jQuery Ajax POST 请求

Question

我正在尝试获取以下代码以通过 POST 将变量发送到 PHP 页面。我不太确定该怎么做。这是我通过 GET 发送数据并通过 JSON 编码接收数据的代码。我需要更改什么才能通过 POST 将变量传递给 process_parts.php？

function imagething(){
    var done = false,
    offset = 0,
    limit = 20;
    if(!done) {
        var url = "process_parts.php?offset=" + offset + "&limit=" + limit;

        $.ajax({
            //async: false, defaults to async
            url: url
        }).done(function(response) {

            if (response.processed !== limit) {
                // asked to process 20, only processed <=19 - there aren't any more
                done = true;
            }
            offset += response.processed;
            $("#mybox").html("<span class=\"color_blue\">Processed a total of " + offset + " parts.</span>");
            alert(response.table_row);
            console.log(response);
            imagething(); //<--------------------------recursive call
        }).fail(function(jqXHR, textStatus) {

            $("#mybox").html("Error after processing " + offset + " parts. Error: " + textStatus);

            done = true;
        });
    }
}
imagething();

Answer 1

默认方法是GET，要更改它，请使用type 参数。您还可以将查询字符串属性作为对象提供，这样它们在 URL 中就不会立即显现出来：

var url = "process_parts.php";

$.ajax({
    url: url,
    type: 'POST',
    data: {
        offset: offset,
        limit: limit
    } 
}).done(function() {
    // rest of your code...
});

Answer 2

试试这个

         $.ajax({
                    url: "URL",
                    type: "POST",
                    contentType: "application/json;charset=utf-8",
                    data: JSON.stringify(ty),
                    dataType: "json",
                    success: function (response) {
                        alert(response);
                    },

                    error: function (x, e) {
                        alert('Failed');
                        alert(x.responseText);
                        alert(x.status);
                    }
                });