从PHP获取多个数组到AJAX

Question

修改

我需要将两个获取的数组放入一个数组中，然后使用echo json_encode将其发送回AJAX。

这是我的PHP代码：

header('Content-Type: application/json');
$arr = array();
$user = $_SESSION['username'];
$id_logged = $_SESSION['login_id'];



$date1 = $_POST['current_year'];

    //$res = array();
    $c = "cash";
    $i = "installment";
    //SUM of cash paid Month
    $sql = "SELECT 
    count(app_id) as patients, 
    t.m as month 
FROM (    
   SELECT 1 AS m UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION
   SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION
   SELECT 9 UNION SELECT 10 UNION SELECT 11 UNION SELECT 12) AS t
LEFT JOIN dentist.appointment ON t.m = month(date_app) AND
          id_logged = :logged AND year(date_app) = :year_now
GROUP BY t.m";
$sqlStmt = $conn->prepare($sql);
$sqlStmt->bindValue(":logged", $id_logged);
$sqlStmt->bindValue(":year_now", $date1);
$exec = $sqlStmt->execute();
$res1 = $sqlStmt->fetchAll();

$sql2 = "SELECT 
    count(app_id)+1 as patients, 
    t.m as month 
FROM (    
   SELECT 1 AS m UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION
   SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION
   SELECT 9 UNION SELECT 10 UNION SELECT 11 UNION SELECT 12) AS t
LEFT JOIN dentist.appointment ON t.m = month(date_app) AND
          id_logged = :logged AND year(date_app) = :year_now
GROUP BY t.m";
$sqlStmt2 = $conn->prepare($sql2);
$sqlStmt2->bindValue(":logged", $id_logged);
$sqlStmt2->bindValue(":year_now", $date1);
$exec2 = $sqlStmt2->execute();
$res2 = $sqlStmt2->fetchAll();

$res['one']=$res1;
$res['two']=$res2;

echo json_encode($res);

这里是XHR的答复：

现在在AJAX脚本中，我需要获取数组的每个部分，以便我可以在图表上显示它：

success:function(res)
{
  var patientData = [];
  $.each(res, function( key, row)
  {
    patientData.push(row['one']);
    console.log(patientData);
  });
  //Bar Chart Script;
 }

但我无法在控制台和条形图上看到任何数据。控制台的结果是undefined：

Answer 1

以下是如何操作：

    success: function(data){
        var patientData = [];
        var res = $.parseJSON(data);
        $.each(res, function(index, element) {
            alert(res[index]);
            patientData.push(res[index]);
        });
    }

Answer 2

您将获得a = [i for i in range(1,11)] #a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] b = [a[i]+1 if i%2==1 else a[i] for i in range(len(a))] #b = [1, 3, 3, 5, 5, 7, 7, 9, 9, 11] 中的密钥（“one”，“two”，...）和key中的行数据

  

由于$ .each在第一个参数中返回对象属性名称（row），在第二个参数中返回属性key，所以我们如下：

value

例如，我们可以像这样检查结果

success:function(res){
  var res = $.parseJSON(res);
  console.log(res);
  var patientData = [];

  $.each(res, function (key, row){
    patientData.push(row);
    console.log(patientData);
  });
  //Bar Chart Script;
}

$.each({"one":[1, 2, 3, 4], "two":[5, 6, 7, 8]}, function (key, value){
  console.log(key, value);
  
  /* write to document */
  document.write("<br>");
  document.write("Key:" + key);
  document.write(", Value:" + value);
});

Answer 3

尝试：

success:function(res)
{
  var res = $.parseJSON(res);
  console.log(res);
  var patientData = [];
  $.each(res, function (key, row)
  {
    patientData.push(row.one);
    console.log(patientData);
  });
  //Bar Chart Script;
 }