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Re: problem saving date fields [message #174080 is a reply to message #174079] Sat, 21 May 2011 06:47 Go to previous messageGo to previous message
Co is currently offline  Co
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On 21 mei, 05:13, Jeff North <jnort...@yahoo.com.au> wrote:
> On Fri, 20 May 2011 22:35:03 -0400, in comp.lang.php Jerry Stuckle
> <jstuck...@attglobal.net>
>
> <ir78cp$53...@dont-email.me> wrote:
>> | On 5/20/2011 6:58 PM, Jeff North wrote:
>> | > On Fri, 20 May 2011 11:38:06 -0700 (PDT), in comp.lang.php Co
>> | > <vonclausow...@gmail.com>
>> | > <2407eb47-b406-432c-9aed-b76abc94a...@z37g2000vbl.googlegroups.com>
>> | > wrote:
>> | >
>> | >> | Hi All,
>> | >> |
>> | >> | I have two date fields.
>> | >> | When I change of them the other one gets set back to default:
>> | >> | 00-00-0000
>> | >> | I have to save both to make it work. Is there a solution for this?
>> | >> |
>> | >> | The arrival_date and departure_date are build up like this:
>
> [snip]
>
>> | >> |<select name="arrival_year" class="formFields" id="arrival_year">
>> | >> |<option value="<?php print "$arrival_year"; ?>"><?php print
>> | >> | "$arrival_year"; ?></option>
>> | >> |<option value="2011">2013</option>
>> | >> |<option value="2011">2012</option>
>> | >> |<option value="2011">2011</option>
>> | >> |<option value="2010">2010</option>  etc......
>> | >
>> | > How are you setting the $arrival_* and $depature_* values?
>> |
>> |  From the input - after is is sent from his form.
>
> Therefore the variables aren't initialised and would cause php errors
> - right?
>
>> | > If this is a new entry then all of the $arrival_* and $depature_* will
>> | > be null or zero - therefore the select options will not be
>> | > automatically selected (View Source of the page to see if there are
>> | > any error messages within these lists).
>> |
>> | No, because he isn't setting them until after the form has been
>> | displayed and the user has made his selection.
>
> Methinks you are making too many assumptions - have you seen the full
> source code of the page?
>
>> | > You might want to initialise the values by:
>> | > $arrival_day   = $depature_day   = date('j');
>> | > $arrival_month = $depature_month = date('n');
>> | > $arrival_year  = $depature_year  = date('Y');
>> | >
>> | > If you read a record from the database then you would over write the
>> | > above values. This will ensure that correct values within the select
>> | > lists are automatically selected.
>> |
>> | No, it is NOT a good idea to add a row every time the page is displayed.
>> |   Wait for the user to respond to the request, then add it to the
>> | database, as he is already doing.
>
> Who said anything about "add a row every time the page is displayed".
> According to your criteria there would never be any edit pages - only
> add pages.
>
> If editing an existing item wouldn't you want to read the already
> stored data from the database?
>
> i.e.
> // initialise variables to meaningful values
> // if editing an existing entry
> //    read record
> //    overwrite initialised variables
> // display the page
>
>> | >> | The table is updated like this:
>
> [snip]
>
>
>
>
>
>
>
>
>
>> | >> |     $departure_month = preg_replace('#[^0-9]#i', '',
>> | >> | $_POST['departure_month']);
>> | >> |     $departure_day = preg_replace('#[^0-9]#i', '',
>> | >> | $_POST['departure_day']); // filter everything but numbers
>> | >> |        $departure_year = preg_replace('#[^0-9]#i', '',
>> | >> | $_POST['departure_year']); //
>> | >> |        $departure_date = "$departure_year-$departure_month-$departure_day";
>> | >
>> | > You are assuming that items within the listboxs have been selected
>> | > (normally this the case only if the initial setup is correct).
>> | > You should use
>> | > $arrival_day = ( isset( $_POST['arrival_day'] ) ?
>> | > intval($_POST['arrival_day']) : 0;
>> | > if( $arrival_day == 0 ) { $pageError = "Invalid arrival day number";}
>> | >
>> | > I also assume that you are checking that you can't arrive before you
>> | > depart :-)
>> |
>> | He needs to do other validation on the data also, but first he needs to
>> | get this part working.
>
> ...and how do you know this, hmmmm?
>
>> | >> |        $sqlUpdate = mysql_query("UPDATE myMembers SET
>> | >> | firstname='$firstname', lastname='$lastname', gender='$gender',
>> | >> | partner='$partner', country='$country', rank='$rank',
>> | >> | service='$service', position='$position',
>> | >> | arrival_date='$arrival_date',departure_date='$departure_date' WHERE
>> | >> | id='$id' LIMIT 1");
>> | >> |
>> | >> | Is there a reason why this happens? I do create the right format to
>> | >> | put the data back in the table.
>> | >
>> | > But are the values your setting the ones you expect?
>> |
>> | That's why the suggestion to print the SQL statement he generated.
>
> The SQL statement should be ok - it's the "When I change of them the
> other one gets set back to default: 00-00-0000" I'm considering.
>
>> | BTW - your sig separator is broken.  It needs to be exactly
>> | hyphen-hyphen-space-newline.
>
> Oh Jerry is now the sig line cop :-P
> Complain to the writers of Thunderbird as I couldn't care less about
> whether or not my sig appears 'correctly' in your news reader.
> -- -------------------------------------------------
> The supplied code is for guideline purposes only.

Did some testing with the print command.
When I update the dates these are the results.

when I change departure:
arrival 2009-0120 departure 11-07-01

when I change arrival:
arrival 2009-07-01 departure 2011-01

it doesn't make any sense to me.

Marco

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