First off, I know about sql injection and that my code is not foolproof, prone to injection etc. Will be working on that next.
Now : from my Android app to my PHP file I submit a JSON array of phone numbers like :
[{"phone_number":"+12345678"},
{"phone_number":"+23456789"},
{"phone_number":"34567890"},
{"phone_number":"45678901"}
etc... etc...
These are contacts in my app user's phone. If these contacts are people who are also users of my app then I want to insert those numbers into my contacts table.
But I can't get it to work. mysqli_fetch_assoc isn't working correctly. I don't know why.
In my contacts table I have 3 columns - an auto increment, user_id and contact_id. The first two values are inserted correctly but the contact_id is always put in as '0', which is wrong.
Here is my code :
require('dbConnect.php');
//this is me, +567890123, my user_id in the user table
$user_id = '20';
//post all contacts in my phone as a JSON array
$json = $_POST['phonenumber'];
$array = json_decode($json);
foreach ($array as $value) {
$phonenumber = $value->phone_number;
$sql = "SELECT username FROM user WHERE username = '$phonenumber'";
$result = mysqli_query($con, $sql);
$num_rows = mysqli_num_rows($result);
if ($num_rows > 0) {
echo "phonenumber is " . $phonenumber . "<br>";
// we want to put $phonenumber in the contacts table, as one of +567890123 contacts
// In the user table get the associated rows of $phonenumber
while ($row = mysqli_fetch_assoc($result)) {
// get the associated user_id in that row, that's what we want to put into the contacts table
$contact_id = $row['user_id'];
$insert_into_contacts_command = "INSERT INTO contacts VALUES(NULL, '$user_id','$contact_id')";
$insert_into_contacts_table = mysqli_query($con, $insert_into_contacts_command);
}
} //if +353864677745 is NOT in the user table...
else {
echo 'not a match.';
}
}
$contact_id = $row['user_id'];
Here $contact_id will be null, because you are trying to access not existing field $row['user_id'] of the $row .
Actually there is only one field username in your results set, as you specified:
$sql = "SELECT username FROM user WHERE username = '$phonenumber'";
Try to change your query to this:
$sql = "SELECT user_id, username FROM user WHERE username = '$phonenumber'";
Your query selects the column username, not userid.
You haven't posted anything about the table user, so it's hard to suggest a new query, but I guess it's the following:
$stmt = mysqli_prepare($con, "SELECT userid FROM user WHERE username = ?");
$stmt->bind_param("s", $phonenumber);
$stmt->execute();
$stmt->bind_result($userid);
while ($stmt->fetch()) {
// Work with $userid
}
You'll note that this uses a prepared statement with a bound parameter. That way, your code is not prone to SQL injections.