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phpmysql
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Reputation: 1844

PHP Function : how do send value to html

i have this function for check news files from MySQL database:

function check_Files($id){

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        foreach($files as $file){
            $url = $file['url'];
        return TRUE;
        }

    } else
    {
        return FALSE;
    }

}

HTML : 

if(check_Files($id) == TRUE){

echo $url;

}

But I see this error:

Notice: Undefined variable: url in ......

How do can I back/send $url from function to html ?!

Upvotes: 0

Views: 77

Answers (5)

hsz
hsz

Reputation: 152294

Instead of 

return TRUE;

in your check_Files function do:

return $url;

And use it in following way:

if (($url = check_Files($id)) !== false){

    echo $url;

}

Or to pass an array of urls from database:

function check_Files($id){

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        $urls = array();
        foreach($files as $file){
            $urls[] = $file['url'];
        }
        return $urls;
    } else {
        return FALSE;
    }

}

if (($urls = check_Files($id)) !== false){

    foreach ($urls as $url) {
        echo $url;
    }

}

The test can be simplified to:

if ($urls = check_Files($id)){}

Upvotes: 4

Anish
Anish

Reputation: 5078

function check_Files($id){

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        foreach($files as $file){
            $url = $file['url'];
        return $url;
        }

    } else
    {
        return FALSE;
    }

}

and in html 

if (($url = check_Files($id)) != false){

    echo $url;

}

Upvotes: 0

Rhythem Aggarwal
Rhythem Aggarwal

Reputation: 356

Is the function defined in the same file that is is being called ?? Because the error seems to be that you are trying to output a variable that has not be defined. A better approach would be

function check_Files($id){

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        foreach($files as $file){
            $url = $file['url'];
        return $url
        }

    } else
    {
        return FALSE;
    }

}

if(check_Files($id) != FALSE){

echo check_Files($id);

}

Upvotes: 0

Alauddin Ansari
Alauddin Ansari

Reputation: 250

Make $url global inside the function and define $url above the function then only you can get the $url;

$url="";

function check_Files($id){
    global $url;

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        foreach($files as $file){
            $url = $file['url'];
        return TRUE;
        }

    } else
    {
        return FALSE;
    }

}

if(check_Files($id) == TRUE){
    echo $url;
}

Check yourself and enjoy!

Upvotes: 0

Fluffeh
Fluffeh

Reputation: 33542

This is about variable scope. Inside the function, $url exists. Outside the function, it isn't around anymore. Your best bet is to return the URL instead of true/false:

function check_Files($id){

    $files = sqlaccess::fetch("SELECT url FROM ".FILES." WHERE news_id=".$id." AND type = 'video' "); 

    if (count($files) > 0) {
        foreach($files as $file){
            $url = $file['url'];
        return $url;
        }

    } else
    {
        return FALSE;
    }

}

Then outside in your code, you can call it like so:

$link=check_Files($id);
echo $link;

Or even simply

echo check_Files($id);

Upvotes: 0

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