Show checkbox with database value - php

Question

I am having trouble with a part of my code. I want the checkbox in the last column to be checked if the corresponding DB value is 1, but something keeps going wrong with my code, anyone who sees what's wrong? It's probably very simple but I can't find it.
<?php
$sql = "SELECT * FROM registered ORDER BY datum";
$myData=mysql_query($sql,$con) ;
?>
<table width="1100" border="1">
<tr>
<th style="text-align:center; padding:0 10px">naam</th>
<th style="text-align:center; padding:0 10px">betaald?</b></th>
</tr>
<?php
$betaald = $record['betaald'];
while($record = mysql_fetch_array($myData)) {
echo "<tr>";
echo "<td>" . $record['naam'] . "</td>";
echo "<td> <input type='checkbox' name='betaald' id='betaald' value='1' ". echo ($betaald==1 ? 'checked' : ''); . " ></td>";
echo "</tr>";
}
mysql_close($con);
?>

Answer 1

<?php
$sql = "SELECT * FROM registered ORDER BY datum";
$myData = mysql_query($sql, $con);
?>
<table width="1100" border="1">
<tr>
<th style="text-align:center; padding:0 10px">naam</th>
<th style="text-align:center; padding:0 10px">betaald?</b></th>
</tr>
<?php while ($record = mysql_fetch_array($myData)) { ?>
<tr>
<td><?php echo $record['naam'] ?></td>
<td>
<input type='checkbox' name='betaald' id='betaald' value='1'
<?php if($record['betaald'] == 1){ ?>
checked="checked"
<?php } ?>
/>
</td>
</tr>
<?php } ?>
</table>

Answer 2

Try this
<?php
$sql = "SELECT * FROM registered ORDER BY datum";
$myData= mysql_query($sql) ;
?>
<table width="1100" border="1">
<tr>
<th style="text-align:center; padding:0 10px">naam</th>
<th style="text-align:center; padding:0 10px">betaald</b></th>
</tr>
<?php
while($record = mysql_fetch_array($myData))
{
?>
<tr>
<td> <?php echo $record['naam'] ?></td>
<td> <input type='checkbox' name='betaald' id='betaald' value='1' <?php echo ($record['betaald']==1 ? 'checked' : '')?>> </td>
</tr>
<?php
}
mysql_close($con);
?>
mysql_query($sql) no need to define $con
Note: This extension is deprecated as of PHP 5.5.0, and has been removed as of PHP 7.0.0

Answer 3

<?php
$sql = "SELECT * FROM registered ORDER BY datum";
$myData=mysql_query($sql,$con) ;
?>
<table width="1100" border="1">
<tr>
<th style="text-align:center; padding:0 10px">naam</th>
<th style="text-align:center; padding:0 10px">betaald?</b></th>
</tr>
<?php
while($record = mysql_fetch_array($myData)) {
$betaald = $record['betaald'];
echo "<tr>";
echo "<td>" . $record['naam'] . "</td>";
echo "<td> <input type='checkbox' name='betaald' id='betaald' value='1' ";
echo $betaald==1 ? 'checked' : '';
echo " ></td>";
echo "</tr>";
}
mysql_close($con);
?>
You have assigned $betaald value outside the while loop so there is no value in $betaald.
I have included $betaald inside while loop.

Answer 4

You need several changes in your php and html code. First of all I've some confusion about your code. Let me clear those.
Point 1 [name='betaald']: If you have a name like this you will not get multiple selected value.
Point 2 [id='betaald']: ID must be unique in a page.
Point 3 [value='1']: How will you differ values if all have same value?
so your code should be something like this:
<?php
$sql = "SELECT * FROM registered ORDER BY datum";
$myData=mysql_query($sql,$con) ;
?>
<table width="1100" border="1">
<tr>
<th style="text-align:center; padding:0 10px">naam</th>
<th style="text-align:center; padding:0 10px">betaald?</b></th>
</tr>
<?php
$betaald = $_POST['betaald'];
$i = 1;
while($record = mysql_fetch_array($myData)) {
echo "<tr>";
echo "<td>" . $record['naam'] . "</td>";
echo "<td> <input type='checkbox' name='betaald[]' id='betaald".$i."' value='".$record['betaald']."' ";
if ($record['betaald'] == $betaald)
{
echo "checked";
}
echo " ></td>";
echo "</tr>";
$i++;
}
mysql_close($con);
?>