Generating single JSON file for two tables - php

Question

I am trying to create a single JSON file displaying data from two different tables. I can manage to fetch them, however I don't understand how to concatenate the two arrays. I am new to JSON. Any suggestions/tutorials would be very helpful. Thanks in advance.
Here the expected JSON:
{
"array1":[
{"days":"1","id":"1","image":"image1.jpg, image2.jpg, image3.jpg, image4.jpg"},{"days":"2","id":"1","image":"elephanta.jpg,image2.jpg,image1.jpg,imagica.jpg"},
{"days":"3","id":"1","image":"image3"},{"days":"4","id":"2","image":"image4"}
],
"array2":[
{"id":"1","image_1":"image1.jpg"},
{"id":"2","image_1":"image2.jpg"},
{"id":"3","image_1":"image3.jpg"}
]
}
but the JSON that I get is this:
{
"array1":[
{"days":"1","id":"1","image":"image1.jpg, image2.jpg, image3.jpg, image4.jpg"},{"days":"2","id":"1","image":"elephanta.jpg,image2.jpg,image1.jpg,imagica.jpg"},
{"days":"3","id":"1","image":"image3"},{"days":"4","id":"2","image":"image4"}
]},
{"array2":[
{"id":"1","image_1":"image1.jpg"},
{"id":"2","image_1":"image2.jpg"},{"id":"3","image_1":"image3.jpg"
}
]}
PHP code:
<?php
//open connection to mysql db
$connection = mysqli_connect("localhost","root","","test") or die("Error " . mysqli_error($connection));
//fetch table rows from mysql db
$sql = "select * from image_data";
$sql1 = "select * from one";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));
$result1 = mysqli_query($connection, $sql1) or die("Error in Selecting " . mysqli_error($connection));
//create an array
$imageArray = array();
$imageArray1 = array();
$imageArray["array1"] = array();
$imageArray1["array2"] = array();
while($row =mysqli_fetch_array($result))
{
$tmp = array();
$tmp["days"] = $row["id"];
$tmp["id"] = $row["place_id"];
$tmp["image"] = $row["image"];
array_push($imageArray["array1"], $tmp);
// $imageArray[] = $row;
}
while($row =mysqli_fetch_array($result1))
{
$tmp = array();
$tmp["id"] = $row["id"];
$tmp["image_1"] = $row["image"];
array_push($imageArray1["array2"], $tmp);
// $imageArray[] = $row;
}
echo json_encode($imageArray);
echo ",";
echo json_encode($imageArray1);
//close the db connection
mysqli_close($connection);
?>

Answer 1

If you want the 2 sub arrays to belong to the same array then use only one array and address the 2 sub arrays like this when loading them
$imageArray["array1"][] = $tmp;
and
$imageArray["array2"][] = $tmp;
Amended code
<?php
//open connection to mysql db
$connection = mysqli_connect("localhost","root","","test") or die("Error " . mysqli_error($connection));
//fetch table rows from mysql db
$sql = "select * from image_data";
$sql1 = "select * from one";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));
$result1 = mysqli_query($connection, $sql1) or die("Error in Selecting " . mysqli_error($connection));
//create an array
$imageArray = array(); // <- removed other arrays
while($row =mysqli_fetch_array($result))
{
$tmp = array();
$tmp["days"] = $row["id"];
$tmp["id"] = $row["place_id"];
$tmp["image"] = $row["image"];
$imageArray["array1"][] = $tmp; // <- address array like this
}
while($row =mysqli_fetch_array($result1))
{
$tmp = array();
$tmp["id"] = $row["id"];
$tmp["image_1"] = $row["image"];
$imageArray["array2"][] = $tmp; // <- address array like this
}
echo json_encode($imageArray); // <- only one json_encode
//close the db connection
mysqli_close($connection);
?>
I would have done it this way using stdObject and selecting the specific data from the tables, and using mysqli_fetch_obect() so there is a lot less you have to do woman-draulically
<?php
//open connection to mysql db
$connection = mysqli_connect("localhost","root","","test") or die("Error " . mysqli_error($connection));
//fetch table rows from mysql db
// only select what you want
$sql = "select id,place_id,image from image_data";
$sql1 = "select id,image from one";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));
$result1 = mysqli_query($connection, $sql1) or die("Error in Selecting " . mysqli_error($connection));
//create an array
$imageObject = new stdObject();
while($row =mysqli_fetch_object($result))
{
$imageObject->array1[] = $row;
}
while($row =mysqli_fetch_object($result1))
{
$imageObject->array2[] = $row;
}
echo json_encode($imageObject);
?>