Please solved it for me

drizdor

Hi,

Can anyone tell me why:

(a) the following 2 codes produced blank page:

$result1 = mysql_query("SELECT * FROM art WHERE a_id ='$a_id'") or die( mysql_error() );

$result1 = mysql_query("SELECT * FROM art WHERE a_id ='$a_id'");

(b) the following code get a "not a valid source" error:

$result1 = mysql_query("SELECT * FROM art WHERE a_id =$a_id");

If you look at all the 3 codes, the difference is at:

a_id=$a_id

Please help me.

halojoy

those with blank pages
are probably maybe correct

but why it is blank is because there is no 'echo'

drizdor

Both are not correct. One results in a balnk page and another an "not a valid source" error. Nothing to do with echo cuz the echo exist in other line sof the code. Im asking cuz I found that there is something wrong with the:

a_id=$a_id

igorlopez

Well since we have no idea what you are trying to accomplish we can just make guesses.

My guess for your first construction (where you say you got blank pages) is that your query turned out to be empty.
Now if you know that column a_id in art has at least one record with what is inside $a_id then we have a different problem.

For you second problem, a_id=$a_id will only work as expected for numerical values in $a_id and where a_id is a numerical type in the table. For columns of string type you need to enclose the compare to string, i.e a_id='thisString' and when used with a variable you end up with a_id='$a_id' for example.
I don't know what the error

"not a valid source" error:

means and exactly where you are getting it.
I am sure others here knows it.