Displaying image from within a php module

Question

I am trying to print the image whose location is saved in my database, I have stored the absolute location in the database and not the relative one , I browsed through a lot of question including this one include a PHP result in img src tag I tried all the options that were given to the respective asker of the question but I didn't get my output, rest everything is being displayed apart from the image, its showing no file found

Here's my code, any help will be appreciated

while($result=@mysql_fetch_array($resul,MYSQL_ASSOC)){
    $image = $result['image'];
    echo $result['company'] . " " . $result['model'] . "<br>" ;
    echo '<img src="$image" height="50" width="50" />';
}

I know I am using mysql functions instead of mysqli but this code is not getting live ever.

Can you provide a sample output generated by this? I'm not going to talk about the mysqli, but please, stop using @'s :) — lsouza
– lsouza, Commented Aug 8, 2013 at 18:28

rink.attendant.6 · Accepted Answer · 2013-08-08 18:50:55Z

3

As watcher said, PHP does not do variable interpolation within single-quoted strings.

The most important feature of double-quoted strings is the fact that variable names will be expanded.

Read more about strings from the PHP manual.

Therefore, when you view the HTML, you will literally see this:

<img src="$image" height="50" width="50" />

Your code should be:

while($result = mysql_fetch_array($resul,MYSQL_ASSOC)) {
    $image = $result['image'];
    echo $result['company'] . " " . $result['model'] . "<br>";
    echo "<img src='$image' height='50' width='50'>";
}

Alternatively, interpolate the array value:

while($result = mysql_fetch_array($resul,MYSQL_ASSOC)) {
    echo $result['company'] . " " . $result['model'] . "<br>";
    echo "<img src='{$result['image']}' height='50' width='50'>";
}

If the filename contains spaces or other special characters, you may need to use rawurlencode(). In this case, you must concatenate the string since you are calling a function that returns a string value:

echo "<img src='" . rawurlencode($result['image']) . "' height='50' width='50'>";

edited Aug 8, 2013 at 18:50

answered Aug 8, 2013 at 18:33

rink.attendant.6

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3 Comments

CAO Over a year ago

<img src="file:///c:/wamp/upload/computer/SIGN 001.jpg" height="250" width="250"> This is what I am getting when I inspect the element for the image but there is nothing being displayed, now not even the error that says no file chosen, Its just the frame for the image

CAO Over a year ago

Yes I am running this locally and the file does exist here, when I inspect the element for this img tag I get the absolute location and when I click on that the image is displayed in a new tab

CAO Over a year ago

I did it but now the code became something like this <img src="file%3A%2F%2F%2Fc%3A%2Fwamp%2Fupload%2Fcomputer%2FSIGN%20001.jpg" height="50" width="50"> but when I moved the image to the local directory of wamp i.e. in www the image is being displayed, thanks for your help, I really appreciate it.

Jeff Lambert · Accepted Answer · 2013-08-08 18:30:02Z

1

PHP will not interpolate variables when you include them within single quotes. For more information, see the manual.

answered Aug 8, 2013 at 18:30

Jeff Lambert

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