[nycphp-talk] Update a table using AJAX
Aniesh joseph
anieshjoseph at gmail.com
Wed Apr 25 01:10:43 EDT 2007
Hello
I have a html form with action to paypal's site. When submitting the page, I
need to update the table with the form values( USING AJAX) .
To do this I write the following code.
<script>
/* Call this function on the onClick of the submit button */
function check_form_values()
{
/* Inside this function I check wether the USER enters the details */
ajaxFunction();
/* IT WORKS PERFECTLY IF WE UNCOMMENT THIS ALERT PAGE */
}
function ajaxFunction()
{
append_url =
'contact_info.php?contact_id='+document.form1.user_id.value+'&contact_name='+document.form1.contact_name.value+'&position='+document.form1.position.value+'&email='+document.form1.email.value+'&telephone='+document.form1.telephone.value+'&mobile='+document.form1.mobile.value+'&company='+document.form1.company.value+'&address='+document.form1.address.value;
var xmlHttp;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("Your browser does not support AJAX!");
return false;
}
}
}
xmlHttp.onreadystatechange=function()
{
if(xmlHttp.readyState==4)
{
//response_txt = xmlHttp.responseText;
}
}
xmlHttp.open("GET",append_url,true);
xmlHttp.send(null);
}
</script>
I am trying to update the table using the contact_id field, I did not return
any values from AJAX action page.
But this code did not work properly.
If I uncomment the alert inside the function "check_form_values()", then
updation will work perfectly.
Can someone help me to fix this problem ?
Regards
Aniesh Joseph
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