Resource ID #6

[aace]

I am trying to pull data out of a mysql database. The data is the date. Right now it's stored as a date type in mysql. So today is 2004-07-04, I need to find the record in the db with that in the date field. Here is my query..
$testingTimeOutTotal = "SELECT TimeOut FROM TestTime WHERE Dat = NOW() AND TimeIn = 0";

I was trying CURDATE(), but that didn't work, gave me the same Resource ID #6 problem.

Right now in the db,, I have one record, Dat = 2004-07-04 , TimeIn = 0, and TimeOut = 1088688433.

I need to get that TimeOut number.

I am making an employee clockin/clockout script, but just learning php and mysql.

Thank you.

[TheDefender]

At the top of your script, define a variable to make your comparison with:

$nowdate = date("Y-m-d");

// Then run your query:
$testingTimeOutTotal = "SELECT TimeOut FROM TestTime WHERE Dat = '$nowdate' AND TimeIn = '0'";

$getit = mysql_query($testingTimeOutTotal);
WHILE($gotit = mysql_fetch_array($getit)){
echo $gotit['TimeOut'];
}

[aace]

Great, thanks. But, is there anyway to get the 'TimeOut' out of the array? I need to work with it out of the array, have to math with it and the array would make it more complicating. Thanks again.

[TheDefender]

This line is pulling it out of the array.

$nowdate = date("Y-m-d"); 

// Then run your query: 
$testingTimeOutTotal = "SELECT TimeOut FROM TestTime WHERE Dat = '$nowdate' AND TimeIn = '0'"; 

$getit = mysql_query($testingTimeOutTotal); 
WHILE($gotit = mysql_fetch_array($getit)){ 
$TimeOut = $gotit['TimeOut']; 

// Now you can manipulate the variable any way you wish.  You don't have to assign the array value to a variable.  You can do your calculations using the array just the same.

}

If this isn't what you meant, then expand on your question a bit, and explain a bit more about what you are wanting to do please.