PHP mySqli not updating table

Question

I am trying to update the table but nothing is changing. The database name and fields are correct.

<?php
require("config.php");
    $forname  = $_POST['name'];
    $newval = "yes";
    mysqli_query($con, "UPDATE pupils SET signin = '$newval' WHERE forname = '$forname'");
    mysqli_close($con); 
?>

Help appreciated! Thanks,

UPDATE

Appears that data is not posting correctly for some reason.

<form class="form-inline" name="markin" role="form" method="POST" action="markin.php">
    <div class="form-group">
    <select class="form-control" name"name" id="name">
    <?php
    $query = "SELECT * FROM `pupils` WHERE signin = 'no'";//Grab the data
    $result = mysqli_query($con, $query);
    while ($row = mysqli_fetch_array($result)) {//Creates a loop to loop through results
    echo "<option>" . $row['forname'] . "</option>";//$row['index'] the index here is a field name
    }
    ?>
    </select>
    </div>
    <button type="submit" class="btn btn-success">Mark in</button>
    </form>

[ahhhhhhhhhhhhhhhh](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) — Alexis King, Jan 28 '15 at 22:17
Ensure that data is actually posted and that the value for `$_POST['forname']` is actually in `pupils`. — Ding, Jan 28 '15 at 22:19
@AlexisKing serious? You forgot at the start one [`h`](http://php.net/manual/en/book.pdo.php) and and at the end one [`h`](http://php.net/manual/en/book.mysqli.php) — Rizier123, Jan 28 '15 at 22:19
@crablab Do you get any errors? Did you checked that the connection didn't failed? (Error reporting: ``) — Rizier123, Jan 28 '15 at 22:20
You're not checking for any MySQL(i) errors at all. Do that. — Charlotte Dunois, Jan 28 '15 at 22:23
Correction: appears that data isn't posting correctly-sorry. Name is same at both ends and form is configured correclty.Not sure what is happening... @Ding — crablab, Jan 28 '15 at 22:40

Phil · Accepted Answer · 2019-12-16T22:47:38.763

1

First, I suggest you read the prepared statements quickstart guide which would lead you to something like this...

$stmt = $con->prepare('UPDATE pupils SET signin = ? WHERE forname = ?');
$stmt->bind_param('ss', $newval, $forname);
$stmt->execute();

As an added bonus, you should set mysqli to throw exceptions on error so you don't have to check return values all the time. In your config.php file, before creating the connection, do this...

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$con = new mysqli(...); // or however you create $con

and as mentioned in the comments, the php.ini file in your development environment should enable full error reporting with the following directives

error_reporting = E_ALL
display_errors = On

To directly answer your question, you're missing an equals sign for the <select> elements name attribute. It should be

<select class="form-control" name="name" id="name">

edited Dec 16 '19 at 22:47

answered Jan 28 '15 at 22:25

Phil

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Is it not possible to continue using it this way? All my other queries are formatted thus and I would prefer not go around fixing them. – crablab Jan 28 '15 at 22:27
1

Of course it's possible but it's hardly a wise decision – Phil Jan 28 '15 at 22:28
I see. I did some debugging and it appears the form isn't posting...even though it was a few minutes ago. Running the query with a harcoded variable for forname edited the table as expected, – crablab Jan 28 '15 at 22:29
Having full error reporting would have alerted you to the fact that `$_POST['forname']` didn't exist – Phil Jan 28 '15 at 22:31
Yeah, I take the point and on future projects I will read up on this. – crablab Jan 28 '15 at 22:34

PHP mySqli not updating table

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