Re: Php if statement in a form

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Sorry.  Forgot to remove the word echo from that line.

On Thu, 5 Aug 2004 21:29:01 -0400, Eric Schwartz
<eric.j.schwartz@xxxxxxxxx> wrote:
> On Fri, 6 Aug 2004 10:21:19 +1000, justin.baiocchi@xxxxxxxx
> <justin.baiocchi@xxxxxxxx> wrote:
> > Thanks Eric,
> >
> > I have changed it somewhat but am just getting a parse error,
> > unexpected T_STRING on that line.
> >
> > My revised code is below:
> >
> >
> > print "<b>Current Staff Working Alone</b>";
> > print "</td></tr>";
> > print "</table>\n";
> > 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 $row["timein"];
> >     print "</td><td>";
> >     if ($row["timeout"] IS NULL);
> >     {
> >     print <input name="timeout" type="text" value="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>
> >
> >
> >
> >
> > -----Original Message-----
> > From: Eric Schwartz [mailto:eric.j.schwartz@xxxxxxxxx]
> > Sent: Friday, 6 August 2004 9:31 AM
> > To: php-db@xxxxxxxxxxxxx
> > Subject: Re:  Php if statement in a form
> >
> > On Fri, 6 Aug 2004 08:53:53 +1000, justin.baiocchi@xxxxxxxx
> > <justin.baiocchi@xxxxxxxx> wrote:
> > > 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
> > >
> > >
> >
> > You seem to have your php tags not around your if statement and you
> > are checking to see if $timeout rather than $row["timeout"] has a
> > value.  I also think the proper way to check for NULL is to say IS
> > NULL or NOT NULL, not $foo = NULL.  Could be wrong about that one.
> >
> > <?php
> > if ($row["timeout"] IS NULL) {
> >  echo date('H:i');
> > } else {
> > print $row["timeout"];
> > }
> > ?>
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
> >
> 
> Try adding a quote on the line and escaping the other quotes, like this:
> 
> print "<input name=\"timeout\" type=\"text\" value=\"".echo date('H:i')."\">";
>

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