我的脚本正确吗?
我想要的是检查id是否存在,然后更新,如果不存在,则插入。
$sqlCheckRow = mysql_query( "SELECT * FROM wctilerack WHERE gameID = '$up_gameID' " ) or die ( mysql_error() );
$rowCounted = mysql_num_rows( $sqlCheckRow );
if ( $rowCounted == '0' ) {
// INSERT wctilerack
$sqlTileINSERT = mysql_query( "INSERT INTO wctilerack VALUE('', '$up_gameID', '$up_email_player1', '$playerRack' ) ") or die ( mysql_error() );
} elseif ( $rowCounted == '1' ) {
// INSERT wctilerack
$sqlTileINSERT = mysql_query( "UPDATE wctilerack SET tiles = '$playerRack' WHERE gamedID = '$up_gameID' ") or die ( mysql_error() );
}
我测试了它的工作原理,只是想确认过程是否正确。
谢谢
最佳答案
你也可以这样
if (empty($rowCounted))
{
$sqlTileINSERT = mysql_query( "INSERT INTO wctilerack VALUE('', '$up_gameID', '$up_email_player1', '$playerRack' ) ") or die ( mysql_error() );
}
else
{
$sqlTileINSERT = mysql_query( "UPDATE wctilerack SET tiles = '$playerRack' WHERE gamedID = '$up_gameID' ") or die ( mysql_error() );
}