0

I am currently working on a php site that pulls in data from a MySql server, Unfortunately the code I'm using to run the query to get the data from the server is causing a nasty parse error on my site, and I am having extreme dificulty sorting it out. Here is my code

$q = = 'SELECT Customers.CustomerID, Customers.OldCustomerID, CONCAT_WS(Lastname, FirstName, MiddleName ) AS CustomerName, CONCAT( LEFT( FirstName, 1 ) , LEFT( MiddleName, 1 ) , LastName ) AS UserName, Customers.IsLaptop AS Laptop, zlu_BirthMonth.Description AS BirthMonth, zlu_Residence.Description AS Residence, zlu_CarColor.Description AS CarColor, zlu_Cars.Description, zlu_Computers.Description AS Computer, zlu_Race.Description AS Race
FROM zlu_Residence INNER JOIN (zlu_Race INNER JOIN (zlu_Computers INNER JOIN (zlu_Cars INNER JOIN (zlu_CarColor INNER JOIN (zlu_BirthMonth INNER JOIN Customers ON zlu_BirthMonth.BirthMonthID = Customers.BirthMonthID) ON zlu_CarColor.CarColorID = Customers.CarColorID) ON zlu_Cars.CarsID = Customers.CarID) ON zlu_Computers.ComputersID = Customers.ComputerID) ON zlu_Race.RaceID = Customers.RaceID) ON zlu_Residence.ResidenceID = Customers.ResidenceID
WHERE customerID >=500
AND CustomerID <=600[...]';

Any help would be greatly appreciated

3
Contributors
2
Replies
3
Views
7 Years
Discussion Span
Last Post by hemgoyal_1990
0

You only need one '=' sign if you are assigning the string to the variable $q.

This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.