php/ajax: form get method not working

Question

i have two php file , test.php and test2.php
in 'test.php' i have a form and i want to get name and echo it with ajax and php here is my code , what is wrong with it?

test.php

<html>
<head>
<script>
function showHint(str)
{
if (str.length==0)
  {
  document.getElementById("txtHint").innerHTML="";
  return;
  }
var xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","test2.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<p><b>Start typing a name in the input field below:</b></p>
<form method="get">
First name: <input type="text" name="fname">
<input type="submit" value="click me" onclick="showHint(fname)">
</form>
<p>Suggestions: <span id="txtHint"></span></p>

</body>
</html>

and test2.php is

<?php
$q = intval($_GET['q']);
echo $q;
?>

You're supposed to tell us what's wrong with it, what errors you get or what happens vs what you expect to happen, and what troubleshooting step has you stuck. — Digital Chris
– Digital Chris, Commented Feb 21, 2014 at 12:50
there is no error , just not working , i mean when i click on button there will be no change in tag <span id="txtHint"> — farbod_chitsaz
– farbod_chitsaz, Commented Feb 21, 2014 at 12:54
Because you are not returning any parameters through url. You are just printing the value of $q. You can use this value on success though instead of getting the url with .open(). Found it. Check this link. It might help you. stackoverflow.com/questions/4971254/… — MentalRay
– MentalRay, Commented Feb 21, 2014 at 13:01

Andrew Mackrodt · Accepted Answer · 2014-02-21 13:12:58Z

0

The button in the form is a submission input which will result in the page being reloaded. Add return false to your onclick method to prevent this from happening:

<input type="submit" value="click me" onclick="showHint(fname); return false">

Additionally, your showHint function expects a string but you are passing an HTML input. You can get the text value by adding the following to the top of the function:

if (typeof str !== 'string' && typeof str.value !== 'undefined') {
    str = str.value;
}

answered Feb 21, 2014 at 13:12

Andrew Mackrodt

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2 Comments

farbod_chitsaz Over a year ago

i made this changes, i think it becames better but not completly ok, it prints just '0' (without quet's) , what is wrong?

Andrew Mackrodt Over a year ago

Your php code forces the output to be an integer as you are using the function intval. You probably want to use htmlspecialchars instead.

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