barkster Posted August 2, 2007 Share Posted August 2, 2007 I have a table called ads and a second table called images that has a list of images for each as by adID I want to do a query for each and and get the first image(imageID) from the images table. If there insn't a image then would be null. How can I make a query like this? I tried a subquery but didn't work for me. --Ads-- AdID AdName --Images-- ImageID AdID ImageName Quote Link to comment https://forums.phpfreaks.com/topic/63074-solved-sub-query/ Share on other sites More sharing options...
akitchin Posted August 2, 2007 Share Posted August 2, 2007 this calls for a JOIN i believe: SELECT DISTINCT Ads.AdID, Ads.AdName, Images.ImageID, Images.ImageName FROM Ads INNER JOIN Images ON Ads.AdID=Images.AdID give that a whirl - if syntax is off, check the manual and you should be able to correct it. Quote Link to comment https://forums.phpfreaks.com/topic/63074-solved-sub-query/#findComment-314126 Share on other sites More sharing options...
barkster Posted August 2, 2007 Author Share Posted August 2, 2007 I've tried that and it gives me duplicate rows if there are more than one image. I need to get all rows in the Ad table and if there is an image get the first occurrence. See below --Sql-- SELECT DISTINCT Ads.AdID, Ads.CatID, images.Image FROM Ads Left Join images ON Ads.AdID = images.AdID --results-- AdID CatID Image 338 40_5 null 337 10_99 null 336 30 null 335 30 null 334 20 null 333 30 TestImage.jpg 333 30 TestImage2.jpg 333 30 TestImage3.jpg 332 30 null Quote Link to comment https://forums.phpfreaks.com/topic/63074-solved-sub-query/#findComment-314219 Share on other sites More sharing options...
spankpup Posted August 2, 2007 Share Posted August 2, 2007 Try this: select a.adid, a.adname, min(i.imageid), min(i.imagename) from ads a left outer join images i on a.adid = i.adid group by 1,2 Quote Link to comment https://forums.phpfreaks.com/topic/63074-solved-sub-query/#findComment-314413 Share on other sites More sharing options...
barkster Posted August 3, 2007 Author Share Posted August 3, 2007 Thanks, that got me in the right place!! Quote Link to comment https://forums.phpfreaks.com/topic/63074-solved-sub-query/#findComment-314543 Share on other sites More sharing options...
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