Jump to content

Image browse & upload not displaying using php

Vinodpv

By Vinodpv
in PHP Coding Help

 Share

Recommended Posts

Vinodpv Newbie

Vinodpv

Posted
Posted

Hi,

 

Below is my image browse upload code , here I can insert image in database, but it is dispaly only icon instead of image, Please help me to display image...many many thanks in advance

 

index.php

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Logo Upload here..</title>
</head>
<body>
<form action="index.php" method="post" enctype="multipart/form-data">
<label>Product Name</label>
<input type="text" name="pname" />
<label>Product image</label>
<input type="file" name="image" >
<input type="submit" name="submit" value="Add Logo">
</form>
<?php
include "db_conexn.php";
if(!isset($_FILES["image"]["tmp_name"]))
echo "Please select an image..";
else
{
$image=addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$image_name=addslashes($_FILES["image"]["name"]);
$image_size=getimagesize($_FILES["image"]["tmp_name"]);

    if($image_size==FALSE)
        echo "It is not an image";
    else
    {
    $insert=mysql_query("insert into images values('','$image_name','$image')");
    echo "<img src='uploadsql.php'>";
    }
}
?>
</body>
</html>
 

uploadsql.php

<html>
<body>
<?php
$result=mysql_query("select * from images");
while($row=mysql_fetch_array($result))
{
header("Content-Type:image/jpeg");
$image=$row['image'];
echo $image;
}
?>
</html>

 

table

table : images

id name image 26 Chrysanthemum.jpg [bLOB - 858.8   KiB]

 

Link to comment
Share on other sites
jazzman1 Member

jazzman1

Posted
Posted

Can you post the output of:

<?php
$result=mysql_query("select * from images where id = 26");
while($row=mysql_fetch_assoc($result))
{
echo $row['image'].'<br />';
echo $row['name'];
}
?>

Use the forum code [ code ] tags when providing code.

Link to comment
Share on other sites
Ch0cu3r Newbie

Ch0cu3r

Posted
Posted (edited)

Don't store the images in the database, this makes things more complex later. You are best of saving the uploaded image to the filesystem, look at using move_uploaded_file to do this.

And then only save the image path in the database.

 

eg

$image_dir = 'images/'; // where images are to be stored

$image_name=mysql_real_escape_string($_FILES["image"]["name"]);
$image_path = $image_dir . $image_name; // image path, this path is saved in the database

// move the uploaded file, to the image dir
if(move_uploaded_file($_FILES["image"]["tmp_name"], $image_path))
{
    $insert=mysql_query("insert into images values('','$image_name','$image_path')");

   if(!$insert)
      echo 'Database error: ' . mysql_error());
}
else
{
    echo 'Unable to move '.$_FILES["image"]["name"].' to '.$image_dir;
}

Now to display the image you can echo the image path into the src attribute for the <img> tag, eg

<?php
$result=mysql_query("select * from images");
while($row=mysql_fetch_array($result))
{
    echo '<img src="'.$row['image'].'" />';
}
?>

Also dont use addslashes to sanitize user input use mysql_real_escape_string or better yet convert your code to mysqli or PDO and use prepared queries

Edited by Ch0cu3r
Link to comment
Share on other sites
Vinodpv Newbie

Vinodpv

Posted
  • Author
Posted

Don't store the images in the database, this makes things more complex later. You are best of saving the uploaded image to the filesystem, look at using move_uploaded_file to do this.

And then only save the image path in the database.

 

eg

$image_dir = 'images/'; // where images are to be stored

$image_name=mysql_real_escape_string($_FILES["image"]["name"]);
$image_path = $image_dir . $image_name; // image path, this path is saved in the database

// move the uploaded file, to the image dir
if(move_uploaded_file($_FILES["image"]["tmp_name"], $image_path))
{
    $insert=mysql_query("insert into images values('','$image_name','$image_path')");

   if(!$insert)
      echo 'Database error: ' . mysql_error());
}
else
{
    echo 'Unable to move '.$_FILES["image"]["name"].' to '.$image_dir;
}

Now to display the image you can echo the image path into the src attribute for the <img> tag, eg

<?php
$result=mysql_query("select * from images");
while($row=mysql_fetch_array($result))
{
    echo '<img src="'.$row['image'].'" />';
}
?>

Also dont use addslashes to sanitize user input use mysql_real_escape_string or better yet convert your code to mysqli or PDO and use prepared queries

thank you friend it is working good, image is uploading in folder and database both. Can u please tell me how to display it

Link to comment
Share on other sites
jazzman1 Member

jazzman1

Posted
Posted

If the binary string is stored in a database as expected, your script should work properly. Try my script and add the header function with content-type to it.

$result=mysql_query("select * from images where id = 26");
$row=mysql_fetch_assoc($result);
header("Content-Type:image/jpeg");
echo $row['image'];

Much better is to use application server store the images instead the database as Ch0cu3r already stated.It's also not recommended to use the old mysql extension for new development, use either the mysqli or pdo_mysql

Link to comment
Share on other sites
Vinodpv Newbie

Vinodpv

Posted
  • Author
Posted

index.php

<tr>
      <td height="31">logo Upload </td><td><input type="file" name="image" /></td>
</tr>

db.php

$image_dir = 'imagess/';
$image_name=mysql_real_escape_string($_FILES["image"]["name"]);
$image_path = $image_dir . $image_name;
if(!move_uploaded_file($_FILES["image"]["tmp_name"], $image_path))
{    echo 'Unable to move '.$_FILES["image"]["name"].' to '.$image_dir;
}
else{

        $insert = mysql_query("INSERT INTO `company`(comp_name,comp_address,branch,user_from,photo) VALUES ('$compName','$compAddress','$branch','$sub','$image_name')");

}

This is my working code .. image will save in database and folder, Can u please help me for updation  image using update query . And display this image ...

 

Link to comment
Share on other sites
Ch0cu3r Newbie

Ch0cu3r

Posted
Posted

Does your company table have an id field? That will be the field you'd reference the row in the WHERE clause for the UPDATE/SELECT query,

 

eg example for select

<?php

$company_id = intval(.. company id val..);

// get the row that matches the company id
$result=mysql_query("select * from company where id=$company_id");
if(mysql_num_rows($result) > 0) {
{
    $row=mysql_fetch_assoc($result);

    // output image path in the src attribute
    echo '<img src="'.$row['photo'].'" />';
}
?>
Link to comment
Share on other sites
This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
 Share
Go to topic listing
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.