PHP Function : how do send value to html

Question

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 ?!

What kind of database library is behind your sqlaccess class? Maybe they return values as properties: $file->url — Daniel W.
– Daniel W., Commented Jun 11, 2014 at 10:06
You are using loop to get the urls and it might be possible to have multiple urls ? — Abhik Chakraborty
– Abhik Chakraborty, Commented Jun 11, 2014 at 10:07
Your code could be vulnerable to SQL injection. Use PDO if possible! — tangrs
– tangrs, Commented Jun 11, 2014 at 10:19
Hmmm, it doesn't look like you're using any of the parameter binding code in the PDO library... Well, as long as $id and FILES is properly sanitised, I guess it's fine. — tangrs
– tangrs, Commented Jun 11, 2014 at 10:28

hsz · Accepted Answer · 2014-06-11 10:13:08Z

4

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)){}

edited Jun 11, 2014 at 10:13

answered Jun 11, 2014 at 10:07

hsz

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Comments

Fluffeh · Accepted Answer · 2014-06-11 10:07:47Z

0

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);

answered Jun 11, 2014 at 10:07

Fluffeh

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Comments

Alauddin Ansari · Accepted Answer · 2014-06-11 10:09:55Z

0

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!

answered Jun 11, 2014 at 10:09

Alauddin Ansari

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Comments

hsz · Accepted Answer · 2014-06-11 10:12:17Z

0

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);

}

edited Jun 11, 2014 at 10:12

hsz

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answered Jun 11, 2014 at 10:09

Rhythem Aggarwal

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Comments

hsz · Accepted Answer · 2014-06-11 10:12:30Z

0

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;

}

edited Jun 11, 2014 at 10:12

hsz

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answered Jun 11, 2014 at 10:08

Anish

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