無法讓函數調用在PHP中工作

Question

我在使用$.getJSON請求獲取JSON數據時遇到問題。在一個PHP文件中，我有我的功能。該代碼是：

function get_topchart($dbc, $var_ziekenhuis_id, $var_department_id) { 

    if($var_department_id == "Not Entered") { 

     $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' ORDER BY afdeling DESC"; 
     $result = mysqli_query($dbc,$query); 

     $data = array(); 
     while ($row = $result->fetch_assoc()) { 
      $data[] = $row; 
     } 
     echo json_encode($data); 
     //return(json_encode($data)); 

    } else { 

     $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' and afdeling='$var_department_id' ORDER BY afdeling DESC"; 
     $result = mysqli_query($dbc,$query); 

     $data = array(); 
     while ($row = $result->fetch_assoc()) { 
      $data[] = $row; 
     } 
     echo json_encode($data); 
     //return(json_encode($data)); 
    } 
} 

我用下面的代碼來調用上面的函數，它是在getcharts.php：

<?php 
session_start(); 

include('../functions/datafunctions.php'); 
include('../config/db.php'); 


if(!isset($_SESSION['username'])) { 
    header('Location: ../login.php'); 
} 

//Determine the function that should be called 
$var_function_id = $_GET['funcid']; 
//Set variables 
$var_ziekenhuis_id = $_GET['hosp']; 
$var_department_id = $_GET['dept']; 

switch ($var_function_id) { 
    case "top_chart": 
     //Call function 
     get_topchart($dbc, $var_ziekenhuis_id, $var_department_id); 
    case "bottom_chart": 
     //New code here 
} 
?> 

的$.getJSON我用的是：

$.getJSON('functions/getcharts.php', { 
    "hosp":hospital, 
    "dept":department, 
    "funcid":functionid}, 

當我跳過整個getcharts.php並直接使用開關功能時，事情確實起作用。這代碼：

<?php 

session_start(); 
include('../config/db.php'); 

//Controleer eerst of gebruiker succesvol is ingelogd: 

if(!isset($_SESSION['username'])) { 
    header('Location: ../login.php'); 
} 

//Determine the function that should be called 
$var_function_id = $_GET['funcid']; 
//Set variables 
$var_hospital_id = $_GET['hosp']; 
$var_department_id = $_GET['dept']; 

switch ($var_function_id) { 
    case "top_chart": 

     //function get_topchart($dbc, $var_ziekenhuis_id, $var_department_id) { 

      if($var_department_id == "Not Entered") { 

       $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' ORDER BY afdeling DESC"; 
       $result = mysqli_query($dbc,$query); 

       $data = array(); 
       while ($row = $result->fetch_assoc()) { 
        $data[] = $row; 
       } 
       echo json_encode($data); 
       //return(json_encode($data)); 

      } else { 

       $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' and afdeling='$var_department_id' ORDER BY afdeling DESC"; 
       $result = mysqli_query($dbc,$query); 

       $data = array(); 
       while ($row = $result->fetch_assoc()) { 
        $data[] = $row; 
       } 
       echo json_encode($data); 
       //return(json_encode($data)); 
      } 
     //} 
} 

?> 

的$getJSON代碼在這種情況下是：

$.getJSON('functions/datafunctions.php', { 
    "hosp":hospital, 
    "dept":department, 
    "funcid":functionid}, 

我在做什麼錯？

Answer 1

更換

echo json_encode($data); 

與

return json_encode($data); 

更換

switch ($var_function_id) { 
    case "top_chart": 
     //Call function 
     get_topchart($dbc, $var_ziekenhuis_id, $var_department_id); 
    case "bottom_chart": 
     //New code here 
} 

有：

switch ($var_function_id) { 
    case "top_chart": 
     //Call function 
     echo get_topchart($dbc, $var_ziekenhuis_id, $var_department_id); 
    case "bottom_chart": 
     //New code here 
} 

Answer 2

取消對return語句的功能： 功能get_topchart（$ DBC，$ var_ziekenhuis_id，$ var_department_id）{

if($var_department_id == "Not Entered") { 

    $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' ORDER BY afdeling DESC"; 
    $result = mysqli_query($dbc,$query); 

    $data = array(); 
    while ($row = $result->fetch_assoc()) { 
     $data[] = $row; 
    } 
    return(json_encode($data)); 

} else { 

    $query = "SELECT afdeling, ligduur as ld FROM datavoorbeeld WHERE ziekenhuis='$var_hospital_id' and afdeling='$var_department_id' ORDER BY afdeling DESC"; 
    $result = mysqli_query($dbc,$query); 

    $data = array(); 
    while ($row = $result->fetch_assoc()) { 
     $data[] = $row; 
    } 
    return(json_encode($data)); 
} 

}

，影響您的功能上，當一個變量return語句你叫它。

變化：

get_topchart($dbc, $var_ziekenhuis_id, $var_department_id);

與

$topchart = get_topchart($dbc, $var_ziekenhuis_id, $var_department_id);