ldap_add（）：地址：Object类的违规错误

Question

当我尝试属性添加到通过PHP了OpenDS我得到以下错误：

ldap_add(): Add: Object class violation

请帮助。

这里是我的代码

<?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'] = "[email protected]"; 
    $ldaprecord['mmmm'] = "77878"; 

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

} else { 

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

} 
?> 

如果我从它工作正常的代码删除$ldaprecord['mmmm'] = "77878";。我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置objectclass到inetOrgPerson，但你也必须设置从哪个inetorgPerson正在扩大其他上层阶级 - 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'] = "[email protected]"; 
$ldaprecord['mmmm'] = "77878";