update json replacing variable input from form in php

Question

Greeting, I'm new to PHP, and currently looking for a way to edit json. I have a form to input the key that wants to edit. I echo the input from form and successfully got the output showing but it didn't pass to json file to update. The code is as below.

function testfun()
{
    // read file
    $data = file_get_contents('Book2.json');

    // decode json to array
    $json_arr = json_decode($data, true);

    foreach ($json_arr as $key => $value) {
        if ($value['CELL_NAME'] == $_POST['in_cell']) {
            $json_arr[$key]['@25'] = $_POST['in_remark'];
        }
    }

    // encode array to json and save to file

    file_put_contents('Book2.json', json_encode($json_arr));
}
//you this post contain test?
//test is the the button when i submit the form

if (array_key_exists('test',$_POST))
{
    testfun();
}

Am I missing something?

seems like your code is okay. are you sure json file didn't change? — david
– david, Commented Sep 26, 2018 at 7:12
I don't manage to understand your exact problem. Those the testfun being called? can you please print the $json_arr before and after the for loop? — dWinder
– dWinder, Commented Sep 26, 2018 at 7:13
Can you please share example (short one) of the $json_arr? — dWinder
– dWinder, Commented Sep 26, 2018 at 7:20

J.Doe · Accepted Answer · 2018-09-26 07:59:38Z

1

Try my code.

 function testfun()
{
// read file
$data = file_get_contents('Book2.json');

// decode json to array
$json_arr = array(json_decode($data, true));
foreach ($json_arr as $key => $value) {
    if ($value['CELL_NAME'] == $_POST['in_cell']) {
    $json_arr[$key]['@25'] = $_POST['in_remark'];
    }
}

// encode array to json and save to file

file_put_contents('Book2.json', json_encode($json_arr));
}


if (array_key_exists('test',$_POST))
{
    testfun();
}

answered Sep 26, 2018 at 7:59

J.Doe

156 bronze badges

Sign up to request clarification or add additional context in comments.

12 Comments

J.Doe Over a year ago

Foreach $json_arr because it's a one-dimensional array $value that only outputs the result directly, so no key value is required

dWinder Over a year ago

What have you changed?

J.Doe Over a year ago

Change here $json_arr = array(json_decode($data, true));@user10417098

dWinder Over a year ago

What? I didn't cast to array - I think his input is corrupted

J.Doe Over a year ago

You just need to change $json_arr = json_decode($data, true); of your code to $json_arr = array(json_decode($data, true)).

|

dWinder · Accepted Answer · 2018-09-26 08:22:05Z

0

As you mention in the comments, the content of the $json_arr is:

{"CELL_NAME":"1234A","@25":"remark value"}

So when you trying to access in:

 foreach ($json_arr as $key => $value) {
     if ($value['CELL_NAME'] == $_POST['in_cell']) {
         $json_arr[$key]['@25'] = $_POST['in_remark'];
     }
 }

it has no key for $value at CELL_NAME key.

I guess your $data from the file should be like that (an array of JSONs):

$data = '[{"CELL_NAME":"1234A","@25":"remark value"}, {"CELL_NAME":"1234B","@25":"remark value"}]';

Now you can do this and it will work:

$arr = json_decode($data, true);
foreach ($arr as $key => $value) {
    if ($value['CELL_NAME'] == "1234A") // you can change with $_POST['in_cell']
        $arr[$key]['@25'] = "test"; // you can change with $_POST['in_remark']
}

edited Sep 26, 2018 at 8:22

answered Sep 26, 2018 at 8:14

dWinder

11.7k3 gold badges27 silver badges40 bronze badges

Collectives™ on Stack Overflow

update json replacing variable input from form in php

2 Answers 2

12 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

12 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related