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我有一个 SQLite 数据库，但是当我尝试动态添加项目时它不起作用。

这是添加脚本：

-(void) addPatientToDatabase:(Patient *)newPatient {
    NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
    NSString *documentsPath = [paths objectAtIndex:0];
    NSString *filePath = [documentsPath stringByAppendingPathComponent:@"cities.sqlite"];

    sqlite3 *database;

    if(sqlite3_open([filePath UTF8String], &database) == SQLITE_OK) {
        const char *sqlStatement = "insert into patients (firstName, surname, dob, homeNumber, mobileNumber, email, address, image) VALUES (?, ?, ?, ?, ?, ?, ?, ?)";
        sqlite3_stmt *compiledStatement;
        if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
            sqlite3_bind_text(compiledStatement, 1, [newPatient.patientName UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 2, [newPatient.patientSurname UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 3, [newPatient.patientDoB UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 4, [newPatient.patientHomeNumber UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 5, [newPatient.patientMobileNumber UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 6, [newPatient.patientEmail UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(compiledStatement, 7, [newPatient.patientAddress UTF8String], -1, SQLITE_TRANSIENT);
            NSData *dataForPicture = UIImagePNGRepresentation(newPatient.patientPicture);
            sqlite3_bind_blob(compiledStatement, 8, [dataForPicture bytes], [dataForPicture length], SQLITE_TRANSIENT);
        }
        if(sqlite3_step(compiledStatement) == SQLITE_DONE) {
            sqlite3_finalize(compiledStatement);
        }
    }
    sqlite3_close(database);
}

我添加了一些断点，并指出如果此 if 语句中有断点：

if(sqlite3_step(compiledStatement) == SQLITE_DONE) {

然后它不会被捡起来。

提前致谢

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1 回答 1

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当 step 返回 SQLITE_DONE 以外的任何内容时，尝试调用： sqlite3_errmsg(database);

它将返回一个指向错误消息的指针（http://www.sqlite.org/c3ref/errcode.html）。显示或记录该消息，以找出问题所在。

于 2013-07-10T13:01:24.183 回答