1.PHP發送variabiles AJAX
...
<script src="/jquery-1.3.1.min.js" type="text/javascript"></script>
<script type="text/javascript">
var a = $a
var b = $b
var c = $c
apclick = function() {
$.ajax({
url: 'a1.php',
data: { a: a, b: b, c: c },
datatype: json,
success: function(results) {
if (results.msg == 'success') {
alert(a)
alert(b)
alert(c)
} else {
alert(results.msg)
}
},
error: function(results) {
alert("Data returned: " + results.msg)
}
});
setTimeout("location.reload(true);", 3000)
return false;
}
</script>
.....
<strong><br><a href="#" onclick="apclick();return false;">Afiseaza </a></strong>
a1.php
<?php
$return = array();
$a = $_POST['a'];
$b = $_POST['b'];
$c = $_POST['c']
if ($a == "hello") {
$return['msg'] = 'success';
$return['a'] = "Buna";
};
if ($b == "say") {
$return['msg'] = 'success';
$return['a'] = "Spune";
};
if ($c == "man") {
$return['msg'] = 'success';
$return['a'] = "Om";
};
header("Content-type: application/json");
echo json_encode($a);
echo json_encode($b);
echo json_encode($c);
?>
的問題是:如何 發送,B,C,以a1.php和1個接收A,B,C .PHP
變量res的作用是什麼? – Neo77 2010-11-16 12:54:56
我收到該消息返回的數據:未定義 – Neo77 2010-11-16 14:44:31
忽略變量「res」。我添加了檢查代碼,應該刪除它。 – rsmoorthy 2010-11-17 13:58:50