ldap_add():添加:对象类违规错误

Question

当我尝试通过PHP向OpenDS添加属性时,我收到以下错误:

ldap_add():添加:对象类违例

请帮忙.

这是我的代码

<?php
$ldapconfig['host'] = 'PC100';
$ldapconfig['port'] = 1389;
$ldapconfig['basedn'] = 'dc=company,dc=com';

$ds=ldap_connect($ldapconfig['host'], $ldapconfig['port']);

$password=1;
$username="cn=Directory Manager";

if ($bind=ldap_bind($ds, $username, $password)) {
  echo("Login correct");
  ldap_set_option($ds, LDAP_OPT_PROTOCOL_VERSION, 3); // IMPORTANT
  $dn = "cn=roshan1,dc=example,dc=com"; 
  //$newuser["objectclass"] = "inetOrgPerson"; 
  //$newuser["cn"] = "new1"; 
  //$newuser["sn"] = "user"; 

  $ldaprecord['cn'] = "roshan1";
  $ldaprecord['givenName'] = "mkljl";
  $ldaprecord['sn'] = "roshan";
  $ldaprecord['objectclass'] = "inetOrgPerson";    
  $ldaprecord['mail'] = "lkl@fh.com";
  $ldaprecord['mmmm'] = "77878";

  // add data to directory
  $r = ldap_add($ds, $dn, $ldaprecord);

} else {

  echo("Unable to bind to server.</br>");

}
?>

如果我$ldaprecord['mmmm'] = "77878";从代码中删除它工作正常.如何添加这样的新属性？

Answer 1

嗯,它看起来像您正在尝试只设objectclass到inetOrgPerson,但你也必须设置从哪个其他上层阶级inetorgPerson正在扩大-这将是top和person也许...

所以:

$ldaprecord['cn'] = "roshan1";
$ldaprecord['givenName'] = "mkljl";
$ldaprecord['sn'] = "roshan";
$ldaprecord['objectclass'][0] = "top";
$ldaprecord['objectclass'][1] = "person";
$ldaprecord['objectclass'][2] = "inetOrgPerson";
$ldaprecord['mail'] = "lkl@fh.com";
$ldaprecord['mmmm'] = "77878";