Question

1.PHP

...
<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);

?>

问题是：如何将a，b，c发送到a1.php并在1.php中接收a，b，c

Answer 1

data: { 'a': 'a', 'b': 'b', 'c': 'c' },
type: 'POST'

尝试（使用数据上的引号和类型设置为POST。）

Answer 2

提供的代码有几个语法错误，您应该在发布之前修复它们。

无论如何，这是适合您的工作代码：

<script type="text/javascript">
    var a = "hello";
    var b = "say";
    var c = "man";
    var res;
    apclick = function() {
        $.ajax({
            url: 'a1.php',
            data: { a: a, b: b, c: c },
            datatype: 'json',
            type: 'POST',
            success: function(results) {
                res = results;
                if (results.msg == 'success') {
                    alert(results.a)
                    alert(results.a)
                    alert(results.a)
                }
                else {
                    alert(results.msg)
                }
            },
            error: function(results) {
                alert("Data returned: " + results.msg );
            }
        });

        setTimeout("location.reload(true);",30000);
        return false;
    };
</script>
</head>
<body>

<strong>
    <br>
    <a href="#" onclick="apclick();return false;">Afiseaza </a>
</strong>

和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($return);

?>

Answer 3

为rsmoorthy竖起大拇指，但我不会使用$ _REQUEST

决定你的请求方法并将类型设置为Gazler相应的建议。

发送变量ajax

3 个答案: