PHP/html/mysql Image display

Question

Hey my php/mysql/html people,

I have this variable that holds a path to an image and then insert it into a db (yes, I am using mysql_real_escape_string) which works perfectly.

$file_name = $_FILES["file"]["name"];
$path='images\ '. $file_name;
insert into...blah blah blah

The path is later pulled from the db and stored in said variable.

$path       = $row['file_path'];

I am trying to display it with:

 // the contents of $path in this case is: images\ cats.jpg
echo "<img src=" . $path . ">";

However the image breaks because the only thing that src pics up is: images\ and not the actualname+extension of the image. I know this probably has to do with the slashes, but I am a newbie and could use some help. Thanks in advance!

have you tried printing the variable to check if it is showing what you think it is showing? try "echo $path" before you insert into. also try "echo $path" after you retrieve it from the db. does it print what you expect it to? — eatonphil
– eatonphil, Commented May 1, 2013 at 3:19
just wondering shouldn't the path be images\cats.jpg instead of images\ cats.jpg — Satya
– Satya, Commented May 1, 2013 at 3:19

Satya · Accepted Answer · 2013-05-01 03:20:13Z

1

your

$path='images\ '. $file_name;

should be

$path='images\'. $file_name;

answered May 1, 2013 at 3:20

Satya

8,9016 gold badges41 silver badges57 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Jonathan Scialpi Over a year ago

agreed, however as soon as i delete that space then the rest of my code in the script goes grey. This is due to the fact that the \ will actually escape the '

Satya Over a year ago

so just escape the \ with \\

Jonathan Scialpi Over a year ago

that worked! , in terms of getting the full file path. However I am still getting broken pic icon displayed :/

Satya Over a year ago

try echoing $path first and see if the file is actually there

Jonathan Scialpi Over a year ago

seems that the path needs to be displayed like: /images/filename.jpg

|

Anoop Pete · Accepted Answer · 2013-05-01 03:50:21Z

0

Print $path variable & see the values.

echo $path;exit;

You will get value on web browser.

Copy the value displayed on browser & paste it in the URL, if you see the image, your saved path is correct, if not, you have gone wrong in specifying the relative path too the resource.

answered May 1, 2013 at 3:50

Anoop Pete

4932 gold badges5 silver badges18 bronze badges

Collectives™ on Stack Overflow

PHP/html/mysql Image display

2 Answers 2

6 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related