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Home/ Questions/Q 7721013
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T03:49:00+00:00 2026-06-01T03:49:00+00:00

this should not be hard but I’m struggling with it so I thought I’d

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this should not be hard but I’m struggling with it so I thought I’d request help as I could not finding anything specific on web.

I am posting from a checkbox on a form. When I look at the querystring in browser window when I freeze output, it shows, the names of the variables = on. Similarly, when I echo the posted values they show up as on. The problem is when inserting into the dbase although inputting TRUE works, on does not. Fieldtype is tinyint(1). Do I have to convert all the ons to TRUEs. I feel like this has never been a problem before.

<form action="processform.php" method="post">
<input type = "checkbox" name="var1" CHECKED>
<input type = "checkbox" name="var2" CHECKED>
<input type="submit" name="submit" value="submit"></form>

script on other end.

$var1 = $_POST['var1'];
$var2 = $_POST['var2'];
echo $var1; //echoes "on"

$sql = "INSERT into table (var1,var2) VALUES (TRUE,TRUE)"; //works
$sql = "INSERT into table (var1,var2) VALUES ($var1,$var2)"; //does not work.
$sql = "INSERT into table (var1,var2) VALUES ('$var1','$var2')"; //does not work.

mysql_query($sql);

Is this a typo? Can’t imagine you really need to change “on” blank to TRUE or FALSE for each variable

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  1. Editorial Team
    Editorial Team
    2026-06-01T03:49:02+00:00Added an answer on June 1, 2026 at 3:49 am

    You need to include a value="" field if you want to return a value other then ON

    <input type = "checkbox" name="var2" value="1" CHECKED>
    
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