將NSString中的匹配項替換爲NSArray中的匹配項

Question

我正在尋找一種在iOS應用程序中調試我的SQLite查詢的快速方法。我有這樣的事情：

NSString *sql = [NSString stringWithString: @"SELECT * FROM table WHERE foo = ? AND bar = ?"]; 
NSArray *params = [NSArray arrayWithObjects: @"baz", @"bat", nil]; 
NSLog(@"%@ %@", sql, params); 

我想知道如果有一個或兩個班輪與PARAMS替換問號，使輸出更具可讀性。我是不是尋找的東西，輸出有效的SQL，而是比我目前的NSLog更容易閱讀。

Answer 1

@interface NSString (SQLQueryDebugging) 
- (NSString *)stringBySubstitutingParameters:(NSArray *)params forInstancesOfPlaceholder:(NSString *)placeholder; 
@end 

@implementation NSString (SQLQueryDebugging) 

- (NSString *)stringBySubstitutingParameters:(NSArray *)params forInstancesOfPlaceholder:(NSString *)placeholder 
{ 
    NSString *composedQuery = self; 
    NSRange substitutionRange = [composedQuery rangeOfString:placeholder]; 
    NSInteger parameterIndex = 0; 
    while(substitutionRange.length != 0) 
    { 
     NSString *currentParam = [params objectAtIndex:parameterIndex]; 
     composedQuery = [composedQuery stringByReplacingCharactersInRange:substitutionRange withString:currentParam]; 
     ++parameterIndex; 
     NSInteger lastSubstitutionIndex = substitutionRange.location + [currentParam length]; 
     NSRange searchRange = NSMakeRange(lastSubstitutionIndex, [composedQuery length] - lastSubstitutionIndex); 
     substitutionRange = [composedQuery rangeOfString:placeholder options:0 range:searchRange]; 
    } 

    return composedQuery; 
} 

下降，在某個地方，那麼你就可以得到你想要什麼：

NSString *completeSQL = [sql stringBySubstitutingParameters:params forInstancesOfPlaceholder:@"?"];