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Mysql - Insert if specific field does not exists

Hi.

I have an INSERT query but I'm getting duplicate content every time it's executed.

How can I avoid to get duplicated content in the column "videourl" ?

I want "videourl" column with unique value so just ignore at insertion if value already exist?

Could you please use my real code attached in order to conform the right fix?

Thanks.

while ($row_rsMainData = mysql_fetch_assoc($rsMainData))
{
$sql_command = " INSERT INTO `video` (`videoid`, `title`, `description`, `videourl`, `videocategoryid`, `dateline`, `userid`, `videoservice`, `videoidservice`, `views`, `timelength`, `commentcount`, `ratingtotal`, `ratingnum`, `rating`, `reportthreadid`, `username`, `cachetags`) 
VALUES(1, '".$row_rsMainData['title']."', '".$row_rsMainData['description']."', '".$row_rsMainData['link']."', ".$row_rsMainData['rss_data_id'].", 1239699262, 1, 'YouTube', '$codelink', ".$row_rsMainData['num_views'].", 373, 0, 0, 0, 0, 0, 'Reportero', '<em>None...</em>')";
$ok= mysql_query($sql_command, $vb_videos_conn);
}

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Roger Baklund

>> How can I avoid to get duplicated content in the column "videourl" ?

Define an unique index on the column:

ALTER TABLE vide ADD UNIQUE(videourl);

The php code will be unchanged, the INSERT will fail when the videourl allready exists in the table.

Roger Baklund

Sorry, a small typo there, "vide" whould be "video":

ALTER TABLE video ADD UNIQUE(videourl);

Fernanditos

ASKER

Thanks. Could you please tell me how would be the implementation in my code:

{
$sql_command = " INSERT INTO `video` (`videoid`, `title`, `description`, `videourl`, `videocategoryid`, `dateline`, `userid`, `videoservice`, `videoidservice`, `views`, `timelength`, `commentcount`, `ratingtotal`, `ratingnum`, `rating`, `reportthreadid`, `username`, `cachetags`)
VALUES(1, '".$row_rsMainData['title']."', '".$row_rsMainData['description']."', '".$row_rsMainData['link']."', ".$row_rsMainData['rss_data_id'].", 1239699262, 1, 'YouTube', '$codelink', ".$row_rsMainData['num_views'].", 373, 0, 0, 0, 0, 0, 'Reportero', '<em>None...</em>')";
$ok= mysql_query($sql_command, $vb_videos_conn);
}

Thanks

ASKER CERTIFIED SOLUTION

Roger Baklund

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