Handle POST request in JSON format

Question

I'm trying to handle a POST request from a web service. It's sending an HTTP POST request like this:

{
"latitude":"12.232",
"longitude":"123.323"
}

It's posting to a PHP file on my server. I know that it is hitting the file for sure. However, I'm not getting the data.

In my PHP, I have this (leaving out a bunch of stuff:

$json = file_get_contents('php://input');
$obj = json_decode($json);
$mine ="sixteen"; //using this for a test

$sql = "INSERT INTO rr_emergency (random) VALUES('$obj');";
$result = $dbh->query($sql)->fetchAll(PDO::FETCH_ASSOC);

This makes no change to my database.

If I do this instead:

$sql = "INSERT INTO rr_emergency (random) VALUES('$mine');";

Then "sixteen" is added in the right spot in a new row in my table each time the webservice calls my PHP. This is how I know I'm receiving data.

NOTE: I was trying to simply add $obj into my table just to see the data format that's returned before I tried to properly parse it and put everything where it belongs.

What am I doing wrong here? I think the problem is here ($json = file_get_contents('php://input');), but not sure what else to try.

Thanks.

You cannot store an PHP object onto a mysql table column. The result of `$obj = json_decode($json);` is a PHP Object not a string — RiggsFolly, Jul 12 '16 at 22:34
Add some error checking to your query code and you will probably see a useful error message. — RiggsFolly, Jul 12 '16 at 22:35
can you not get the values with `$_POST['latitude'];`? I guess JSON won't allow it? — ntgCleaner, Jul 12 '16 at 22:38
@ntgCleaner Thought this might work, it does successfully add a new entry to the database, but it's completely empty. — jonmrich, Jul 12 '16 at 22:41

score 3 · Accepted Answer · answered Jul 12 '16 at 22:40

3

So there's a few problems

$obj = json_decode($json);

This will return an object. You want an array

$obj = json_decode($json, true);

Then your PDO is incorrect

$sql = "INSERT INTO rr_emergency (random) VALUES(:val);";
$prep = $dbh->prepare($sql);
foreach($obj as $row) $prep->execute([':val' => $row]);

This will insert your data correctly (using a prepared statement) and loop over the JSON return data

answered Jul 12 '16 at 22:40

Machavity

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This completely got me where I needed to go. The issue was this: ```$obj = json_decode($json, true);```, which then allowed me to do ```$first =$obj['latitude'];``` and then use that variable to put the right data in my table. The loop wasn't necessary, but I can see where you got that based on my question. Thanks! – jonmrich Jul 12 '16 at 22:45

BeetleJuice · Answer 2 · 2016-07-12T22:54:17.897

You're trying to insert an object, when you really need a string. use:

$obj = json_decode($json, true)
$obj_str = implode(", ", $obj);
$sql = "INSERT INTO rr_emergency (random) VALUES('$obj_str');";

After I posted the above, you added:

I was trying to simply add $obj into my table just to see the data format

Objects do not inherently convert to strings, so putting $obj within your query doesn't work. The way I store objects in my DB when I've needed to, is to store the JSON notation directly.

$json = file_get_contents("php://input");
$sql = "INSERT INTO rr_emergency (random) VALUES('$json')";

You lose the ability to perform filtering and selecting operations within the object, but it's an effective way to pack away data that you won't need the DB to parse through.

If you need well formatted, easy to read structure:

$obj = json_decode($json);
$obj_str = print_r($obj,true); //store formatted string
$sql = "INSERT INTO rr_emergency (random) VALUES('$obj_str');";

If as you said, all you need to do is "just see the data format", I suggest echoing to the screen or writing to a log file; do one of the following. To print to screen:

print_r($obj);

To write to file:

$filepath = "/path/to/file.txt"
file_put_contents($filepath,print_r($obj,true));

Important note

Entering text directly into your DB queries without escaping it makes you vulnerable to SQL injection attacks. Use prepared statements instead.

Handle POST request in JSON format

2 Answers2