Question

我有一个文件search.php，我从表单中获取输入，然后创建一个url。我希望数据（url_api）将其传递给AJAX脚本，我可以在其中请求json。如何将变量api_url传递给ajax数据？

这是我的代码：

if(filter_input(INPUT_POST, 'submit')){

$area=filter_input(INPUT_POST, 'm', FILTER_SANITIZE_STRING);



// The access url is created with data from the form
$api_url = "http://wwww/api/v1/wwww?";

if ($m !== "") {
    $api_url = $api_url . "m=" . $m;
}



$api_url = $api_url . "&api_key=wxaaaaaaaaaaaaaaaaaaaaaaaa";

和ajax脚本

 $.ajax({
    url: 'search.php', //This is the current doc
    type: "POST",
    dataType:'jsonp', // add json datatype to get json
    data: ?,
    success: function(data){
        console.log(data);
    }
});

Answer 1

在你的JS脚本中：

$.ajax({
    url: 'search.php', //target PHP script
    type: 'POST', //data will be send with POST method
    dataType: 'json', //data will be sent as JSON
    data: { //data sent to PHP script
        key1: 'val1', //keys / values
        key2: 'val2'
    },
    success: function(data) {
        //get the url sent back by PHP and do whatever you want with it
        console.log(data.url);
    }
});

在PHP脚本中：

//get data post by Ajax as POST parameters
$key1Val = $_POST['key1']; // === 'val1'
$key2Val = $_POST['key2']; // === 'val2'

//build your $url

//send back built URL to your JS script as JSON
echo json_encode([
    'url' => $url
]);

PHP输入表单到ajax请求json

1 个答案: