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value pass from one script to another script

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thirusvga

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Newbie Poster

value pass from one script to another script

Apr 23rd, 2008

Hi.. I am new in php +ajax.....Now i am doing project in php+ajax+linux environment...i create a login page using php,ajax,mysql,, i have mysql tables are. register,slideshow,,,,,In register table having following field...
1.uid (autoincrement)2.first (firstname)3.last(surname)4.user(username),5.pass(password)
In slide show table having following field,,,
1.uid 2.pid(presentation id,autoincrment) 3.slideno 4.description 5.location(this is what image file is store that particular location)..
my login programs are..
//login.html

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<html>
<head>
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction()
     {
	var ajaxRequest;  // The variable that makes Ajax possible!
	 
	try
           {
		// Opera 8.0+, Firefox, Safari
		ajaxRequest = new XMLHttpRequest();
	    }
           catch (e)
                {
		// Internet Explorer Browsers
		try
                  {
			ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
		   } 
              catch (e)
                      {
			try
                          {
				ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
			  } 
              catch (e)
                      {
				// Something went wrong
				alert("Your browser broke!");
				return false;
			 }
		    }
	         }
            
	// Create a function that will receive data sent from the server
	ajaxRequest.onreadystatechange = function()
             {
		if(ajaxRequest.readyState == 4) 
                        {
                        var ajaxDisplay = document.getElementById('ajaxDiv');
                        ajaxDisplay.innerHTML = ajaxRequest.responseText;
                        document.getElementById('message').style.visibility = 'hidden'; 
   		     }
		}

  var user= document.getElementById('user').value;
  var pass= document.getElementById('pass').value;
  var queryString = "?user=" + user + "&pass=" + pass;
document.getElementById('message').style.visibility = 'visible';
  ajaxRequest.open("GET","login.php" + queryString, true);
  ajaxRequest.send(null);
 
}
</script>
</head>
<body>
<div id="message"  style="position:absolute; top:1%; left:95%;  margin-left:-100px; font-size:14;background-color: red ;color: white;width: 165px; height: 25px; overflow: auto;visibility: hidden">Executing...</div>
<form>
<center>
<img src="" style="visibility:hidden" width="0%" height="0%">
<table border="0" bgcolor="#CCCCFF" cellspacing="1" cellpadding="3" width="287">
<tr>
<td align="left" colspan="2" width="275"><b><font size="5" color="#000080">Login</font></b></td>
</tr>
<tr>
<td align="right" width="81"><b><font color="#000080">User
Name:</font></b></td>
<td width="184">
<input type="text" id="user">
</td>
</tr>
<tr>
<td align="right" width="81"><b><font color="#000080">Password:</font></b></td>
<td width="184">
<input type="password" id="pass">
</td>
</tr>
<tr>
<td colspan="2" align="center" width="275"><input type="button"  onclick="ajaxFunction()" value="Login" ></td>
</tr>
</table>
<div id="ajaxDiv">&nbsp;
</center>
</form>
</body>
</html>
//login.php
<?php
$user=$_GET['user'];
$pass=$_GET['pass'];
$conn = mysql_connect("localhost","root","") or die("could not connect server");
$db = mysql_select_db("thiru",$conn) or die("could not connect database");
$query= "select * from register";
$res = mysql_query($query) or die("query failed" . mysql_error());
$num_rows=mysql_num_rows($res);
while($rr=mysql_fetch_array($res))
{
if($user==$rr[3] && $pass==$rr[4])
 {
//get uid 
$insert=$rr[0];
//echo"welcome" .$user;
$flag=true;
}
}
if($flag==false)
{
echo"userid and password mismatch";
}
mysql_close($conn);
?>

<html>
<head>
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction()
     {
	var ajaxRequest;  // The variable that makes Ajax possible!
	 
	try
           {
		// Opera 8.0+, Firefox, Safari
		ajaxRequest = new XMLHttpRequest();
	    }
           catch (e)
                {
		// Internet Explorer Browsers
		try
                  {
			ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
		   } 
              catch (e)
                      {
			try
                          {
				ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
			  } 
              catch (e)
                      {
				// Something went wrong
				alert("Your browser broke!");
				return false;
			 }
		    }
	         }
            
	// Create a function that will receive data sent from the server
	ajaxRequest.onreadystatechange = function()
             {
		if(ajaxRequest.readyState == 4) 
                        {
                        var ajaxDisplay = document.getElementById('ajaxDiv');
                        ajaxDisplay.innerHTML = ajaxRequest.responseText;
                        document.getElementById('message').style.visibility = 'hidden'; 
   		     }
		}

  var user= document.getElementById('user').value;
  var pass= document.getElementById('pass').value;
  var queryString = "?user=" + user + "&pass=" + pass;
document.getElementById('message').style.visibility = 'visible';
  ajaxRequest.open("GET","login.php" + queryString, true);
  ajaxRequest.send(null);
 
}
</script>
</head>
<body>
<div id="message"  style="position:absolute; top:1%; left:95%;  margin-left:-100px; font-size:14;background-color: red ;color: white;width: 165px; height: 25px; overflow: auto;visibility: hidden">Executing...</div>
<form>
<center>
<img src="" style="visibility:hidden" width="0%" height="0%">
<table border="0" bgcolor="#CCCCFF" cellspacing="1" cellpadding="3" width="287">
<tr>
<td align="left" colspan="2" width="275"><b><font size="5" color="#000080">Login</font></b></td>
</tr>
<tr>
<td align="right" width="81"><b><font color="#000080">User
Name:</font></b></td>
<td width="184">
<input type="text" id="user">
</td>
</tr>
<tr>
<td align="right" width="81"><b><font color="#000080">Password:</font></b></td>
<td width="184">
<input type="password" id="pass">
</td>
</tr>
<tr>
<td colspan="2" align="center" width="275"><input type="button"  onclick="ajaxFunction()" value="Login" ></td>
</tr>
</table>
<div id="ajaxDiv">&nbsp;
</center>
</form>
</body>
</html>
//login.php
<?php
$user=$_GET['user'];
$pass=$_GET['pass'];
$conn = mysql_connect("localhost","root","") or die("could not connect server");
$db = mysql_select_db("thiru",$conn) or die("could not connect database");
$query= "select * from register";
$res = mysql_query($query) or die("query failed" . mysql_error());
$num_rows=mysql_num_rows($res);
while($rr=mysql_fetch_array($res))
{
if($user==$rr[3] && $pass==$rr[4])
 {
//get uid 
$insert=$rr[0];
//echo"welcome" .$user;
$flag=true;
}
}
if($flag==false)
{
echo"userid and password mismatch";
}
mysql_close($conn);
?>

The above program is successfully logged .. and i fetch uid in $insert variable...This is my quetion...
uid,pid,slideno,description,location will be stored automaticallyin slideshow table..while i upload file....
my upload proram is given below...

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//upload.html file
<html>

<head>

<script type="text/javascript" language="javascript">
function upload()
{

var oForm = document.uploadform;
oForm.submit();
}
</script>

</head>

<body>

<div id="message" class="drag" style="position:absolute; top:50%; left:50%;  margin-left:-100px; font-size:12;background: transparent;

          width: 75px; height: 75px; overflow: auto; visibility: hidden">

</div>

<center>

<form name="uploadform" action="upload.php" method="POST" ENCTYPE="multipart/form-data" target="hiddenFrame">
Open :
  <input style="font:normal 10px Verdana" size="35" align="left" type="file" name="code" id="code">&nbsp;&nbsp;&nbsp;&nbsp;<br><button  onClick="upload();return false">Upload</button>

</form>

<iframe src="about:blank" name="hiddenFrame" width="400" height="400" 

frameborder="1" ></iframe> 

</center>

</body>

</html>

//upload php file
<?php 
$uploaddir='/var/www/html/upload/';
$uploadfile=$uploaddir.basename($_FILES['code']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['code']['tmp_name'], $uploadfile))
 {
    echo "File was successfully uploaded.\n";
  print "</pre>";
$conn=mysql_connect("localhost","root","");
$db=mysql_select_db("thiru");
$query = "insert into slide(sloc)values('$uploadfile')";
$result=mysql_query($query);
echo"<br>";
mysql_close($conn);
$filename=$_FILES['code']['name'];
echo"filename:";
echo $filename;
echo"<br>";
}
 else 
{
    echo "please choose a correct file!!";
}

?>

please send me the correct coding ..

Last edited by ~s.o.s~ : Apr 23rd, 2008 at 1:46 pm. Reason: Added code tags, learn to use them.