Giriş PHP kodlama doğru

1 Cevap

can someone please check the PHP coding is correct for this script. It needs a fresh pair of eyes to check.

Bu komut Geçen üzerinde çalıştı beri biraz eski olduğundan, ben bunu düzeltmek için yapılan küçük bir değişiklik ne hatırlayamıyorum ...

Peki bunun nesi var? Checkbox izinler şu anda kaydedilmiyor - emin değilim neden: (

if ($_POST) { 
$upd = mysqli_query($db, "UPDATE `tbl_user`  SET `username` = '".mysqli_real_escape_string($db,$_POST['username'])."', `userfullname` = '".mysqli_real_escape_string($db,$_POST['fullname'])."', `useremail` = '".mysqli_real_escape_string($db,$_POST['email'])."', `userlevel`= '".mysqli_real_escape_string($db,$_POST['userlevel'])."', `usertitle` = '".mysqli_real_escape_string($db,$_POST['title'])."' WHERE `userid` = '".mysqli_real_escape_string($db,$_GET['userid'])."'");
$insert = mysqli_query($db,"UPDATE `tbl_perms` SET
`1` ='" . mysqli_real_escape_string($db,$_POST['permsA_1']) . "',    
`2` ='" . mysqli_real_escape_string($db,$_POST['permsA_2']) . "',    
`3` ='" . mysqli_real_escape_string($db,$_POST['permsA_3']) . "',    
`4` ='" . mysqli_real_escape_string($db,$_POST['permsA_4']) . "',    
`5` ='" . mysqli_real_escape_string($db,$_POST['permsA_5']) . "', 
`6` ='" . mysqli_real_escape_string($db,$_POST['permsA_6']) . "',
`7` ='" . mysqli_real_escape_string($db,$_POST['permsA_7']) . "',
`8` ='" . mysqli_real_escape_string($db,$_POST['permsA_8']) . "',
`9` ='" . mysqli_real_escape_string($db,$_POST['permsA_9']) . "',
`10` ='" . mysqli_real_escape_string($db,$_POST['permsA_10']) . "',
`11` ='" . mysqli_real_escape_string($db,$_POST['permsA_11']) . "',
`12` ='" . mysqli_real_escape_string($db,$_POST['permsA_12']) . "',
`13` ='" . mysqli_real_escape_string($db,$_POST['permsA_13']) . "',
`14` ='" . mysqli_real_escape_string($db,$_POST['permsA_14']) . "',
`15` ='" . mysqli_real_escape_string($db,$_POST['permsA_15']) . "',
`16` ='" . mysqli_real_escape_string($db,$_POST['permsA_16']) . "',
`17` ='" . mysqli_real_escape_string($db,$_POST['permsA_17']) . "',
`18` ='" . mysqli_real_escape_string($db,$_POST['permsA_18']) . "',
`19` ='" . mysqli_real_escape_string($db,$_POST['permsA_19']) . "',
`20` ='" . mysqli_real_escape_string($db,$_POST['permsA_20']) . "',
`21` ='" . mysqli_real_escape_string($db,$_POST['permsA_21']) . "',
`22` ='" . mysqli_real_escape_string($db,$_POST['permsA_22']) . "'
WHERE `userid` = '$id' ")or die(mysqli_error($db));

$insert = mysqli_query($db,"UPDATE `tbl_usrdepts` SET
`1` ='" . mysqli_real_escape_string($db,$_POST['dept_1']) . "',    
`2` ='" . mysqli_real_escape_string($db,$_POST['dept_2']) . "',    
`3` ='" . mysqli_real_escape_string($db,$_POST['dept_3']) . "'    
WHERE `userid` = '$id' ")or die(mysqli_error($db));
$updated = "1";} 

if(!empty($_GET['userid'])) {       
 $sql = mysqli_query($db, "SELECT * from tbl_user WHERE userid = $id LIMIT 1");    
 if(mysqli_affected_rows($db) == 0) {        
   $noid = "1";    
 } else {        
   $current = mysqli_fetch_assoc($sql);        
   $currentperms  = mysqli_fetch_assoc(mysqli_query($db, "SELECT * from tbl_perms WHERE userid = $id"));
   $currentdepts  = mysqli_fetch_assoc(mysqli_query($db, "SELECT * from tbl_usrdepts WHERE userid = $id"));                        
  }} else {    
    $noidentered = "1";}
// Set Permissions
 $permissionid_select = mysqli_query($db,"SELECT * FROM `tbl_perms` WHERE `userid` = '$id'")or die(mysqli_error($db));
 $permissionid = mysqli_fetch_array($permissionid_select);
 $deptid_select = mysqli_query($db,"SELECT * FROM `tbl_usrdepts` WHERE `userid` = '$id'")or die(mysqli_error($db));
 $deptid = mysqli_fetch_array($deptid_select);
 $get_perms = mysqli_query($db,"SELECT * FROM `perm_sets` WHERE `status` = '1' ORDER BY `id` ASC")or die(mysqli_error($db));

Ve şimdi ekran kodu:

<tr valign="top">
<td class="alt2">Application Permissions</td>
<td class="alt2"><table cellpadding="0" cellspacing="0" border="0" width="100%"><tr valign="top"><td>
<div id="ctrl_user[membergroupids]" class="smallfont">
<?
   while($i = mysqli_fetch_array($get_perms)){

$pname = $i[pname];
$id = $i[id];
?>
	<div>
<input type="checkbox" tabindex="1" name="permsA_<? echo $id;?>" value="1" <? if($permissionid[$id] == '1') {echo ' checked="checked" ';}?> /><?echo htmlspecialchars($pname);?></div>
<? } ?> 	
</tr>
</table>
<br />
</td>
</tr>

<tr valign="top">
<td class="alt1">User Departments</td>
<td class="alt1"><table cellpadding="0" cellspacing="0" border="0" width="100%"><tr valign="top"><td><div id="ctrl_user[membergroupids]" class="smallfont">
<?
$get_depts = mysqli_query($db,"SELECT * FROM `tbl_depts` ORDER BY `id` ASC")or die(mysqli_error($db));
while($i = mysqli_fetch_array($get_depts)){

$dname = $i[name];
$id = $i[id];
?>  	
	<div><input type="checkbox" tabindex="1" name="dept_<? echo $id ;?>" value="1" <? if($deptid[$id] == '1') {echo ' checked="checked" ';}?> /><?echo htmlspecialchars($dname);?></div>
<? } ?> 
 </tr>
</table>
<br />
</td>
</tr>

Teşekkürler!

1 Cevap

Birkaç standart hata ayıklama teknikleri ve yorumlar:

  • comment out presentational code. It just obscures things.
  • print_r($var) immediately after each $var that isn't behaving. Start at the immediate problem and backtrack from there.
  • don't use php short tags (<?). They are deprecated.
  • echo out your SQL queries so you can see if they look the way you expect them to.

Deftere kod dev kazık hemen açık bir şey yoktur. Olmayan çok önemli şeyleri temizlemek ve size (ve biz) daha kolay oluyor göreceksiniz.

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