My php code is not inserting into database

Question

So, i was working on my website. And after 10 minutes I was ready. I tried it. I have 4 column in my database. id, times, left_over and bruh. When I tried it Everything was getting inserted but not bruh, I don't know what I've did wrong. Here is my script:

$link = mysqli_connect("localhost", "root", "", "testang");
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
$channelid = mysqli_real_escape_string($link, $_POST['channelid']);
$subs = mysqli_real_escape_string($link, $_POST['subs']);
$points = $subs*20;
$lell = $userRow['userid'];
$lel = $userRow['userpoints'];
$pointsreal = $lel - $points;
if ($points > $lel) {
header("Refresh:0; url=NO.php");
}
$bew = "INSERT INTO yt (times, left_over, bruh)
VALUES ('0', '$subs', '$channelid')";
if ($link->query($bew) === TRUE) {
    echo "";
} else {
    echo "Error: " . $bew . "<br>" . $link->error;
}
mysqli_close($link);

And I don't get any messages that there's gone something wrong. Everything is getting inserted but bruh not. Does some one know why?

This is my form:

<form action="insert.php" method="post">
    <p>
        <label for="firstName">Youtube channel ID</label><br>
        <input type="text" name="id" id="channelid">
    </p><br><br>
    <p>
        <label for="lastName">What is your email?</label><br>
        <input type="text" name="subs" id="subs">
    </p><br><br>
    <input type="submit" value="Submit">
</form>

Show us your HTML form. Also, you should consider using prepared statements. — Bart Friederichs
– Bart Friederichs, Commented Sep 24, 2016 at 11:14
But I have this: ` <label for="firstName">Youtube channel ID</label><br><input type="text" name="id" id="channelid">` — Aaron Jonk
– Aaron Jonk, Commented Sep 24, 2016 at 11:15

Gynteniuxas · Accepted Answer · 2016-09-24 11:18:39Z

1

Now that I know from comments what's wrong, let me clarify a bit what's wrong in here. In your field:

<input type="text" name="id" id="channelid">

You have ID: channelid and name: id. To be able to get $_POST['channelid'], you need to set attribute name with the same string as in your $_POST index. Therefore:

<input type="text" name="channelid" id="channelid">

answered Sep 24, 2016 at 11:18

Gynteniuxas

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1 Comment

Drew Over a year ago

@Aaron How does accepting an answer work? ... that's how we roll here.

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