I'm trying to upload a varchar image from my mysql database and it's not working. Any suggestions?

Question

I'm using 000webhost for my php server and phpmyadmin for my database. My image row looks like

image   varchar(100)    utf8_unicode_ci

What i'm trying to do is someone inputs a username and it outputs the image associated with that username. Here's my html:

<!DOCTYPE html>
<html>

<body>

<form action="showimage.php" method="GET">
<h2>Show Image</h2>
        Username: <input type="text" name="username" /><br>

        <input type="submit" value="Show Image" />
    </form>

</body>
</html>

and my php:

<!DOCTYPE html>
<html>
<body>


<?php


$host='localhost';
$user='id1783920_123456';
$pass='';
$db='id1783920_mydb';

$con=mysqli_connect($host,$user,$pass,$db);
if($con) {
//echo 'connected successfully to id1783920_mydb database<br><br>';
}

$id = mysqli_real_escape_string($_GET['id']);
$query = mysqli_query("SELECT * FROM 'blob' WHERE 'id'='$id'"); 
while($row = mysqli_fetch_assoc($query) {

$imagedata = $row[`image`];  

}

echo"$imagedata"; 
mysqli_close($con);

?>

<img src="showimage.php?username=$_GET['username']">

</body>
</html>

Can someone please give me some suggestions? Thanks so much! I also keep getting this error

Parse error: syntax error, unexpected ';' in /storage/h11/920/1783920/public_html/Complete/showimage.php on line 23

Double quotes are what I'm using and that's what I'm supposed to use. So that's not the issue. — Jean Luc Picard, Jun 05 '17 at 19:45
you dont use single qoutes around a table name and column name never ever.. — Raymond Nijland, Jun 05 '17 at 19:46
Yes, it is the issue. Please read the dup link. You're supposed to use backticks for column names and table names, not single or double quotes. — aynber, Jun 05 '17 at 19:53
...what...where are my single quotes around a table or column name? — Jean Luc Picard, Jun 05 '17 at 19:55
Ok, thanks. I'm still getting this error: Parse error: syntax error, unexpected ';' in /storage/h11/920/1783920/public_html/Complete/showimage.php on line 23 — Jean Luc Picard, Jun 05 '17 at 19:59
your code is failing on too many levels. Best you go read the manuals on those functions. You will see the proper syntax. — Funk Forty Niner, Jun 05 '17 at 20:00
You are missing a closing parenthesis on the line with the where statement. — Sloan Thrasher, Jun 05 '17 at 20:01
lol. I've never got that before:) Can I use that quote? @Fred-ii- — Jean Luc Picard, Jun 05 '17 at 20:01
I'm gonna close this question...Fred is right :( I just want to add a profile pic to my users home page but looks like that wont work because "My code is failing on too many levels" - Fred P.S How do you close a question? — Jean Luc Picard, Jun 05 '17 at 20:09

score 0 · Answer 1 · answered Jun 05 '17 at 21:34

The error:

Parse error: syntax error, unexpected ';' in /storage/h11/920/1783920/public_html/Complete/showimage.php on line 23

is because:

$imagedata = $row[`image`];

should be:

$imagedata = $row['image'];

Backticks are very useful in MySQL queries. But in the PHP code it is normally either single or double-quotes. For a constant text value it makes no real difference, but if you want to have variables inside the string (e.g., $row["image{$counter}"]) then you need to use double-quotes.

I'm trying to upload a varchar image from my mysql database and it's not working. Any suggestions?

1 Answers1