Update value in mysql database column with checkbox in php

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Reference - What does this error mean in PHP? 30 answers

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The problem is your echo on line 7. Your doing <?php echo ... ?> inside of an echo. You should instead concatenate the values into the string.

Instead of:

 echo '<td><form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"><input type="hidden" name="id" value="<?php echo $id; ?>" /><input type="checkbox" name="checkinsat" value="1"<?php if($data['checkinsat'] == '1') echo 'checked'; ?>/><input type="submit" ></form></td>';

Try this:

$checked = ($data['checkinsat'] == '1') ? 'checked' : '';
echo '<td><form method="post" action="'.$_SERVER['PHP_SELF'].'"><input type="hidden" name="id" value="'. $id.'" /><input type="checkbox" name="checkinsat" value="1" '.$checked.'/><input type="submit" ></form></td>';

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Don't echo the parameters into the form using php tags, use a string concatenation instead.

Change:

echo '<td><form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"><input type="hidden" name="id" value="<?php echo $id; ?>" /><input type="checkbox" name="checkinsat" value="1"<?php if($data['checkinsat'] == '1') echo 'checked'; ?>/><input type="submit" ></form></td>';

to:

echo '<td><form method="post" action="'. $_SERVER['PHP_SELF'] .'"> <input type="hidden" name="id" value="'. $id .'" /><input type="checkbox" name="checkinsat" value="1"'. ($data['checkinsat'] == '1'? 'checked' : '') .'<input type="submit" ></form></td>';

Further, do not use mysql_* functions - it's deprecated (see red box) and vulnerable to sql-injection. Use PDO or MySQLi.

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you have typo here

echo '<td><form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"><input type="hidden" name="id" value="<?php echo $id; ?>" /><input type="checkbox" name="checkinsat" value="1"<?php if($data['checkinsat'] == '1') echo 'checked'; ?>/><input type="submit" ></form></td>';

should be

echo '<td><form method="post" action="'.$_SERVER['PHP_SELF']'"><input type="hidden" name="id" value="'.$id.'" /><input type="checkbox" name="checkinsat" value="1"';
 if($data['checkinsat'] == '1') 
  echo 'checked />';
 echo '<input type="submit" ></form></td>';

you are writing <?PHP ... ?> in php tag and that is causes this error

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