I am having the hardest time getting a select as statement right. Here is the full query:
select SUM(IF(image='EE0840D00E2ED8A317E5FA9899C48C19',1,0)) as EE0840D00E2ED8A317E5FA9899C48C19, SUM(IF(image='235C7987796D5B7CEBF56FBDA2BF7815',1,0)) as 235C7987796D5B7CEBF56FBDA2BF7815, SUM(IF(image='96DC0562ED6E6F7FE789A18E09BC5889',1,0)) as 96DC0562ED6E6F7FE789A18E09BC5889, SUM(IF(image='D8B0EA710D2EF408391132F451AE724A',1,0)) as D8B0EA710D2EF408391132F451AE724A, SUM(IF(image='018C4DB7229D7D2BEB040D241739B784',1,0)) as 018C4DB7229D7D2BEB040D241739B784, SUM(IF(image='98DE1FCA50AC9CE6E0FEA25BAB0177FE',1,0)) as 98DE1FCA50AC9CE6E0FEA25BAB0177FE, SUM(IF(image='4E5664736F400E8B482EA7AA67853D13',1,0)) as 4E5664736F400E8B482EA7AA67853D13, <--offending line SUM(IF(image='FEB810A43A1B275605BD6B69F444700C',1,0)) as FEB810A43A1B275605BD6B69F444700C from dsrssfeed If I remove that one line, the query works fine. If I do: select SUM(IF(image='4E5664736F400E8B482EA7AA67853D13',1,0)) as 4E from dsrssfeed ; it works. But these fail: select SUM(IF(image='4E5664736F400E8B482EA7AA67853D13',1,0)) as 4E5664736F400E8B482EA7AA67853D13 from dsrssfeed ; select SUM(IF(image='4E5664736F400E8B482EA7AA67853D13',1,0)) as 4E5 from dsrssfeed ; It can't be field name length, since even 4E5 fails, the field name can start with a number since 4E succeeds. Any ideas? The goal is to see what arbitrary images have information associated with them. The table has two fields: image is a UID that is the primary key, and caption which is a varchar(255) that has information about the image. Images are added and deleted from the table as they are changed on a web page. The UID is generated by a third party program that I have to interface with and have no control over. An array of image UIDs is sent to the php script and the script needs to determine which UIDs are present in the table. Rather than make N number of individual queries as I iterate through the array, I iterate through the array and build the query on the fly to make one query. Then I iterate through the array again and check the value in the field. 1 means the UID has an entry, 0 means it doesn't. I thought doing 1 mysql call would be more efficient than lots of calls as I iterate through the array. But since there will probably never be more than 100 images in the table at any one time, it may not make any difference. But now I'm just curious as to why this is happening. Thanks, Bob
