Php if statement in a form

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



Hello, I wonder if someone could point me in the right direction here.

I have a table that is displayed that is also a form, and allowed a
person to select a record to update using a radio button. With one of
the fields of the form/table however, I would like it to display the
value in the db (if there is one). If there is no value then I want the
field to display the current time and then submit it to the db as a
variable.

I can get it to display the current time and submit it to the db, but I
don't know how to do the IF bit of the form/table.

Anyway, the code is below.
Thanks



<?
$dbcnx = @mysql_connect( "localhost", "user", "password");
mysql_select_db("movements"); 
$result = mysql_query("SELECT id, name, location, timein, timeout FROM
workalone WHERE date=CURRENT_DATE()") or die (mysql_error());
// Display the results of the query in a table
print "<b>Current Staff Working Alone</b>";
//below is the table/form with the id, name, location, timein and
timeout taken from the db
print "<table border=\"1\" cellpadding=\"3\" cellspacing=\"0\">\n";
print
"<tr><td></td><td><b>Name</b></td><td><b>Location</b></td><td><b>Time
in</b></td><td><b>Time Out</b></td>/tr>";
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
    print "<tr><td>";
    print "<INPUT TYPE='RADIO' NAME='id' VALUE='".$row["id"]."'>";
    print "</td><td>";
    print $row["name"];
    print "</td><td>";
    print $row["location"];
    print "</td><td>";
    print $row["timein"];
    print "</td><td>";
//below is the IF thing I am having problems with
    ?><input name="timeout" type="text" value="if {$timeout=null <?php
echo date('H:i');?>} else {<?print $row["timeout"];?>"><?
    print "</td></tr>\n";
}
print "</table>\n";

?><input type="submit" name="submit2" value="Select your name and click
here to record your timeout"></form>   <?

if ($submit2)
{
$dbcnx = @mysql_connect( "localhost", "user", "password");
mysql_select_db("movements"); 
$result = mysql_query("UPDATE workalone SET timeout='$timeout' WHERE
id='$id'") or die (mysql_error());
echo "<b>Thank you, your Time Out has been recorded.</b>";

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php



[Index of Archives]     [PHP Home]     [PHP Users]     [Postgresql Discussion]     [Kernel Newbies]     [Postgresql]     [Yosemite News]

  Powered by Linux