mercredi 28 janvier 2015

PHP Ajax error, get 404 when type first char


I see error in my console it says: showHint is not defined, while I defined it already, also this live search doesnt return anything when I type it. It assume should work like this.


gethint.php



// get the q parameter from URL
$q = $_REQUEST["q"];

$hint = "";

// lookup all hints from array if $q is different from ""
if ($q !== "") {
$q = strtolower($q);
$len=strlen($q);
foreach($a as $name) {
if (stristr($q, substr($name, 0, $len))) {
if ($hint === "") {
$hint = $name;
} else {
$hint .= ", $name";
}
}
}
}

// Output "no suggestion" if no hint was found or output correct values
echo $hint === "" ? "no suggestion" : $hint;
?>


index.php



<html>
<head>
<script>
function showHint(str) {
if (str.length == 0) {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","gethint.php?q=" + str, true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<p><b>Start typing a name in the input field below:</b></p>
<form>
First name: <input type="text" onkeyup="showHint(this.value)">
</form>
<p>Suggestions: <span id="txtHint"></span></p>
</body>
</html>


The error when typing a character enter image description here





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