F
foraci
Dear group,
I have tried for the whole week to get the following done. But all I
face is failure. My task is to fetch data from mysql through php and
display the fetched row in javascript with xmlhttprequest object
without using xml. Directly using the open function with GET and POST.
What I get is all the data are displayed. But I want to process
each of mysql in the javascript with the help of xmlhttprequest object.
Is it possible or I am going in a wrong direction. Any help is
appreciated. Since I am a beginner I could not able to do more. If this
is not the group then pls direct me.
Following is the code:
The .html file: (I did not chk for IE and onreadystatechange)
---------------
<html>
<head>
<script language="javascript">
var xmlhttp, tmp;
function start() {
var row_0,row_2,row_1;
xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET",'noxml.php',false);
xmlhttp.send(null);
tmp=xmlhttp.responseText;
var mybody = document.getElementsByTagName("body")[0];
mytable = document.createElement("table");
mytablebody = document.createElement("tbody");
for(var j = 0; j < 3; ++j)
{
mycurrent_row = document.createElement("tr");
for(var i = 0; i < 3 ; ++i)
{
mycurrent_cell = document.createElement("td");
currenttext = document.createTextNode(tmp) ;
mycurrent_cell.appendChild(currenttext);
mycurrent_row.appendChild(mycurrent_cell);
}
mytablebody.appendChild(mycurrent_row);
}
mytable.appendChild(mytablebody);
mybody.appendChild(mytable);
mytable.setAttribute("border","2");
}
</script>
</head>
<body onLoad="start()">
</body>
</html>
The .php file:
-------------
<?php
$user = "sathya";
$passwd = "sathya";
$dbname = "ajax";
$conn = mysql_connect("localhost",$user,$passwd);
global $row_0 ,$row_1,$row_2;
if(!$conn)
die("error in mysql connection".mysql_error());
$db = mysql_select_db($dbname);
$quere = "select * from product";
$countris = mysql_query($quere);
if(!$db)
die("error in mysql db".mysql_error());
if(!$countris)
die("error table".mysql_error());
while($row = mysql_fetch_row($countris))
{
$row_0 = $row[0];
$row_1 = $row[1];
$row_2 = $row[2];
$_POST['row_0'] = $row_0;
$_POST['row_1'] = $row_1;
$_POST['row_2'] = $row_2;
echo $row_0.' ', $row_1.' ', $row_2.'<br>';
}
?>
The .sql has 3 col
Thanks for any help.
I have tried for the whole week to get the following done. But all I
face is failure. My task is to fetch data from mysql through php and
display the fetched row in javascript with xmlhttprequest object
without using xml. Directly using the open function with GET and POST.
What I get is all the data are displayed. But I want to process
each of mysql in the javascript with the help of xmlhttprequest object.
Is it possible or I am going in a wrong direction. Any help is
appreciated. Since I am a beginner I could not able to do more. If this
is not the group then pls direct me.
Following is the code:
The .html file: (I did not chk for IE and onreadystatechange)
---------------
<html>
<head>
<script language="javascript">
var xmlhttp, tmp;
function start() {
var row_0,row_2,row_1;
xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET",'noxml.php',false);
xmlhttp.send(null);
tmp=xmlhttp.responseText;
var mybody = document.getElementsByTagName("body")[0];
mytable = document.createElement("table");
mytablebody = document.createElement("tbody");
for(var j = 0; j < 3; ++j)
{
mycurrent_row = document.createElement("tr");
for(var i = 0; i < 3 ; ++i)
{
mycurrent_cell = document.createElement("td");
currenttext = document.createTextNode(tmp) ;
mycurrent_cell.appendChild(currenttext);
mycurrent_row.appendChild(mycurrent_cell);
}
mytablebody.appendChild(mycurrent_row);
}
mytable.appendChild(mytablebody);
mybody.appendChild(mytable);
mytable.setAttribute("border","2");
}
</script>
</head>
<body onLoad="start()">
</body>
</html>
The .php file:
-------------
<?php
$user = "sathya";
$passwd = "sathya";
$dbname = "ajax";
$conn = mysql_connect("localhost",$user,$passwd);
global $row_0 ,$row_1,$row_2;
if(!$conn)
die("error in mysql connection".mysql_error());
$db = mysql_select_db($dbname);
$quere = "select * from product";
$countris = mysql_query($quere);
if(!$db)
die("error in mysql db".mysql_error());
if(!$countris)
die("error table".mysql_error());
while($row = mysql_fetch_row($countris))
{
$row_0 = $row[0];
$row_1 = $row[1];
$row_2 = $row[2];
$_POST['row_0'] = $row_0;
$_POST['row_1'] = $row_1;
$_POST['row_2'] = $row_2;
echo $row_0.' ', $row_1.' ', $row_2.'<br>';
}
?>
The .sql has 3 col
Thanks for any help.