Simple mysql_query Question

[gospabode2]

Hello,

I am still new to PHP, but am getting better. The Problem is, I have read all the tutorials and everything and successfully accomplished this at other places so I don't know what the issue is. I bet it is small, as it always seems to be an oversight that causes issues.

 

I am trying to do a MySQL Update Query with variables and I keep getting this error message:

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 'LastName = 'Smith' BirthDate = '1993-12-05' Gender = 'Male' De

Here is the code:

mysql_query("UPDATE orphans SET FirstName = '$firstname' LastName = '$lastname' BirthDate = '$birthdate' Gender = '$gender' Description = '$decrip' SponsorCount = '$sponsorcount' Created_On = 'createdon' Changed_On = '$changedon' PictureOne = 'pictureone' PictureTwo = 'picturetwo' PictureThree = '$picturethree' PictureFour = $picturefour' PictureFive = '$picturefive'
WHERE IDa = '$IDa'") or die(mysql_error());

The variables seem to be working, I just don't know what the issue is.  Thanks for the help in advance!

 

 

Edited to Fix Stupid Mistakes

[fugix]

you need to insert commas between your fields

[perky416]

As fugix says,

 

Try something like.....

 

mysql_query("UPDATE orphans SET FirstName = '$firstname',LastName = '$lastname',BirthDate = '$birthdate',Gender = '$gender',Description = '$decrip',SponsorCount = '$sponsorcount',Created_On = 'createdon',Changed_On = '$changedon',PictureOne = 'pictureone',PictureTwo = 'picturetwo',PictureThree = '$picturethree',PictureFour = $picturefour',PictureFive = '$picturefive' WHERE IDa = '$IDa'") or die(mysql_error());

[gospabode2]

Thanks Solved My Issue.

I could have sworn I typed thos in last night. I remember thinking about it. Oh well, All's well that ends well. Sorry it was such an easy fix, hope you didn't waste time on it.