How to get JSON data in Correct format using PHP

Question

I'm developing an IOS app, and for the API I'm sending my requests to a URL that should return JSON data with PHP I am Getting like

[{"Child Care":"After Scool,Breakfast Club\n"},{"Child Care":"Breakfast Club"}]

But I want to get like

[{
"Childcare":[
"All of Childcare",
"After school",
"Breakfast Club"
]}

My code is

 <?php
session_start();
$connection=mysqli_connect('localhost','root','','testing') or die(mysqli_error());
 $sql="select `Child Care` from Activity_type ";
 $result = mysqli_query($connection,$sql) or die("Error in Selecting " . mysqli_error($connection));
 $emparray=array();
 while($row =mysqli_fetch_assoc($result)){
 array_push($emparray,$row);
 } 
 header('Content-Type: application/json');
 echo json_encode($emparray); 
 ?>

• Pathik Vejani: post here what you got
• Sai Kumar: [{"Child Care":"After Scool,Breakfast Club\n"},{"Child Care":"Breakfast Club"}]

Answer 1

Simply put it in your JSON array properly.

$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
 $emparray['Child Care'][] = $row['Child Care'];
}
header('Content-Type: application/json');
echo json_encode($emparray); 

---

To answer your second question from the comments:

$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
 foreach($row as $key => $val) {
 $emparray[$key][] = $val;
 }
}
header('Content-Type: application/json');
echo json_encode($emparray); 

• Darren: Pleasure mate @SaiKumar
• Sai Kumar: If I have 7 Coloumns like 'Childcare' what can i do @Darren
• Darren: How exactly do you want the data to be passed through?
• Sai Kumar: I have 7 Columns in a table i want get like [{ "Childcare":[ "All of Childcare", "After school",] "Classes":[ "All of Classes", "Antenatal",].........}
• Sai Kumar: Ya Got it tnq @darren

Answer 2

This will give the format you want.Let me know if this works for you.

while($row =mysqli_fetch_assoc($result)){
 array_push($emparray,$row['Child Care']);
} 
header('Content-Type: application/json');
echo json_encode(array("Childcare" => $emparray)); 

• Sai Kumar: No its getting error like <b>Notice</b>: Undefined offset: 0 in <b>C:\xampp\htdocs\activitytype.php</b> on line <b>9</b><br /> {"Childcare":[null,null]}

Answer 3

Try something like,

$sql="select `Child Care` from Activity_type ";
$result = mysqli_query($connection,$sql) or die("Error in Selecting " . mysqli_error($connection));
$emparray=array();
while($row =mysqli_fetch_assoc($result)){
 $cares = explode(',' $row['Child Care']);
 foreach($cares as $care){
 $emparray[] = $care;
 }
} 
header('Content-Type: application/json'); 
echo json_encode($emparray); 

Use php explode function