Nested NSArray / NSDictionary issue

I have an NSArray containing several NSDictionaries.

I have an NSString containing an ID string.

What I'm trying to do is iterate through the NSDicti开发者_Go百科onaries until I find the object that matches the NSString. Then I want to return the entire NSDictionary.

I'm fairly new to iPhone development, so I'm probably missing something obvious... Thanks in advance. :)

Edit, here's the .plist file that gets saved out:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
<plist version="1.0">
<array>
    <dict>
        <key>date</key>
        <string>2011/05/20</string>
        <key>id</key>
        <string>1282007740</string>
        <key>name</key>
        <string>Test item name</string>
        <key>niceDate</key>
        <string>May 20, 2011</string>
    </dict>
    <dict>
        <key>date</key>
        <string>2010/08/15</string>
        <key>id</key>
        <string>1282075925</string>
        <key>name</key>
        <string>Test. Nothing to see here.</string>
        <key>niceDate</key>
        <string>Aug 15, 2010</string>
    </dict>
</array>
</plist>

Let's say my NSString is "1282075925". I want to get the dictionary with the id key that matches.

---

This code should do what you want - at the end, dict will either point to the NSDictionary you are searching for, or it will be nil if there was no matching dictionary.

// (These are assumed to exist)
NSArray *dictionaries;
NSString *idToFind; 



NSDictionary *dict = nil;
for (NSDictionary *aDict in dictionaries) {
    if ([[aDict objectForKey:@"id"] isEqualToString:idToFind]) {
        dict = aDict;
        break;
    }
}

---

If you are searching for @"12345", with a key of @"id"

for (NSDictionary* dictionary in array1) {
   id value = [dictionary valueForKey:@"id"];
   if ([value isKindOfClass:[NSString class]]) {
      NSString* string = (NSString*) value;
      if ([string isEqualToString:@"12345"]) {
          return dictionary;
      }
   }
}
return nil;