PHP: Submit checkbox to database

问题

I am having trouble submitted checkbox values and details to my database.

This is my HTML:

<form method="get" action="check.php">
            <input type="checkbox" name="checkAccount"/>
            <input type="hidden" name="table_name" value="masterChart" />
            <input type="hidden" name="column_name" value="account" />
            <p><a href='check.php'><input type="submit" class="btn btn-primary" value="Submit" /></a></p>
    </form>

This is the check.php:

$table = $_GET['table_name'];
$column = $_GET['account'];

$dbc = mysqli_connect('localhost', 'root', 'root', 'database') or die('Connection error!');

if ($value = 1) {
 $checkbox = "INSERT INTO login_table_display(`user`, `table`, `column`, `value`) VALUES(`:user`, '$table', '$column', `$value`)";
 mysqli_query($dbc, $checkbox) or die('Database error, check!');
 }

header('location:index.php');

As you can see above, I used variables to get other details for that checkbox to insert into the table as well.

After I press submit if the checkbox is checked, this is what's seen in the url:

http://localhost/admin/check.php?checkAccount=on&table_name=masterChart&column_name=account

Any suggestions or help will be appreciated!

回答1:

The classic way to submit data is to add the value attribute to your checkboxes element in your form. On server side you have to ckeck the value for "null".

<input type="checkbox" name="checkAccount" value="putyourvaluehere"/>

回答2:

Your Html is not ok

It should be

<form method="get" action="check.php">
            <input type="checkbox" name="checkAccount"/>
            <input type="hidden" name="table_name" value="masterChart" />
            <input type="hidden" name="column_name" value="account" />
            <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
    </form>

Also
if(isset($_POST['checkAccount']) {

Should Be
if( isset($_POST['checkAccount']) ) {

回答3:

Checkbox value will be submitted only when it's checked. Use isset($_GET['checkAccount']) for this:

$var= isset($_GET['checkAccount']) ? 1 : 0; // Or whatever values you use in DB

回答4:

Try this:

First you have to edit your html code as below;

<form method="get" action="check.php">
    <input type="checkbox" name="checkAccount" value='cool'/>
    <input type="hidden" name="table_name" value="masterChart" />
    <input type="hidden" name="column_name" value="account" />
    <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
</form>

you are not giving value to check box and using submit button inside a tag, it's not good practice.

回答5:

Replace:

 if(isset($_POST['checkAccount'])

to:

 if(isset($_GET['checkAccount'])

回答6:

// your html code shoul be like this

<form method="get" action="check.php">

        <input type="checkbox" name="checkAccount"/>

        <input type="hidden" name="table_name" value="masterChart" />

        <input type="hidden" name="column_name" value="account" />

        <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
</form>



<?php

$table = $_GET['table_name'];

$column = $_GET['account'];

$value = isset($_GET['checkAccount']) ? 1 : 0;

$dbc = mysqli_connect('localhost', 'root', 'root', 'database') or die('Connection error!');

if ($value == 1) {

$checkbox = "INSERT INTO login_table_display('user', 'table', 'column', 'value')

VALUES(':user', '$table', '$column', '$value')";

mysqli_query($dbc, $checkbox) or die('Database error, check!');

}

header('location:index.php');

来源：https://stackoverflow.com/questions/17041268/php-submit-checkbox-to-database

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