WHERE table_name = $date_entered?

djalecc

I'm having problems with this line of code, and was wondering (after 6 hours pulling my hair out) if I'm just beeing a noobly nob!

$result=mysql_query("select * from ".$mysql_table_prefix."test WHERE indexdate = '$dateentered' ORDER BY site_id DESC");

index date is stored like 2005-08-04

there are about 12 in the table with this date.

I've made a form that you enter the date to list the results, the form passes the var as bydate which I then convert in the script the form goes to like this...

if ($bydate ==  0) {
        die("No Date was entered");
        } else {
        $dateentered = $bydate;
        print "<h3>$dateentered Testing</h3>";
        index_alldate();

    print "<h1>$bydate - $url</h1>";
    die("It got to here!");

    }

So I convert $bydate to $dateentered, they both return the same result, the result is what i typed in the text box on the form.

However, the function index_alldate(); needs the value of $dateentered put in the WHERE clause so only the results with the value I entered are returned.

For some reason it does not work, however if I change the code to ...

$result=mysql_query("select * from ".$mysql_table_prefix."test WHERE indexdate = '2005-08-04' ORDER BY site_id DESC");

It works !...

What i need it to do, is grab the data from the table from the date I enter in the form!

Anyone? Help!!!!! 🙁 😕

djalecc

Got it ! ... didn't $global the $bydate ! :o

jcasamitjana

Sorry, but at the first look at the code you should pass the variable '$dateentered' to the function 'index_alldate(); '

I mean this:

if ($bydate == 0) {
die("No Date was entered");
} else {
$dateentered = $bydate;
print "<h3>$dateentered Testing</h3>";
index_alldate($dateentered);

    print "<h1>$bydate - $url</h1>";
    die("It got to here!");

    }

This will solve your problem for sure.
Luck!

http://jcasamitjana.blogspot.com/