Filters can be created by calling
BundleContext#createFilter(String) or
FrameworkUtil#createFilter(String) with a filter string.
A Filter can be used numerous times to determine if the match
argument matches the filter string that was used to create the Filter.
Some examples of LDAP filters are:
"(cn=Babs Jensen)"
"(!(cn=Tim Howes))"
"(&(" + Constants.OBJECTCLASS + "=Person)(|(sn=Jensen)(cn=Babs J*)))"
"(o=univ*of*mich*)"
| Modifier and Type | Method and Description |
|---|---|
| public boolean | |
| public int | |
| public boolean | Returns: true if the service's properties match this
Filter; false otherwise.The reference to the service whose properties are used
in the match. reference)Filter using a service's properties. |
| public boolean | Returns: true if the Dictionary's values match this
filter; false otherwise.The dictionary)Dictionary whose key/value pairs are used
in the match.Filter using a |
| public boolean | Returns: true if the Dictionary's values match this
filter; false otherwise.The dictionary)Dictionary whose key/value pairs are used
in the match.Filter using a |
| public boolean | |
| public String |
| equals | back to summary |
|---|---|
| public boolean equals(Object obj) Compares this
This implementation returns the result of calling
| |
| hashCode | back to summary |
|---|---|
| public int hashCode() Returns the hashCode for this
This implementation returns the result of calling
| |
| match | back to summary |
|---|---|
| public boolean match(ServiceReference<?> reference) Filter using a service's properties.
This
| |
| match | back to summary |
|---|---|
| public boolean match(Dictionary<String, ?> dictionary) Filter using a
| |
| matchCase | back to summary |
|---|---|
| public boolean matchCase(Dictionary<String, ?> dictionary) Filter using a
| |
| matches | back to summary |
|---|---|
| public boolean matches(Map<String, ?> map) Filter using a | |
| toString | back to summary |
|---|---|
| public String toString() Returns this The filter string is normalized by removing whitespace which does not affect the meaning of the filter. | |