Notice: Array to string conversion. without array in code

Question

the problem= in my code i get the id of a person out of the database, then i fetch it and then i put it on another table in the database. so first this i get the id out of the database:

$patientid = mysqli_query($connection, "SELECT id FROM patient WHERE     `naam` = '$naam' AND `adres` = '$adres'");

then i fetch it:

$patientid1 = mysqli_fetch_assoc($patientid);

then i put it in another table:

$toevoegen = mysqli_query($connection, "INSERT INTO factuur (`soort`, `hoelang`, `titel`, `omschrijving`, `datum_aangemaaktfactuur`, `patient_id`, `artsid`)
        VALUES ('$soort', '$hoelang', '$titel', '$omschrijving', '$datum', '$patientid1', '$artsid1')");

then when i run the code i get this notice: Notice: Array to string conversion. how do i solve this? i have looked at other questions but they have the same error but different code. i dont even have an array in my code! thanks

ps: if i replace $patientid1 with an number it works perfectly.

Please don't use this code at all. It is wide open for all kind of sql-Injections. — Oliver
– Oliver, Commented Feb 18, 2017 at 12:41
@Oliver i'm new to this, we will learn how to program safe in the next year. — Niek Brouwer
– Niek Brouwer, Commented Feb 18, 2017 at 12:49
@NiekBrouwer if this code never comes live, everything is fine. Just keep it in mind :-) — Oliver
– Oliver, Commented Feb 18, 2017 at 12:58
"we will learn how to program safe in the next year" - It might be too late by then; do it now while you still have a database to work with. — Funk Forty Niner
– Funk Forty Niner, Commented Feb 18, 2017 at 14:05

DonCallisto · Accepted Answer · 2017-02-18 12:44:38Z

2

mysqli_fetch_assoc returns an associative array with selected fields as keys and column values as values.

You should modify code as follows

$patientRow = mysqli_fetch_assoc($patientid);
$patientId = $patientRow['id'];

Remember that for non unique results, you should iterate over whole result set like

while ($patientRow = mysqli_fetch_assoc($patientid)) {
    $patientId = $patientRow['id'];
    // other actions here
}

Last but not least, pay attention to $naam and $naam: use prepared statement instead will be definitely safer

edited Feb 18, 2017 at 12:44

answered Feb 18, 2017 at 12:39

DonCallisto

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