Checkbox和Isset用PHP更新Mysql

Question

I can't get my update checkbox function to work. I need to be able to remove or add a value which I have chosen to call checked in to my table. The code looks like following.

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Name <input type"text" name="inputName" value="<?php echo $hemsida['Namn']; ?>" /> </br>
Commentar <input type"text" name="inputComment" value="<?php echo $hemsida['Comment']; ?>" />
<br/>
</br><input type="checkbox" name="all" value="<?php echo $hemsida['All']; ?>" 
<?php if($hemsida['All'] == 'checked') echo " checked"; ?> /> Alla



<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" />
<input type="submit" name="submit" value="Redigera">
</form>

and the Update PHP looks like this

    if(isset($_POST['submit'])) {
    $all = ($_POST['All'] == 1) ? "checked" : "";
    $u = "UPDATE hemsida SET `Namn`='$_POST[inputName]', `Comment`='$_POST[inputComment]', `ALL`=$all WHERE ID = $_POST[id]";
    mysql_query($u) or die(mysql_error());

    echo "User has been modified";
    header("Location: ..//sokh.php");
} 

The error is Undefined variable: hemsida on all parts. And also You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE ID = 33' at line 1. But I have no problem getting the data in to the Modifier or what I should call it but unable to get it out.

ANSWER !!!

I got it to work but cant answer my own question so i write it down here , i added and removed code until everything broke down. Remove the "$all = ($_POST['All'] == 1) ? checked : ;" part and now it works. I will copy the code underneath it there is an interest

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Name <input type"text" name="inputName" value="<?php echo $hemsida['Namn']; ?>" /> </br>
Commentar <input type"text" name="inputComment" value="<?php echo $hemsida['Comment']; ?>" />
<br/>
<input type="checkbox" name="all" value="checked"   <?php if($hemsida['All'] == 'checked') echo "checked=\"checked\""; ?>/> Alla
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" />
<input type="submit" name="submit" value="Redigera">
</form>

the new php

    if(isset($_POST['submit'])) {
    $u = "UPDATE hemsida SET `Comment`='$_POST[inputComment]', `Namn`='$_POST[inputName]', `All`='$_POST[all]' WHERE ID = $_POST[id]";
    mysql_query($u) or die(mysql_error());

    echo "User has been modified";
    header("Location: ..//sokh.php");
} 

Answer 1

1. It seems that $hemsida is not available at that point in your code, but you are using it.
2. An input type="checkbox" does notshow up in $_POST when it is not checked. , use isset.

sidenote: you're using XHTML? the correct checked type = checked="checked"