I want to update my database entry with an 1 or 0 depends on it is checked or not.
My current code looks like this (only one checkbox, the others are the same just other name):
<form method="post" action="" name="form">
<label class="switch"><input type="checkbox" name="csgo" value="1"><span class="slider"></span></label> CSGO;
<input type="submit" value="Save" class="btn btn-primary" name="submit">
<?php
$csgoac = isset($_POST["csgo"]) ? $_POST["csgo"] : 0;
$sql = "UPDATE Users SET csgoactive='".$csgoac."' WHERE ID='1'";
mysqli_query($db, $sql);
?>
</form>
I tested it with echo $csgoac;
(before the query) and the output was checked = 1
and not checked = 0
.
With the query now, it is only updating "0" to the database, what I did wrong on there?
Of course, the first time the page is rendered, $csgoac
is 0 because isset($_POST["csgo"])
is false.
If the checkbox is checked and then the form is submitted, $csgoac
will be 1.
So the question becomes: What is the result of mysqli_query($db, $sql);
?
Are you getting/checking all available feedback? If display_errors
is Off in php.ini, there will not be feedback in the browser, but perhaps an error is in the php errorlog. (For instance, in the snippet, $db
is not defined, so mysqli_query($db, $sql);
fails with Undefined variable: db
.).
If the UPDATE is not successful, mysqli_query($db, $sql);
will return FALSE
and $db->error
should tell you what error was encountered.
You could also use mysqli_stmt_affected_rows
for additional feedback.
Perhaps add some feedback/error checking to get as much information as possible so you can know what is really happening.
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