I am an absolute beginner at PHP and MySql so please be kind.

My problem is I am trying to pass on some information using a form and use this information to isolate the details required using the SELECT command. I know that I am getting the information from my form but am unsure on how to retrieve the necessary data from the database.



Where I have "Dr" i wish to replace this with the selected information from my form.

Help please.

Thanks

Hmm lets say in your form you have a select box with clan names, something like:



Then in your script you might do this:



Note the use of ". You must use double quotes when using variables in strings, if you use single quotes php assumes its all text and does not look for variables.

ditch the

...it's useless.

Thanks for that, but for some reason I get this now



At least now I am getting the variable into the statment.

Post the code you used please. We may need a peek at your form code as well, not sure. Lets also clarify the structure of your database. It sounds like the sql statement has a value in the wrong place though.

Here is the code I used for MySql minus the mysql_connect stuff



and here is an extract of the code I used for the form.



and finally here is the structure of the database.

try

Please note that after the equal sign, there is a single quote, then double quotes. Similarly, after the final concatenator, there is a set of double quotes, a single quote, and then double quotes again.

Actually he shouldn't have to concatenate the string. That is the whole point of using double quotes in php, so you can insert variables straight into a string without extra syntax. So this should work: (but then again you never know )



My next question is: You gave us the field names of a table in your database. What is the table name? If it is the same as a field in the table I suggest you change it to something else to eliminate possible confusion. Now lets look at the sql statement again.



means (this is not code, just for explanation):

$sql = "SELECT allvalues FROM tablename WHERE fieldname isexactlythesameas $clan_select";

So you might wanna go back to using LIKE in case the value in the database is not exactly the same as the value assigned in the form. Also lets make sure about the table name.

PS READ THIS:
I just noticed that I don't see where you actually made your query in the code you supplied. Do you actually have:



in your code? Because it definitly won't work if you don't execute the query.

This post has been edited by DanceInstructor: 22 Apr, 2005 - 12:45 PM

Yeah, the only reason I suggested concatenating the strings was due to his error message. It seems clear that 'Dr' is being interpreted as a column name, not as a string variable.

Good spot, never even noticed myself...

Thank you very much that works a treat.



Ops My mistake, in copying and pasting I accidently left it out.

Thanks Amadeus and DanceInstructor for your help.

This post has been edited by Adsa: 22 Apr, 2005 - 03:10 PM

Thanks guys for helping out, and welcome to dream.in.code Adsa!

And me. I optimized your query by 1/100000ms at least.