我正在尝试通过ajax提交数据,这是我的信息:
var jsondata =
{"address" : [
{ "id": addid, "streetaddress": streetaddress, "city": city, "state": state, "zipcode": zipcode, "latitude": latitude},
]
};
var jsontosend = JSON.stringify(jsondata, null, 2);
ajax函数:
$.ajax({
type: "POST",
url: "process.php",
contentType: "application/json; charset=utf-8",
dataType: "JSON",
data: jsontosend,
success: function(msg){
alert(msg);
}
});
return false;
alert('Data sent');
}
array(0) {
}
我正在使用(alert)并且通过POST方法在firebug中显示所有正确的参数,但唯一可以传递任何数据的方式是使用GET方法。
非常感谢任何建议!
编辑:从firebug添加POST数据。 这是从警报函数中警报的内容:
{"address":[{"id":1473294,"streetaddress":"3784 Howard Ave","city":"Washington DC","state":"DC","zipcode":20895,"latitude":39.027820587}]}
这是Firebug在使用POST方法时显示的传递内容:
myData=%7B%0A++++%22address%22%3A+%5B%0A++++++++%7B%0A++++++++++++%22id%22%3A+76076%2C%0A++++++++++++%22streetaddress%22%3A+%223784+Howard+Ave%22%2C%0A++++++++++++%22city%22%3A+%22Washington+DC%22%2C%0A++++++++++++%22state%22%3A+%22DC%22%2C%0A++++++++++++%22zipcode%22%3A+20895%2C%0A++++++++++++%22latitude%22%3A+39.027820587%0A++++++++%7D%0A++++%5D%0A%7D
这是对 $_POST 进行 var_dump 的响应:
array(0) {
}
这是 $_POST['myData'] 的 var_dump 结果。
NULL