sqlite_current
(PHP 5)
sqlite_current(no version information, might be only in CVS)
SQLiteResult->current(no version information, might be only in CVS)
SQLiteUnbuffered->current -- Fetches the current row from a result set as an array說明
array sqlite_current ( resource result [, int result_type [, bool decode_binary]] )Object oriented style (method):
class SQLiteResult {array current ( [int result_type [, bool decode_binary]] )
}class SQLiteUnbuffered {
array current ( [int result_type [, bool decode_binary]] )
}
sqlite_current() is identical to sqlite_fetch_array() except that it does not advance to the next row prior to returning the data; it returns the data from the current position only.
參數
- result
The SQLite result resource. This parameter is not required when using the object-oriented method.
- result_type
可選的 result_type 參數接受一個常量並決定返回的陣列如何索引。用 SQLITE_ASSOC 只會返回關聯索引（有名稱欄位）而 SQLITE_NUM 只會返回數字索引（有序欄位數）。SQLITE_BOTH 會同時返回關聯和數字索引。 SQLITE_BOTH 是本函數的預設值。
- decode_binary
當 decode_binary 參數設為 TRUE（預設值）時，PHP 將解碼那些由 sqlite_escape_string() 編碼的資料。通常應保留此值為其預設值，除非在動作其它支援 sqlite 程式建立的資料庫時。
返回值
Returns an array of the current row from a result set; FALSE if the current position is beyond the final row.
由 SQLITE_ASSOC 和 SQLITE_BOTH 返回的列名會根據 sqlite.assoc_case 配置選項的值來決定大小寫。