parsing - PHP parse code error -
please me error. please see code..
the error is:
"parse error: syntax error, unexpected '=' in\\...delete_user.php on line 21" line 21 has this:
if ($_server['request_method'] = = 'post') { pls help. dont know how check syntax new php codes..
thanks, jane
<?php # script 1.7 delete_user.php $page_title = 'delete user'; //include ('includes/header.html');// echo '<h1>delete user</h1>'; // check valid user id value:// if ( (isset($_get['id'])) && (is_numeric($_get['id'])) ) { $id = $_get['id']; } elseif ( (isset($_post['id'])) && (is_numeric($_post['id'])) ) { $id = $_post['id']; } else { echo '<p class="error">this page has been accessed in error.</p>'; //include ('includes/footer.html');// exit( ); } require_once ('../mysqli_connect.php'); if ($_server['request_method'] = = 'post') { if ($_post['sure'] = = 'yes') { $q = "delete users10 user_id=$id limit 1"; $r = @mysqli_query($dbc, $q); if (mysqli_affected_rows($dbc) = = 1) { echo '<p>the user has been deleted.</p>'; } else { echo '<p class="error">the user not deleted due system error. </p>'; echo '<p>' . mysqli_error($dbc) . '<br />query: ' . $q . '</p>'; } } else { echo '<p>the user has not been deleted.</p>'; } } else { $q = "select concat(last_name, ',', first_name) users10 user_id=$id"; $r = @mysqli_query($dbc, $q); if (mysqli_num_rows($r) = = 1) { $row = mysqli_fetch_array($r, mysqli_num); echo "<h3>name: $row[0]</h3> sure want delete user?"; echo '<form action="delete_user.php" method="post"> <input type="radio" name="sure" value="yes" /> yes <input type="radio" name="sure" value="no" checked="checked" /> no <input type="submit" name="submit" value="submit" /> <input type="hidden" name="id" value="' . $id . '" /> </form>'; } else { echo '<p class="error">this page has been accessed in error.</p>'; } } mysqli_close($dbc); ?>
i think may have space here:
if ($_server['request_method'] = = 'post') { try (changing "= =" "=="):
if ($_server['request_method'] == 'post') {
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