| new to ajax and need help with json
|
nicknick1
msg:4530875 | 1:09 am on Dec 25, 2012 (gmt 0) |
hi, ive written my first ajax script and after a night mare it works perfectly on my local web server, however when i upload to my web host is does nothing.
ii suspect its because im using getjson and get is dissabled on web host. so its xmas day and i feel very poorly with flu and i need to get this working, could somone help turn this into a post string for me please.
xmlhttp.open("GETJSON","lookup.php?sname="+str1+"®="+str,true);
this is my get but i need to send a post if that makes sense,
this is the lookup.php
<?php
require"config.inc";
if (isset($_POST[reg]) && ($_POST[sname]))
{
$lookup = mysql_query('SELECT * FROM cust WHERE reg = '$_POST[reg]'AND sname LIKE '%$_POST[sname]%'")or die(mysql_error());
if(mysql_num_rows($lookup)==1){
$row = mysql_fetch_array($lookup);
//echo"$row[acc]<br>$row[2]<br>$row[3]<br>$row[4]<br>$row[5]<br>$row[6]";
$test=array(
accno_d => $row[0],
sname_d => $row[4],
reg_d => $row[1],
make_d => $row[2],
model_d => $row[3]
);
$json_d = json_encode($test);
echo"$json_d";
}
else{echo"Not Found";}
}
?>
and this is part of the main form
<script>
function showUser(str,str1)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
if (xmlhttp.responseText==="Not Found")
{document.getElementById("notfound").innerHTML="<font color =red size=1>Details Not Found, please enter manually</font>";}
else
{ var jsonText = xmlhttp.responseText;
var jsonObject= eval('('+jsonText+')');
eval('JSONOBJECT='+jsonText);
var sname=jsonObject.sname_d;
var reg=jsonObject.reg_d;
var make=jsonObject.make_d;
var model=jsonObject.model_d;
var acc=jsonObject.accno_d;
document.getElementById("reg").value=(reg);
document.getElementById("sname").value=(sname);
document.getElementById("make").value=(make);
document.getElementById("model").value=(model);
document.getElementById("add1").value="Account Number "+(acc)+" no address details required";
document.getElementById("reg").readOnly=true;
document.getElementById("sname").readOnly=true;
document.getElementById("make").readOnly=true;
document.getElementById("model").readOnly=true;
document.getElementById("add1").readOnly=true;
document.getElementById("notfound").innerHTML="";
}
}
}
xmlhttp.open("GETJSON","lookup.php?sname="+str1+"®="+str,true);
xmlhttp.send();
}
</script>
please bear in mind that the most ive done with javascript before now is open windows and go back.
thanks in advance
nick
|
daveVk
msg:4530912 | 5:02 am on Dec 25, 2012 (gmt 0) |
xmlhttp.open("GETJSON","lookup.php?sname="+str1+"®="+str,true);
Try GET in place of GETJSON as the method, I far as know GETJSON is not a standard method. Failing that try POST and/or check server log.
Hope you shake the flu.
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nicknick1
msg:4530926 | 6:52 am on Dec 25, 2012 (gmt 0) |
works perfectly thank you very much indeed.
p.s merry Christmas.
p.ps flu still getting the better of me but now im smiling
|
swa66
msg:4531023 | 4:15 pm on Dec 25, 2012 (gmt 0) |
Code like
if (isset($_POST[reg]) && ($_POST[sname]))
{
$lookup = mysql_query('SELECT * FROM cust WHERE reg = '$_POST[reg]'AND sname LIKE '%$_POST[sname]%'") |
|
is an open invitation to getting hacked. SQL injection.
Suggest you switch away from the obsolete mysql interface to the mysqli (note the i) and use prepared statements or at the very least escape the data
Also improve your input filtering.
Otherwise you risk this: [xkcd.com...]
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