iphone - 不规则地获取sqlite3_prepare_v2中的Exc_Bad_Access

Question

在插入 livecategories 之前，我会检查此 id 是否存在，如果不存在，则插入该值。但有时它会崩溃并获得 exc_bad_access。为什么我会得到这个？请参阅附图。

当我将鼠标悬停在 &statement 上时，它显示 0x000000:

Answer 1

检查下面的示例代码:

// Setup the database object
    sqlite3 *database;

    // Init the animals Array
    animals = [[NSMutableArray alloc] init];

    // Open the database from the users filessytem
    if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
        // Setup the SQL Statement and compile it for faster access
        const char *sqlStatement = "select * from animals";
        sqlite3_stmt *compiledStatement;
        if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
            // Loop through the results and add them to the feeds array
            while(sqlite3_step(compiledStatement) == SQLITE_ROW) {
                // Read the data from the result row
                NSString *aName = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
                NSString *aDescription = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)];
                NSString *aImageUrl = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 3)];

                // Create a new animal object with the data from the database
                Animal *animal = [[Animal alloc] initWithName:aName description:aDescription url:aImageUrl];

                // Add the animal object to the animals Array
                [animals addObject:animal];

                [animal release];
            }
        }
        // Release the compiled statement from memory
        sqlite3_finalize(compiledStatement);

    }
    sqlite3_close(database);

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