename'];
$filesize=$_POST['filesize'];
$filetype=$_POST['filetype'];
$category=$_POST['category'];
$adapt_flag=$_POST['adapt_flag'];
$name=$_POST['name'];

@MYSQL_CONNECT("localhost","user","pass");
@mysql_select_db("dbname");
$query="UPDATE binary_data SET description='$description', bindata='bindata', filename='$filename', filesize='$filesize', filetype='$filetype', name='$name' WHERE id='$id'";
mysql_query($query);
echo "Record Updated";
mysql_close();
?>

The following is my table structure.

id INT(4) NOT NULL AUTO_INCREMENT PRIMARY KEY,
description varchar(255),
bin_data LONGBLOB,
filename CHAR(50),
filesize CHAR(50),
filetype CHAR(50),
name varchar(50)

Please help to update the above table records.

Thanks,
Jene.

bastien

what do you mean by update records? does this include the image?

Jene200

Sorry bastien.

I just want to update desciption and name.

This is the new code.

<html><head><title></title></head>
<body>
<?
@MYSQL_CONNECT("localhost","user","pass"); 
@mysql_select_db("dbname"); 
$result = mysql_query( "select id,name,description from binary_data" )
or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
print "There are $num_rows records.<P>";
print "<table width=400 border=1>\n";
while ($get_info = mysql_fetch_row($result)){ 
print "<tr>\n";
foreach ($get_info as $field) 
print "\t<td><font face=arial size=1/>$field</font></td>\n";
print "</tr>\n";
}
print "</table>\n";
mysql_close();
?>
<br>
<form method="POST" action="change_form.php">
<pre>
Enter Line Number to Edit: <input type="text" name="id" size="5">
<input type="submit" value="Submit"><input type="reset">
</pre>
</form>
</body>
</html>

change_form.php

<?php  

<?
$id=$_POST['id'];
@MYSQL_CONNECT("localhost","user","pass"); 
@mysql_select_db("dbname"); 

$query=" SELECT * FROM binary_data WHERE id='$id'";
$result=mysql_query($query);
$num=mysql_num_rows($result);

$i=0;
while ($i < $num) {
$description=mysql_result($result,$i,"description");
$name=mysql_result($result,$i,"name");
?>
<table width="300" cellpadding="10" cellspacing="0" border="2" align="center">
<tr  valign="top">
<td align="center" colspan="1" rowspan="1" bgcolor="#FFFFFF">
<h3>Edit and Submit</h3>
<form action="change_record.php" method="post">
<input type="hidden" name="id" value="<? echo "$id" ?>">
Description: <input type="text"  size="100" name="description" value="<? print "$description"?>"> 
Name?:    <input type="text" name="name" value="<? echo "$name"?>"><br>
<input type="Submit" value="Update">
</form>
</td></tr></table>

<?
++$i;
}
?>

change_record.php

<?
$description = $_POST['description'];
$name = $_POST['name'];

@MYSQL_CONNECT("localhost","user","pass"); 
@mysql_select_db("dbname");
mysql_query(" UPDATE binary_data SET description='$description', name='$name'  WHERE id='$id'");
commit;
echo "Record Updated";
mysql_close();
?>

First two part is working fine.But its not updating the records in the database.But its showing record updated.

I have one more question related to displaying image.

Is there any way i can display 5-6 images per page.Using " next prev last".

Thanks,
Jene.

bastien

try this, it'll report any errors back to you so you can fix them:

<?
$description = $POST['description'];
$name = $POST['name'];

@MYSQL_CONNECT("localhost","user","pass");

@mysql_select_db("dbname");

$sql = "UPDATE binary_data SET description='$description', name='$name' WHERE id='$id'";

$result = mysql_query($sql) or die ("Can't complete query because ".mysql_error());

if (($result) && (mysql_errno == 0))
echo "Record Updated";
}//end if

mysql_close();
?>

paging: http://hotscripts.com/Detailed/25949.html

Jene200

Thanks bastien.

I tried the code which you gave me.Its not showing any error.But after executing the code, it replaces the field value which i am trying to update, with null value. I mean the field is empty after updating the field.

Please help me.
Thanks,
Jene.

bastien

I got it to work fine with

<?

$description = $_POST['description'];
$name        = $_POST['name'];
$id          = $_POST['id'];


@MYSQL_CONNECT("localhost","user","pass");
@mysql_select_db("dbname");

$sql = "UPDATE binary_data SET description='$description', filename='$name' WHERE id='$id'";
 echo $sql;
$result = connect($sql) or die ("Can't complete query because ".mysql_error());

if ($result) {
echo "Record Updated";
}//end if

mysql_close();
?>

and ensure that you haven't change the field names. I notices from the original posts the name field in these later posts was originally filename

Jene200

Sorry for the name confusion bastien.I added new field name to the binary_data table.Thats why i gave

name='$name'

Now after execting the update statement i t showing me the following.

UPDATE binary_data SET description='asdad', name='abcd' WHERE id='13' Can't complete query because

I tried this in database and it worked fine.

I don't know whats wrong.
Please help me.

Thanks,
Jene.