PHP <b>Warning</b>: Illegal string offset 'image_name' from Android - php

Question

Trying to send arraylist data to server using php as bridge from Android (using Volley), and successfully getting data into database table (and also getting response as submit, I mentioned in php) in Android
So everything looks perfect, if we talk about db connection, storage and response, but one unusal issue always getting 2 in table field, instead the original value
::::imageDataInString::::: {"b":"2020-01-01_00:01:11.jpg"}
::::imageDataInString::::: {"b":"2020-01-01_00:01:21.jpg"}
::::imageDataInString::::: {"b":"2020-01-01_00:01:31.jpg"}
::::Response::::: <br />
<b>Warning</b>: Illegal string offset 'image_name' in <b>C:\xampp\htdocs\test\send_data.php</b> on line <b>15</b><br />
<br />
submit
Here is the foreach, I'm using in php to upload all the data available in a list
$content = $_POST['my_images'];
$json = json_decode($content, true);
foreach ($json as $key => $value) {
$image_name = $value["image_name"];
}
android code
ImageData imageData = new ImageData();
imageData.setImageName(array[0]);
String imageDataInString = new Gson().toJson(imageData);
Log.d("::::imageDataInString::::", imageDataInString);
....
params.put("my_images", imageDataInString);
in my csv, available in raw folder of res in Android, I have these data:
2020-01-01_00:01:11.jpg
2020-01-01_00:01:21.jpg
2020-01-01_00:01:31.jpg

Answer 1

There are a few points to note:
GSON converts the variable names of the class (ImageData) as keys and appends the values, to form the JSON. Hence, you need to use the same variable names, as you have used in the ImageData class to fetch the data, in your PHP Code.
ImageData imageData = new ImageData();
imageData.setImageName(array[0]);
String imageDataInString = new Gson().toJson(imageData);
Hence, the array key should be array["ImageName"]
In your PHP code, the second line, where you are JSON decoding the $content variable, the variable $json is already converted in an Array. You don't need to iterate and fetch corresponding keys.
Hence, you should do something like this:
$content = $_POST['my_images'];
$json = json_decode($content, true);
$image_name = $json["ImageName"];
With this, the variable $image_name will hold the value which you are looking for.