我的一个朋友给了我这个片段的js文件,可对我来说是有用的:返回值以JSON/AJAX
$.ajax({
type: "POST",
url: "../CheckPerson.php",
data: "{'lastName':'" + _lname + "','firstName':'" + _fname + "','middleName':'" + _mname + "'}",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var res = response.d;
if (res == true) {
jAlert('Patient Name already exists!', 'Error');
return;
}
else {
$.ajax({
type: "POST",
url: "../NewPerson.php",
data: "{'lastName':'" + _lname + "','firstName':'" + _fname + "','middleName':'" + _mname + "','gender':'" + _gender + "','birthDate':'" + _bday + "','ssn':'" + _ssn + "'}",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var _id = response.d;
if (_id.length != 0) {
$('#patientName').removeAttr('disabled');
$('#patientName').val(_lname + ", " + _fname + " " + _mname);
$('#patientId').val(_id[0].patientID);
$('#dateOfBirth').val(_bday);
$('#referringDoctor').removeAttr('disabled');
$('#referringDoctor').focus();
$('#patientAge').val(_id[1]);
$('#ptLastName').val('');
$('#ptFirstName').val('');
$('#ptMiddleName').val('');
$('#ptGender').val('');
// $('input[name="birthdate"]').val(); // $('#ptBirthDate').val();
$('#ptSSN').val('');
}
// if (_id == true) {
// }
insertCallback(_id);
// $('#diagnosis tbody>tr:last').attr('dinfo', '_IEDiagnosis|' + _id);
},
failure: function (msg) {
alert(msg);
}
});
}
}
});
我使用PHP,但我是新来使用JSON。有没有办法在我的PHP文件返回“response.d”一true
值:
success: function (response) {
var _id = response.d;
}
这里是我的逻辑,但不知道该代码使用方法:
$lastname = isset($_REQUEST['lastName'])?$_REQUEST['lastName']:'';
$firstname = isset($_REQUEST['firstName'])?$_REQUEST['firstName']:'';
$middlename = isset($_REQUEST['middleName'])?$_REQUEST['middleName']:'';
$response = array();
mysql_connect ("localhost", "root") or die ('Error: ' . mysql_error());
mysql_select_db ("healthpal");
$query = "SELECT Lastname, Firstname, MiddleName FROM db_patients WHERE Lastname = '$lastname' || Firstname = '$firstname' || MiddleName = '$middlename'";
$qrytest = mysql_query($query);
if (isset($qrytest)) {
//"response.d" will be true if the query return not NULL
}
感谢您的帮助。已经使代码工作。 我现在的问题是如何在我的php文件中发布我的Json数据。 类似于: $ var = $ _POST ['jsondata1']; //该变量将具有Json数据的值1 我在尝试json_decode();还有 isset($ _ REQUEST ['lastName'])?$ _ REQUEST ['lastName']:''; 但没有奏效。 任何帮助将真正被赞赏。谢谢.. –