CODEWITHSUNDEEP
phpmysql
Shibbir Ahmed
Shibbir Ahmed

Reputation: 95

Php image show issue

I'm confused! Basically i've a form with 3 field ex: subject, image, message. So,

(1) If i upload image it's should be show image,
(2) if i don't upload image it's should be show just only subject and message no image box and
(3) if i don't upload image and if there are a youtube link in message box then it's should be show the vedio. **

Following is my php code, but i can't solved the number 3 point! **

<?php 

if(empty($imgname))
{
echo '';
}
elseif(!empty($imgname))
{

echo "<span class='vedio'>";            
echo '<img src="' . $upload_path . '/' . $imgname . '" width="195" height="130" 
style=" background-color:f2f4f6;"  />'; 
    echo "</span>";
}
else
{
    echo "<span class='vedio'>";                
require_once("func.php");   
if (preg_match('%(?:youtube\.com/(?:[^/]+/.+/|(?:v|e(?:mbed)?)/|.*[?&]v=)|youtu\.be/)
    ([^"&?/ ]{11})%i', $msg, $match))
        {
            //    echo $video_id = $match[1];
        }       
        echo @get_youtube_embed($video_id = $match[1]);     
        echo "</span>";


}
?>

Any Idea or Solution that would be better for me.
Shibbir.

Upvotes: 0

Views: 103

Answers (1)

Josh
Josh

Reputation: 8191

You'll never get into your else statement because of your preceding if conditions. In ALL cases, $imgname will either be empty, or not empty.

To demonstrate further:

if (empty($imgname)) {
    // ... First condition
} else if (!empty($imgname)) {
    // ... Last possible condition
} else {
    // Will never reach here because $imgname will 
    // fall into one of the two of the above conditionals.
}

It's difficult to add to your code without full context, but it seems you may want to include the following after the above code (not with):

if(!empty($msg) && preg_match('%(?:youtube\.com/(?:[^/]+/.+/|(?:v|e(?:mbed)?)/|.*[?&]v=)|youtu\.be/)
([^"&?/ ]{11})%i', $msg, $match)) {

    require_once("func.php"); 
    echo "<span class='vedio'>"; 
    echo @get_youtube_embed($match[1]);     
    echo "</span>";
}

Upvotes: 1

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