[英]getting error in sqlite Update in ios
我在sqlite更新中遇到錯誤,這里正確給出了查詢,但是值未更新。這里我需要更新描述值在給定的地方,下面我添加了我的代碼,請幫幫我
-(void) update:(NSString *)filePath withName:(NSString *)place description:(NSString*)description
{
sqlite3* db = NULL;
int rc=0;
sqlite3_stmt* stmt =NULL;
rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
if (SQLITE_OK != rc)
{
sqlite3_close(db);
NSLog(@"Failed to open db connection");
}
else
{
NSString * query = [NSString stringWithFormat:@"UPDATE places SET description=%@ WHERE place=%@",description,place];
sqlite3_bind_text(stmt, 1, [place UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 2, [description UTF8String], -1, SQLITE_TRANSIENT);
NSLog(@"query %@",query);
char * errMsg;
rc = sqlite3_exec(db, [query UTF8String] ,NULL,&stmt,&errMsg);
if(SQLITE_OK != rc)
{
NSLog(@"Failed to insert record rc:%d, msg=%s",rc,errMsg);
}
else{
NSLog(@"Sucessfully updated");
}
sqlite3_finalize(stmt);
sqlite3_close(db);
}
}
表格創建
-(int) createTable:(NSString*) filePath
{
sqlite3* db = NULL;
int rc=0;
NSLog(@"create");
rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE | SQLITE_OPEN_CREATE, NULL);
if (SQLITE_OK != rc)
{
sqlite3_close(db);
NSLog(@"Failed to open db connection");
}
else
{
char * query ="CREATE TABLE IF NOT EXISTS places ( id INTEGER PRIMARY KEY AUTOINCREMENT,place TEXT ,locationname TEXT,time TEXT,description TEXT,notifyTime TEXT,radius TEXT,lat DOUBLE,lon DOUBLE ,notify TEXT,selection INTEGER)";
char * errMsg;
rc = sqlite3_exec(db, query,NULL,NULL,&errMsg);
if(SQLITE_OK != rc)
{
NSLog(@"Failed to create table rc:%d, msg=%s",rc,errMsg);
}
else{
NSLog(@"Sucessfully Created ");
}
sqlite3_close(db);
}
return rc;
}
首先打開數據庫連接。 然后寫下面的代碼。
NSString * query = [NSString stringWithFormat:@"UPDATE places SET description=%@ WHERE place=%@",description,place];
SQL=[NSString stringWithString:query];
sqlite3_stmt *insert_statement=nil;
if (sqlite3_prepare_v2(database, [SQL UTF8String], -1, &insert_statement, NULL) != SQLITE_OK) {
//NSAssert1(0, @"Error: failed to prepare statement with message '%s'.", sqlite3_errmsg(database));
NSLog(@"Error: failed to prepare statement with message '%s'.", sqlite3_errmsg(database));
}
int result=sqlite3_step(insert_statement);
sqlite3_finalize(insert_statement);
然后關閉數據庫連接
試試這個...檢查數據庫是否已打開並更改此代碼...
-(void) update:(NSString *)filePath withName:(NSString *)place description:(NSString*)description
{
sqlite3* db = NULL;
int rc=0;
sqlite3_stmt* stmt =NULL;
rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
if (SQLITE_OK != rc)
{
sqlite3_close(db);
NSLog(@"Failed to open db connection");
}
else
{
/*
NSString *strMQueryupdate=[NSString stringWithFormat:@"update br set TOTAL='%@',QUTY='%@' WHERE PID='%@'",[@(total )stringValue],[@(lblQtn )stringValue],produdId];
*/
NSString * query = [NSString stringWithFormat:@"UPDATE places SET description='%@' WHERE place='%@'",description,place];
const char *sql = [query UTF8String];
if (sqlite3_prepare_v2(db, sql, -1, & stmt, NULL) != SQLITE_OK)
{
NSLog(@"update fails");
}
else
{
sqlite3_bind_text(stmt, 1, [place UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 2, [description UTF8String], -1, SQLITE_TRANSIENT);
int success = sqlite3_step(stmt);
sqlite3_reset(stmt);
if (success == SQLITE_ERROR)
{
}
else
{
NSLog(@"update success");
}
sqlite3_finalize(update_statement);
sqlite3_close(db);
}
}
您使用的准備語句不正確。 您將值硬編碼為字符串(不好的主意),然后設置占位符值-這沒有任何意義,因為SQL中沒有占位符。 嘗試類似:
NSString * query = @"UPDATE places SET description= ? WHERE place= ?";
sqlite3_bind_text(stmt, 1, [place UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 2, [description UTF8String], -1, SQLITE_TRANSIENT);
您不想直接在查詢字符串中包含值的原因:想象有人在為“倫敦”,放置表格位置;等位置傳遞值。 (不知道這是否行得通,但是它應該使您對用戶可能會冒犯的惡作劇有所了解。)
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