Saving checkbox state

weeie

Hi,

i need to preserve the state of previously selected checkbox when the user didn't complete the form or when the form has error after validation.

im able to do it successfully for the radio button, but not for the checkbox because of the array used. How am i going to do the same effect for the checkbox? Someone please help..

form.php

if (isset($POST['gender']))
$gender = $POST['gender'];
else
$gender = '';

form.php (html - radio codes)

<label>
<input type="radio" name="gender" value="male" <? if ($gender == "male") { echo "checked"; } ?>>Male
</label>
<label>
<input type="radio" name="gender" value="female" <? if ($gender == "female") { echo "checked"; } ?>>Female
</label>

form.php (html - checkbox codes)

<input name="private[]" type="checkbox" value="dog">Dog<br>
<input name="private[]" type="checkbox" value="cat">Cat<br>
<input name="private[]" type="checkbox" value="mouse">Mouse

khaine

You will need to check the array private to see if the current value is in it.

weeie

i've managed to do it with a

<? if ( in_array("dog",$_POST['private'])) { echo "CHECKED"; } ?>

but when the checkbox is not checked, i got a error on the same line:

Warning: in_array(): Wrong datatype for second argument in D:\Apache2\htdocs\ps\regAnimal.php on line 429

khaine

Thats not an error its a warning. I wouldn't worry about it. You can remove it from screen by tweaking you php.ini file. (I would recommend doing it anyway) so that errors and warnings don't appear on screen just in the error logs.

cahva

if (!empty($_POST['private']))
    $private=$_POST['private'];
else
    $private=array();

Now you can use this to check if its checked or not:

if ( in_array("dog",$private)) { echo "CHECKED"; }

weeie

thanks cahva and all. now it works well.

weeie

btw, is there any shortcut way to do for country list menu so that i do not need to put in echo "checked"; for every country.

batman

Yeah in the country dropdown, only give it two values.....

United States, and International. 😃