CODEWITHSUNDEEP
php
Abdul Qadeer
Abdul Qadeer

Reputation: 372

displaying uploaded images

I am trying to create a code which obtains uploaded images, stores them and also displays the image preview and confirms that the image was successfully uploaded.

<?php
$name=$_FILES['myfile']['name'];
$tmp=$_FILES['myfile']['tmp_name'];
$error=$_FILES['myfile']['error'];
$path='myweb/';
if(move_uploaded_file($tmp,$path.$name)==1){echo 'success';}else{echo $error;};
echo ('<img src="$path.$name" height="100px" width="100px"/>');
<?php

The problem is that images are not displaying.

I have also tried

echo ('<img src="$path$name" height="100px" width="100px"/>');

but it still doesn't work.

How can I get the images to display?

Upvotes: 0

Views: 83

Answers (5)

Abdul Qadeer
Abdul Qadeer

Reputation: 372

i replaced line 13 in above code like so.

echo (" <img src=$path.$name height=100px width=100px/>");

And its working now. Seems like double quotes were the problem here. I'l read m0re about it. Thank u s0 much guys u all've been very helpful.:)

Upvotes: 0

DoXicK
DoXicK

Reputation: 4812

There is a difference between using double "" and single '' quotes.

Double quotes are getting parsed, which means that

$variable = 10;
echo "$variable";

will output:

10

Single quotes don't get parsed:

$variable = 10;
echo '$variable';

will output:

$variable

you use variables within '' which means they don't get parsed.

Upvotes: 0

Prasanth Bendra
Prasanth Bendra

Reputation: 32710

<img src="<?php echo $path,$name; ?>" height="100px" width="100px"/>

OR

echo '<img src="' . $path.$name . '" height="100px" width="100px"/>';

Upvotes: 1

Alvaro
Alvaro

Reputation: 41595

You have to include the variable in the printed string like this:

echo ('<img src="'.$path.$name.'" height="100px" width="100px"/>');

You can read more about it in the documentation.

If you don't do so, PHP will think that you want to print the text $path.$name instead the variables content.

Upvotes: 1

Yogesh Suthar
Yogesh Suthar

Reputation: 30488

problem was with single and double quotes.

echo '<img src="' . $path.$name . '" height="100px" width="100px"/>';

You have used single quote and because of that it was not taking the variable name.

Upvotes: 1

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