dflow Posted September 21, 2011 Share Posted September 21, 2011 how can i perform this directly in phpmyadmin : UPDATE `apartments` SET mainImage = ID-SupplierID-0.jpg WHERE InternalSupplierID=7'; ID SupplierID are table fields in apartments it's actually a foreach loop Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/ Share on other sites More sharing options...
Muddy_Funster Posted September 21, 2011 Share Posted September 21, 2011 what's the problem when you try that in PHPMyAdmin? Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271344 Share on other sites More sharing options...
dflow Posted September 21, 2011 Author Share Posted September 21, 2011 what's the problem when you try that in PHPMyAdmin? #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'jpg WHERE InternalSupplierID=7' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271361 Share on other sites More sharing options...
awjudd Posted September 21, 2011 Share Posted September 21, 2011 You are missing the quotes around your string ... and you have 1 too many quotes at the end ... UPDATE `apartments` SET mainImage = CONCAT(CAST(ID-SupplierID-0 AS VARCHAR(50)),'.jpg') WHERE InternalSupplierID=7; Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271364 Share on other sites More sharing options...
dflow Posted September 21, 2011 Author Share Posted September 21, 2011 You are missing the quotes around your string ... and you have 1 too many quotes at the end ... UPDATE `apartments` SET mainImage = CONCAT(CAST(ID-SupplierID-0 AS VARCHAR(50)),'.jpg') WHERE InternalSupplierID=7; still quote error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'VARCHAR(50)),'.jpg') WHERE InternalSupplierID=7' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271375 Share on other sites More sharing options...
awjudd Posted September 21, 2011 Share Posted September 21, 2011 But it is now at a different spot ... Looking at the definition for CONCAT (http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_concat) you can give it a number. UPDATE `apartments` SET mainImage = CONCAT(ID-SupplierID-0,'.jpg') WHERE InternalSupplierID=7; Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271378 Share on other sites More sharing options...
dflow Posted September 21, 2011 Author Share Posted September 21, 2011 But it is now at a different spot ... Looking at the definition for CONCAT (http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_concat) you can give it a number. UPDATE `apartments` SET mainImage = CONCAT(ID-SupplierID-0,'.jpg') WHERE InternalSupplierID=7; this updated, but only ID.jpg without -0 output: 34567.jpg instead of 34567-12-0.jpg thanks ill read about CONCAT Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271408 Share on other sites More sharing options...
Muddy_Funster Posted September 21, 2011 Share Posted September 21, 2011 so what you're looking for is this ?: UPDATE `apartments` SET mainImage = CONCAT(ID, '-', SupplierID, '-', 0 ,'.jpg') WHERE InternalSupplierID=7; Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271417 Share on other sites More sharing options...
dflow Posted September 21, 2011 Author Share Posted September 21, 2011 so what you're looking for is this ?: UPDATE `apartments` SET mainImage = CONCAT(ID, '-', SupplierID, '-', 0 ,'.jpg') WHERE InternalSupplierID=7; cheers! Quote Link to comment https://forums.phpfreaks.com/topic/247581-quick-update/#findComment-1271425 Share on other sites More sharing options...
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