PHP checkbox always returns on

Question

I have this checkbox in my form:

<input type="checkbox" name="private" id="private" />

When I submit my form, the checkbox always returns on, even when I didn't check it.

echo $_POST['private'] //result is always returns on

$private = isset($_POST['private']) ? 1 : 0; //result is 1 always

What am I doing wrong?

score 1 · Accepted Answer · answered Jan 26 '18 at 15:54

1

Give the input a value and check that:

<input type="checkbox" name="private" id="private" value="1"/>

$private = isset($_POST['private']) && $_POST['private'] == 1 ? 1 : 0;

answered Jan 26 '18 at 15:54

Danny Thompson

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I tried adding a value, still the same issue, when I have my checkbox unchecked it returns 1 – user979331 Jan 26 '18 at 15:59
And when I have it checked, it still returns 1 – user979331 Jan 26 '18 at 16:00
can you update your question with what you tried when giving the input a value – Danny Thompson Jan 26 '18 at 16:38

score 0 · Answer 2 · answered Jan 26 '18 at 15:54

0

isset($_POST['private']) will always return true, because $_POST['private'] is exists.

Add value to checkbox <input type="checkbox" name="private" id="private" value="1" /> and use $private = $_POST['private'] ? 1 : 0; or if you are use PHP 7+ $private = $_POST['private'] ?? 0;

answered Jan 26 '18 at 15:54

Levente Otta

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Still returns 1 regardless. – user979331 Jan 26 '18 at 16:01

score 0 · Answer 3 · answered Jan 26 '18 at 16:09

0

When you submit an HTML form, the $_POST['input_name'] for checkbox elements are always set, so your isset($_POST['private']) always returns true.

You should instead be directly checking the value of the checkbox elemenet like so:

$private = $_POST['private'];

This will give you your desired result, if the checkbox is checked, $private will equal 1, otherwise 0.

answered Jan 26 '18 at 16:09

coderodour

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Still returns 1 – user979331 Jan 26 '18 at 16:12

PHP checkbox always returns on

3 Answers3