Base 64 encode image. Upload twitter avatar - php

I'm trying to make an image upload form for twitter to change avatar. Twitter says the image has to be base 64 encoded before I upload it. So here's a sample
https://dev.twitter.com/docs/api/1/post/account/update_profile_image
How can I do this so far I've been getting errors
PHP Code:
<?php // process form data
if ($_POST['image']){
// post profile update
$post_data = array(
"image" => $_POST['image']
);
$post_data = base64_encode($post_data);
echo $post_data;
}
else{
echo
" <form action='post.php' method='POST' enctype='multipart/form-data'>
<br><input type='hidden' name='image' value='update_profile_image'>
<input type='file' name='image' size='30'/><br>
<input type='submit' value='Upload' class='button'/>
</form>";}
?>
Please help

From the looks of this, you aren't retrieving the image from the POST at all. The image itself will be available in $_FILES
// See what's in $_FILES
var_dump($_FILES);
// You need the temporary name of your image from $_FILES
$filedata = file_get_contents($_FILES['image']['tmp_name']);
$post_data = base64_encode($filedata);
// Now you should have a base64 ascii string
echo $post_data;

This should do it.
Move the uploaded file to a directory because you cannot get required results by encrypting the [tmp_name]
if ($_FILES['image']){
move_uploaded_file($_FILES["image"]["tmp_name"], "path" .basename($_FILES["image"]["name"]));
$meh = file_get_contents("path/" .basename($_FILES["image"]["name"]));
$results = base64_encode($meh);
}

Related

Post multiple URL

I want to post multiple images by posting the url's but I don't know how to do this because I am new to PHP.
<?PHP
if(isset($_POST['post_image']))
{
$image_url=$_POST['image_path'];
$data = file_get_contents($image_url);
$new = '../images/myimage.jpg';
$upload =file_put_contents($new, $data);
if($upload) {
echo "<img src='../images/myimage.jpg'>";
}else{
echo "Please upload only image files";
}
}
?>
The problem with your script is, that you are not saving the file anywhere. So you have no access to it when trying to return it to the client.
In order to save an image using a POST you need to make sure you have set the following enctype inside your form.
<form action="upload.php" method="post" enctype="multipart/form-data">
Then you can use this function to save the img
move_uploaded_file($_FILES["myImg"], "/path/to/imgDir")
And later show it to the client using:
echo "<img src='/path/to/imgDir/myImg'>"
For more information about the topic i advice you to consult the following URL
https://www.w3schools.com/php/php_file_upload.asp

php base64 iphone picture to webservice

i need your help please and i haven't found some answers for my problem.
I want to take a picture (or take it from the gallery) on the iphone/ipad and then encode the picture to a base64 string with php and send the base64 string to a webservice.
If i try it on the pc, everything is alright.
if i want to do this on the ipad, it seems like nothing or a incorrect base64 string is send to the webservice but i have no idea why or so?!
picture.php
<form action="picture.php?action=upload" method="post" enctype="multipart/form-data">
<input type="file" name="datei" accept="capture=camcorder">
<br/><br/>
<input type="submit" value="up">
</form>
<?
if(isset($_GET['action']))
{
$tmp_name = $_FILES["datei"]["tmp_name"];
$name = $_FILES["datei"]["name"];
$name = substr($name,0,-4);
$name.="_".time().".jpg";
move_uploaded_file($tmp_name, "upload/".$name);
$content = file_get_contents ( "upload/".$name );
$imageBase = base64_encode( $content );
$success = savePicture($imageBase);
}
function savePicture()
function savePicture($pic_base64)
{
ini_set("soap.wsdl_cache_enabled", "0");
$pageURL = 'http://....';
$page = new NTLMSoapClient($pageURL);
$params = array("pRecordID" => "1",
"pFieldID" => 70000,
"pUserID" => "153",
"pContent" => $pic_base64
);
$result = $page->SetBLOBValue($params);
if($result->return_value != "ERROR")
{
return true;
}
else
{
return false;
}
}
Any Ideas why this didn't work on ipad/iphone?
i hope anybody could help me.
I think that the iPhone/iPad has some problems with capture=camcorder.
Please try this:
<input type="file" name="datei" accept="image/*;capture=camera">
instead of
<input type="file" name="datei" accept="capture=camcorder">
For more information check this:
http://www.html5rocks.com/de/tutorials/getusermedia/intro/
SOLVED
The Problem is that the SOAP Service had a Max-Size for the Files for 1MB!
That was the Problem why i can't upload Base64 Strings who are bigger than 1MB!
Now we increase the Max-Size for Files to 10MB and now it works!
And that was the reason why i cant upload some pictures because some pictures was bigger than 1mb in the base64 string.
Thx for help to everybody!

PHP, using $_FILES and $_POST at the same time

I have a problem in using $_FILES and $_POST at the same the because I have a form to upload an image and some data bus when I use one of them it works but when I used the other one id doesn't work.
my code is :
<?php
include 'debugging.php';
//phpinfo();
echo '<br />';
echo '<h1>Image Upload</h1>';
//create a form with a file upload control and a submit button
echo <<<_END
<br />
<form method='post' action='uplaodex.php' enctype='multipart/form-data'>
Select File: <input type='file' name='picName' size='50' />
name: <input type='text' name='usName' size='50' />
username : <input type='text' name='usUsername' size='50' />
pass: <input type='password' name='usPass' size='50' />
email: <input type='text' name='usEmail' size='50' />
<br />
<input type='submit' value='Upload' />
<input type="hidden" name="submit" value="1" />
</form>
<br />
_END;
//above is a special use of the echo function - everything between <<<_END
//and _END will be treated as one large echo statement
//$_FILES is a PHP global array similar to $_POST and $_GET
if (isset($_FILES['picName'])and isset($_POST['submit'])) {
//we access the $_FILES array using the name of the upload control in the form created above
//
//create a file path to save the uploaded file into
$name = "images//" . $_FILES['picName']['name']; //unix path uses forward slash
//'filename' index comes from input type 'file' in the form above
//
//move uploaded file from temp to web directory
if (move_uploaded_file($_FILES['picName']['tmp_name'], $name)) {
// Create the file DO and populate it.
include 'Do_profile.php';
$file = new Do_profile();
//we are going to store the file type and the size so we get that info
$type = $_FILES['picName']['type'];
$size = $_FILES['picName']['size'];
$usName = trim($_POST['usName']);
$usUsername = trim($_POST['usUsername']);
$usPass = trim($_POST['usPass']);
$usEmail = trim($_POST['usEmail']);
$file->FileName = $name; //$name is initialised previously using $_FILES and file path
$file->FileSize = $size;
$file->Type = $type;
$file->usName = $usName;
$file->usUsername = $usUsername;
$file->usPass = $usPass;
$file->usEmail = $usEmail;
if ($file->save()) {
//select the ID of the image just stored so we can create a link
//display success message
echo "<h1> Thankyou </h1><p>Image stored successfully</p>";
//this above line of code displays the image now stored in the images sub directory
echo "<p>Uploaded image '$name'</p><br /><img src='$name' height='200' width='200'/>";
//create alink to the page we will use to display the stored image
echo '<br><a href="Display.php?id=' . $fileId . '">Display image ' .
$file->FileName . '</a>';
} else
echo '<p class="error">Error retrieving file information</p>';
}
else {
echo '<p class="error"> Oh dear. There was a databse error</p>';
}
} else {
//error handling in case move_uploaded_file() the file fails
$error_array = error_get_last();
echo "<p class='error'>Could not move the file</p>";
// foreach($error_array as $err)
// echo $err;
}
echo "</body></html>";
?>
I don't know what is the problem, any help??
Everything inside that if (isset($_FILES['picName'])and isset($_POST['submit'])) doesn't work because the superglobal $_FILES is probably not having a key named picName. To check this out, try var_dump-ing the $_FILES, like var_dump($_FILES);
By the output of the var_dump you'd get to know if there is any content inside $_FILES. And if it is populated, just see what the key name is and, access the file by using that key.
But if the array is empty, there may be some mis-configurations in PHP, or APACHE.
One possible fix would be setting file_uploads = On in the ini file for php.
Hope it helps!
You have to validate the size of the file if you want to do an isset. I don't know if this is works, but the better way for do that is check first the size for validate if isset or was send to the server.
Then, in your <form method='post' action='uplaodex.php' enctype='multipart/form-data'> you have to create another PHP file with the name uplaodex.php where you'll send al the data. So, your code with be like the below code following and considering the step 1. This will be your uploadex.php
$name_file = $_FILES['picName']['name'];
$type = $name_file['type'];
$size = $name_file['size'];
$tmp_folder = $name_file['tmp'];
$usName = trim($_POST['usName']);
$usUsername = trim($_POST['usUsername']);
$usPass = trim($_POST['usPass']);
$usEmail = trim($_POST['usEmail']);
if ( $size > 0 ) {
//REMOVE another slash images//
$name = "images/" . $name_file; //unix path uses forward slash
//'filename' index comes from input type 'file' in the form above
//
//move uploaded file from temp to web directory
if ( move_uploaded_file($tmp_folder, $name) ) {
// Create the file DO and populate it.
include 'Do_profile.php';
$file = new Do_profile();
$file->FileName = $name; //$name is initialised previously using $_FILES and file path
$file->FileSize = $size;
$file->Type = $type;
$file->usName = $usName;
$file->usUsername = $usUsername;
$file->usPass = $usPass;
$file->usEmail = $usEmail;
if ($file->save()) {
//USE PRINTF
printf('<h1> Thankyou </h1><p>Image stored successfully</p><br>
<p>Uploaded file: %s</p>. <img src="%s" height="200" width="200" />',
$name_file, $name );
#WHAT IS $fileId? In which moment was define?
//echo '<br><a href="Display.php?id=' . $fileId . '">Display image ' .
$file->FileName . '</a>';
}
else
echo '<p class="error">Error retrieving file information</p>';
}
else {
echo '<p class="error"> Oh dear. There was a databse error</p>';
} //ENDIF OF if (move_uploaded_file($_FILES['picName']['tmp_name'], $name))
} //ENDIF OF if ( $size > 0 )
#ELSE OF YOUR if ( $size > 0 )
else {
//error handling in case move_uploaded_file() the file fails
$error_array = error_get_last();
echo "<p class='error'>Could not move the file</p>";
// foreach($error_array as $err)
// echo $err;
}
I solved the problem, you can't perform $_FILES and $_post at the same time or one of them inside the other.
start with $_Post and then $_FILES and outside the $_FILES run your saving function
thats it

Displaying an image from an SQL BLOB with PHP

I have two files, one to upload an image and another to retrieve it from the database and present it on the web page. However the image only appears as broken image icon. I do not understand why.
Thanks for any help in advance.
The HTML form:
<form action="index.php" method="POST" enctype="multipart/form-data">
<label for="image">File:</label>
<input type="file" name="image">
<input type="submit" value="Upload">
</form>
The php to upload it(The connection to the DB is left out below but it works perfectly):
//file stuff
$file= $_FILES['image']['tmp_name'];
if(!isset($file))
echo "Please select an image";
else {
$image=addslashes(file_get_contents($_FILES['image']['tmp_name']));
$imageName=addslashes($_FILES['image']['name']);
$imageSize=getimagesize($_FILES['image']['tmp_name']);
if(!$imageSize)
echo "Thats not an image you mong";
else {
//upload
$query="INSERT INTO store VALUES('','$imageName','$image')";
$sendQuery=mysql_query($query);
if(!$sendQuery)
echo "This is embarressing. It didn't work";
else {
$lastid=mysql_insert_id();
echo "Image was uploaded. <br>Your image:";
echo "<img src=get.php?id=$lastid/>";
}
}
}
The PHP to retrieve the image(Again the DB connection is left out below but works perfectly):
$id=addslashes($_REQUEST(['id']));
$imageQuery="SELECT * FROM store WHERE id=$id;";
$sendImageQuery=mysql_query($imageQuery);
$image=mysql_fetch_assoc($sendImageQuery);
$image=$image['image'];
header("Content-type: image/jpeg");
echo $image;
Make sure the the opening php tag is the very first line of the file. If the opening php tag is not on the first line of the file then content has already been written to the response causing calls to the header() function to be ignored.
<?php // this has to be on line 1
// do database connections stuff, no output can be sent to the response.
$id=addslashes($_REQUEST(['id']));
$imageQuery="SELECT * FROM store WHERE id=$id;";
$sendImageQuery=mysql_query($imageQuery);
$image=mysql_fetch_assoc($sendImageQuery);
$image=$image['image'];
header("Content-type: image/jpeg");
echo $image;
?>
Maybe one of the magic quote setting (magic_quotes_gpc), try:
$image=stripslashes($image['image']);
You should be using mysql_real_escape_string() instead of addslashes() and only if magic quotes are not enabled.

How to serve multiple images which reside above the www root within a single page?

I am hoping to offer users a user submitted image gallery. I have written a upload script which saves the file above the www root. I know I can serve the file by specifying the page header and then using readfile, however I am planning to throw the images within a table to be displayed with other information and dont think the header/readfile is the best solution. I am thinking maybe symlinks but I am not sure.
What is the best method to achieve this?
You'll need a script like getimage.php which sends the image headers and echo's out its contents. Then in your HTML, you just utilize it as the <img src=''> in your HTML. The only purpose of getimage.php is to retrieve and output the image. It remains separate from whatever PHP you use to generate the HTML sent to the browser.
Additionally, you can check if the user has a valid session and permission to view the image in getimage.php and if not, send a some kind of access-denied image instead.
The contents of getimage.php are small and simple:
// Check user permissions if necessary...
// Retrieve your image from $_GET['imgId'] however appropriate to your file structure
// Whatever is necessary to determine the image file path from the imgId parameter.
// Output the image.
$img = file_get_contents("/path/to/image.jpg");
header("Content-type: image/jpeg");
echo($img);
exit();
In your HTML:
<!-- as many as you need -->
<img src='getimage.php?imgId=12345' />
<img src='getimage.php?imgId=23456' />
<img src='getimage.php?imgId=34567' />
It then becomes the browser's job to call getimage.php?imgId=12345 as the path to the image. The browser has no idea it is calling a PHP script, rather than an image in a web accessible directory.
If the script is running on a Unix server, you might try to create a symlink in your web root that links to the directory outside of your web root.
ln -s /webroot/pictures /outside-of-webroot/uploads
If you're using an Apache server you could also have a look at mod_alias.
I've heard that there are a few issues when using mod_alias and configuring it through .htaccess. Unfortunately I don't have any experience with mod_alias whatsoever.
Something that always has worked well for me is to have users upload their images directly into my mysql db. The PHP will encode into base64 and store into a blob. Then you do something similar to what michael said to retrieve and display the image. I've included some code from a project I was working on in 2008. I wouldn't copy it exactly if it's a method you're interested in using since it's old code.
This is the PHP to upload and store into a DB. Obviously replace your info and connect to your own DB.
<?php
include("auth.php");
// uploadimg.php
// By Tyler Biscoe
// 09 Mar 2008
// Test file for image uploads
include("connect.php");
include("include/header.php");
$max_file_size = 786432;
$max_kb = $max_file_size/1024;
if($_POST["imgsubmit"])
{
if($_FILES["file"]["size"] > $max_file_size)
{
$error = "Error: File size must be under ". $max_kb . " kb.";
}
if (!($_FILES["file"]["type"] == "image/gif") && !($_FILES["file"]["type"] == "image/jpeg") && !($_FILES["file"]["type"] == "image/pjpeg"))
{
$error .= "Error: Invalid file type. Use gif or jpg files only.";
}
if(!$error)
{
echo "<div id='alertBox'> Image has been successfully uploaded! </div>";
$handle = fopen($_FILES["file"]["tmp_name"],'r');
$file_content = fread($handle,$_FILES["file"]["size"]);
fclose($handle);
$encoded = chunk_split(base64_encode($file_content));
$id = $_POST["userid"];
echo $_FILES["file"]["tmp_name"];
$default_exist_sql = "SELECT * FROM members WHERE id='".$id."'";
$default_result = mysql_query($default_exist_sql);
$results = mysql_fetch_array($default_result);
if(!$results["default_image"])
{
$insert_sql = "UPDATE members SET default_image = '$encoded' WHERE id='". $id ."'";
mysql_query($insert_sql);
}
$sql = "INSERT INTO images (userid, sixfourdata) VALUES ('$id','$encoded')";
mysql_query($sql);
}
else
{
echo "<div id='alertBox'>". $error . "</div>";
}
}
?>
<br />
<font class="heading"> Upload images </font>
<br /><br />
<form enctype = "multipart/form-data" action = "<?php $_SERVER['PHP_SELF']; ?>" method = "post" name = "uploadImage">
<input type = "hidden" name="userid" value = "<?php echo $_GET["userid"]; ?>" >
<input id="stextBox" type="file" name="file" size="35"><br />
<input type="submit" name="imgsubmit" value="Upload">
</form>
<?php include("include/footer.php"); ?>
This next one displays the file:
<?php
// image.php
// By Tyler Biscoe
// 09 Mar 2008
// File used to display pictures
include("connect.php");
$imgid = $_GET["id"];
$result = mysql_query("SELECT * FROM images WHERE imgid=" . $imgid . "");
$image = mysql_fetch_array($result);
echo base64_decode($image["sixfourdata"]);
echo $image["sixfourdata"];
?>
Then:
<img src="image.php?id=your_img_id">

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