Re: Php Include bug in firefox [message #169583 is a reply to message #169569] |
Sat, 18 September 2010 12:25 |
sheldonlg
Messages: 166 Registered: September 2010
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Senior Member |
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On 9/17/2010 10:05 PM, GarryJones wrote:
> Is there another way of including a secondary file to a first?
>
> I have just started to use ajax, it loads in php pages that are
> written on the server.
>
> My ajax script is at the bottom of this for info only, there is not
> problem with the ajax code, it works perfectly except for one little
> problem caused by php.
>
> One of the documents I am reading from the server calls another file
> with php like this.
>
> <? include ("ct04a.php") ?>
>
> This works on msie. This does not work on firefox!! Yeah I know I can
> tell my users to stop using firefox (and mac and iphone and linux and
> crome and safari), but I WOULD like this to work in firefox.
>
> Is the problem a bug in php or ajax or javascript?
>
> Is there another php command that I can test to include a secondary
> php file
I don't have any better answers for you than have already been posted,
but here are a few items:
1 - Never use short tags. If you want a php area, use <?php ?>.
2 - Another one that does the job of "include" is "require".
3 - Always use "include_one" or "require_once" unless you specifically
want to include the same code in multiple places. If your include has a
function in it, then your script will throw an error if you don't have
the "_once" at the end (multiply defined).
>
> <script language=JavaScript>
> var xmlhttp;
> function loadXMLDoc(url)
> {
> xmlhttp=null;
> if (window.XMLHttpRequest)
> {// code for Firefox, Opera, IE7, etc.
> xmlhttp=new XMLHttpRequest();
> }
> else if (window.ActiveXObject)
> {// code for IE6, IE5
> xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
> }
> if (xmlhttp!=null)
> {
> xmlhttp.onreadystatechange=state_Change;
> xmlhttp.open("POST",url,true);
> xmlhttp.send(null);
> }
> else
> {
> alert("Email if it is a big problem");
> }
> }
>
> function state_Change()
> {
> if (xmlhttp.readyState==4)
> {// 4 = "loaded"
> if (xmlhttp.status==200)
> {// 200 = "OK"
> document.getElementById('T1').innerHTML=xmlhttp.responseText;
> }
> else
> {
> alert("Problem med datahämtning:" + xmlhttp.statusText);
> }
> }
> }
> </script>
--
Shelly
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