How to Save a image recieved via HTTP Post into mySQL - php

Question

I want to write an image received via http into my database using PHP. I access the image with
$inputImage = file_get_contents('php://input');
Echoing the $inputImage works fine, so the transport to the server doesn't seam to be a problem.
I now tried to insert the image with
$sqlRequest="INSERT INTO Image(time, data) SET (NOW(), '$inputImage')";
mysqli_query($connection, $sqlRequest) or die("Error in Inserting " . mysqli_error($connection));
But it doesn't work and i recieve the following error
Error in Inserting You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SET (NOW(), '����' at line 1
Can someone give me a hint
thanks
edit:
okay changed the sytax problem, got to look for the blob problem

Answer 1

Use VALUES() instead of SET(). SET is meant for updating (using UPDATE) whereas VALUES() is meant for inserting (using INSERT).
See this for INSERT syntax and this for UPDATE syntax.

Answer 2

Looking at the funny characters ����, this sounds like you're trying to upload a binary file into a column that isn't capable of handling that type.
Sidenote edit: seeing your comment now, you still need to escape that variable.
By "escape", I mean that you need to use mysqli_real_escape_string().
If this isn't set, then you will need to ALTER your column to either be a BLOB or LONGBLOB.
More importantly, you need to escape the contents of $inputImage because that will be binary data and could contain bytes that will cause MYSQL to assume it is shorter than it actually is.
$inputImage = mysqli_real_escape_string($connection, $inputImage);
$sqlRequest="INSERT INTO Image(time, data) VALUES (NOW(), '$inputImage')";
And of course as I already stated in comments under your question, you're using the wrong syntax in your query.
Use VALUES() and not SET().
Add or die(mysqli_error($connection)) to mysqli_query() also.
Reference:
http://php.net/manual/en/mysqli.real-escape-string.php
Using a prepared statement would also work:
http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php

Answer 3

you want to make sure no data obstructing your sql string is stored so do this:
$inputImage = base64_encode(file_get_contents('php://input'));
before storing the data and then
$myImage = base64_decode( myGetImageFromDBFunction( $myImageID ) );