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); 

    ?>

[{"Child Care":"After Scool,Breakfast Club\n"},{"Child Care":"Breakfast Club"}] — Sai Kumar
– Sai Kumar, Commented Dec 24, 2015 at 4:57

Darren · Accepted Answer · 2015-12-24 05:22:46Z

3

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);

edited Dec 24, 2015 at 5:22

answered Dec 24, 2015 at 5:06

Darren

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6 Comments

Darren Over a year ago

Pleasure mate @SaiKumar

Sai Kumar Over a year ago

If I have 7 Coloumns like 'Childcare' what can i do @Darren

Darren Over a year ago

How exactly do you want the data to be passed through?

Sai Kumar Over a year ago

I have 7 Columns in a table i want get like [{ "Childcare":[ "All of Childcare", "After school",] "Classes":[ "All of Classes", "Antenatal",].........}

Sai Kumar Over a year ago

Ya Got it tnq @darren

|

Murali krishna · Accepted Answer · 2015-12-24 05:10:44Z

1

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));

edited Dec 24, 2015 at 5:10

answered Dec 24, 2015 at 5:05

Murali krishna

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1 Comment

Sai Kumar Over a year ago

No its getting error like Notice: Undefined offset: 0 in C:\xampp\htdocs\activitytype.php on line 9 {"Childcare":[null,null]}

Double H · Accepted Answer · 2015-12-24 05:05:39Z

0

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

answered Dec 24, 2015 at 5:05

Double H

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