how verify a duplicate uid to before the entry is created

[Anonymous]

I want validate if a uid is duplicated before the entry is created. for this i make a template.

<template>
  <name>uid_user</name>
  <description>uid user. 
  <author>JJF</author>
  <version>1.0</version>
  <email>jose.fernandez@open-world.com.ve</email>

  <extends>user</extends>

  <attribute>
    <name>objectclass</name>
    <value>inetOrgPerson</value> 
    <value>shadowAccount</value>
  </attribute>

<control type="panel">
    <caption>Informacion del empleado</caption>
    <bevel>lowered</bevel>
 <attribute type="text">
     <name>employeeNumber</name>
     <description>Numero de Empleado</description>
 </attribute>

 <attribute type="text">
     <name>employeeType</name>
     <description>Cargo</description>
 </attribute>             
</control>

<script>
entry.OnWrite = entryOnWrite;

function entryOnWrite(Sender) {
  attributesByName['gecos'].value = attributesByName['uid'].value
  busca='uid='+ attributesByName['uid'].value;
  sr = session.search(busca, 'dc=bolipuertos,dc=gob,dc=ve', 2, 'uid', false );
  if (sr.count > 0 )
   {
    alert("Login repetido. Cambiar por favor");
   }
}
</script>
</template>

The template tell me if one uid if duplicated, but allways created the entry.
How do I can do to make the function if the uid is repeated display the alert and return to the form for the user to change the uid ?
in html i can return false to the function submit.

and the other thing if that i create the entry then return me this message
"Zugriffsverletzung bei Adresse 00417CE2 in modul 'LdapAdmin.exe'. Lesen Von Adresse 00000017."

how can i correct that message?

thanks for the help!!!

[Tihomir Karlovic]

Give me a few days to see into it...

[Tihomir Karlovic]

Sorry, I have to admit that I forgot about this issue. I apologize! Could you find a solution or is this still a problem?

Cheers,
Tihomir