Php/ajax help

[Markus_Newhart]

I need some help. Im trying to call a php function in ajax. The data variable needs to be inside the called function.

Like: 

_("statusarea").innerHTML = '<div><?php showBBcodes('+data+'); ?></div>';
				

But no matter what i do it wont call it and if it does call it it wont show the data.

[Skewled]

Provide some code otherwise this is too vague to even answer.

[Markus_Newhart]

function showBBcodes($text) {


// BBcode array

$find = array(

'~\[b\](.*?)\[/b\]~s',

'~\[i\](.*?)\[/i\]~s',

'~\[u\](.*?)\[/u\]~s',

'~\[quote\](.*?)\[/quote\]~s',

'~\[size=(.*?)\](.*?)\[/size\]~s',

'~\[color=(.*?)\](.*?)\[/color\]~s',

'~\[url\]((?:ftp|https?)://.*?)\[/url\]~s',

'~\[img\](https?://.*?\.(?:jpg|jpeg|gif|png|bmp))\[/img\]~s'

);


// HTML tags to replace BBcode

$replace = array(

'<b>$1</b>',

'<i>$1</i>',

'<span style="text-decoration:underline;">$1</span>',

'<pre>$1</'.'pre>',

'<span style="font-size:$1px;">$2</span>',

'<span style="color:$1;">$2</span>',

'<a href="$1">$1</a>',

'<img src="$1" alt="" />'

);


// Replacing the BBcodes with corresponding HTML tags

return preg_replace($find,$replace,$text);

}

[Markus_Newhart]

Bump. I still need help.

[Ch0cu3r]

You cannot mix javascript and PHP code together. 

 

Is data the response from an ajax request? if it is, then call the showBBcodes PHP function when you receive the ajax request in PHP.

[mac_gyver]

code in a browser cannot CALL a function on a server. the only thing that a browser can do is make a http request to the server, where the server-side code takes the submitted data (post, get, files, or cookie) and uses those values to perform some action.

 

i recommend you read up on what ajax stands for and read some ajax/php examples.

[Markus_Newhart]

Alright, thanks. Guess ill just reload the page when ever the player posts stuff...