Question

I am trying to do something I know is probably simple, but I am having the worst time.
I have functioning so far:
1.Script to upload image files to server
2. write the image file names to the database
3. I want to retrieve the image filename from the db and add it to the img src tag
here is my retrieval script

 <?php

 $hote = 'localhost';
 $base = 'dbasename';
 $user = 'username';
 $pass = '******';
 $cnx = mysql_connect ($hote, $user, $pass) or die(mysql_error ());
 $ret = mysql_select_db ($base) or die (mysql_error ());
 $image_id = mysql_real_escape_string($_GET['ID']);
 $sql = "SELECT image FROM image_upload WHERE ID ='$image_id'";
 $result = mysql_query($sql);
 $image = mysql_result($result, 0);

 header('Content-Type: text/html');
 echo '<img src="' $image'"/>';

 ?>

I was trying to pass the Value through image.php?ID=2 but no luck

The PHP script successfully returns the filename, but I cannot for the life of me get it to print it to the html

Any suggestions, please and thank you very much 🙂

OK, it does return the proper tag, but now it seems as though the script doesnt run to generate the tag.
I have tried two ways:

<div class="slides">
    <div class="slide">
        <div class="image-holder">
            <?php
    include ("image.php?ID=2");
    ?>
        </div>

and:

 <img src="image.php?ID=2" alt="" />     

but neither one will insert the filename…
I need to identify each img src by the primary key, so I was passing it the ID from each image src location
but alas, my PHP ninja skills need to be honed.

Just to clarify: I am uploading images to the server, recording the filenames in a DB and calling that filename in an HTML doc…there are several in each one so I need to pass the ID (i.e. 1,2,3 ) to correspond to the primary key in the table.
But I cant get the script to process the tag first.
If I go to view source, I can click the script and get the proper result…

Thanks again, you guys and girls are very helpful

Answer 1

You can do it as you did but you had the single quotes in twice, (unless you were meaning to use concatenation – which is unnecessary – if you want this see the other answer).

echo "<img src=\"$image\"/>";

Or the longer form with braces if you need to embed inside text.

echo "<img src=\"${image}\"/>";

I’d recommend using heredoc syntax for this if you’re doing lots of HTML. This avoids the need to have lots of echo lines.

echo <<< EOF
   <div class="example">
   <img src="$image" />
   </div>
EOF;