[SOLVED] wtf?

[Niccaman]

I have some php code which javascript connects to in order to return to a table to a div with ajax.

 

Then i insert some code ive been using as i always have with other pages, and nothing gets inserted to the div. No tables ...nada. Just as if there's a syntax error. however if i access this php page seperately, i can see the page fine with no syntax error.

 

The code that makes the ajax stop working is this:

"

// ** Connect to Database **

$con = (mysql_connect("localhost", "xxx", "xxx"));

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

// ** Select MySQL table "agents" **

mysql_select_db("xxx",$con);

"

 

I know this code works. How can inserting this code prevent inserting anything in the div????

[Niccaman]

Also, when using IE, the javascript never works, and there is a yellow triangle with "error on page" at bottom left hand corner of screen. So it must be the javascript? Here is my javascript:

 

<script type="application/javascript" language="javascript">

var xmlHttp;

 

function manageFlights(str)

{

xmlHttp=GetXmlHttpObject();

if (xmlHttp==null)

{

alert ("Browser does not support HTTP Request");

return;

}

var url="/business/airline.php";

url=url+"?b="+str+"&i=1";

url=url+"&sid="+Math.random();

xmlHttp.onreadystatechange=stateChanged;

xmlHttp.open("GET",url,true);

xmlHttp.send(null);

}

function buyPlane(bus)

{

xmlHttp=GetXmlHttpObject();

if (xmlHttp==null)

{

alert ("Browser does not support HTTP Request");

return;

}

var url="/business/airline.php";

var num = document.form.abc;

for (i=0; i<num.length; i++)

{

  if (num.checked == true)

  num = num.value;

}

url=url+"?b="+bus+"&n="+num+"&i=1";

url=url+"&sid="+Math.random();

xmlHttp.onreadystatechange=stateChanged;

xmlHttp.open("GET",url,true);

xmlHttp.send(null);

}

 

function stateChanged()

{

if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")

{

document.getElementById("manage").innerHTML=xmlHttp.responseText;

}

}

function GetXmlHttpObject()

{

var xmlHttp=null;

try

{

// Firefox, Opera 8.0+, Safari

xmlHttp=new XMLHttpRequest();

}

catch (e)

{

//Internet Explorer

try

  {

  xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");

  }

catch (e)

  {

  xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");

}

}

return xmlHttp;

}

</script>

 

[Niccaman]

The IE error on page provides "Expected Object" error when double clicking it. (didn't know you could double click). It referes to a  line where i call the function. Here is that specific line:

 

  <tr class="foundation"><td align="center">Rogue Airways</td><td align="center">Airline</td><td align="center">2009-03-25 17:22:45</td><td align="center"><a href="#" onclick="manageFlights(1);">Manage</a></td></tr></table>

 

Please can someone work out what isnt syntactically correct?

[Niccaman]

lol, i didn't realise these forums were blogs. People much quicker at anwering topics on php help than this.

 

Anyway i realised that my problems were because i have type="application:javascript" rather than text.javascript or something or other.

 

STILL PROBLEMS WITH:

 

executing these lines of code prevents ajax, unless i place in a function and dont call it. Therefore no syntax error but why happening. Here code:

 

function connect()

{

// ** Connect to Database **

$con = (mysql_connect("localhost", "niccaman_bond", "007"));

if (!$con)

{

die('Could not connect: ' . mysql_error());

}

// ** Select MySQL table "agents" **

mysql_select_db("niccaman_agents",$con);

}

 

Cmon guys tell me! Why does it prevent php page returning a table?

[Zane]

function connect()
{
   // ** Connect to Database **
   $con = (mysql_connect("localhost", "niccaman_bond", "007"));
   if (!$con)
   {
   die('Could not connect: ' . mysql_error());
   }
   // ** Select MySQL table "agents" **
   mysql_select_db("niccaman_agents",$con);
}

 

There is nothing wrong with the above code...it is actually the most generic of scripts there are.

Like you mentioned earlier...you forgot text/javascript

 

it most definitely is an external problem....show your full code..with code tags...

[Zane]

Actuall you should shorten that function to 2 lines

 

function connect($dbServer, $user, $pass, $dbName) {
    $c = mysql_connect($dbServer, $user, $pass) or die(mysql_error());
    mysql_select_db($dbName);
}

 

mysql_select_db("niccaman_agents",$con);

You only need to provide the connection link if you are using more than one MySQL server...so nix that second argument next time you go about that

[Niccaman]

Ok, Thanks for the advice Zanus.

 

I found my problem in the php page. Did some old fashioned trial and error. pasted 'die("!");' after every x amount of code till i found where the code failed. I eventually narrowed it down to a

 

while ($row = mysql_fetch_array($res))

{...}

 

situation. Within that loop i called upon a function which i had removed. Thats why it only failed when i connected to a server. It was the only time the while loop condition was met! How SILLY!

 

Thanks anyway for help (even though i solved it myself) 

[corbin]

Niccaman, if you turned on displaying errors, you would have seen a "Call to undefined function" error.

 

 

 

 

Zanus, what if he wanted to add a second link later?  Also, to be really picky, I would imagine it is a tiny bit faster to pass the second parameter since then the mysql_select_db function (or what ever it's called on the PHP core level) does not have to look for the last opened connection.