How to output a NULL field?
in [General]
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From: Shawn McKenzie on 25 Aug 2009 16:02 Shawn McKenzie wrote: > First off, if the value is NULL in the database then in PHP it will be > the string "NULL" and not a null value as far as I remember. Second, > you cant use a function/method in empty(). Thirdly, the string "NULL" > is not empty. I'm not sure what DB class you're using here or what the > Fields() method actually returns. You should do a > var_dump($rs->Fields(22)) to see. If it returns a string (especially > "NULL"), then this or some variation should work: > > $q4 = $rs->Fields(22); > > if($q4 == "NULL"){ > $q4 = ""; > } > > If it returns an empty string or false then you may have nothing to do. > DOH! My bad. MySQL stores an empty value based upon the field type for NULL. So if the field type is int then it sets it to 0 and if its varchar, etc, it sets it to an empty string "". The MySQL functions seem to return everything as a string, so for an int field it returns the string "0", so: $q4 = $rs->Fields(22); if(empty($q4)){ $q4 = ""; } But I'm not sure what benefit you get from this except that "0" is changed to "". -- Thanks! -Shawn http://www.spidean.com
From: Paul M Foster on 25 Aug 2009 16:38 On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote: > $rs->Fields(22) equals a NULL in the database > > My Code: > > if(empty($rs->Fields(22))){ > $q4 = ""; > }else{ > $q4 = $rs->Fields(22); > } > > Produces this error: > Fatal error: Can't use method return value in write context in > D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32 > > Line 32 is the "if" line... > > If I switch the code to (using is_null): > if(is_null($rs->Fields(22))){ > $q4 = ""; > }else{ > $q4 = $rs->Fields(22); > } > > It produces this error: > Catchable fatal error: Object of class variant could not be converted to > string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196 > > Line 196 is: <?php echo $q4;?> > > What am I doing wrong? > > Thanks! Just a thought... do you really mean $rs->Fields(22) or do you mean $rs->Fields[22]? The former is a function call and the latter is an array variable. Paul -- Paul M. Foster
From: "David Stoltz" on 26 Aug 2009 07:29 Paul, This all started because when I try this: <?php echo $rs->Fields(22);?> It work fine, as long as there is a non-null value there, otherwise it produces an error. Also, I'm working with a Microsoft SQL 2000 database, not MySQL....not sure if that matters.... But "echo $rs->Fields(22)" works perfectly for dumping values out of my $rs recordset...that is, unless the value is NULL is the database - then I get: Catchable fatal error: Object of class variant could not be converted to string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176 -----Original Message----- From: Paul M Foster [mailto:paulf(a)quillandmouse.com] Sent: Tuesday, August 25, 2009 4:39 PM To: php-general(a)lists.php.net Subject: Re: [PHP] How to output a NULL field? On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote: > $rs->Fields(22) equals a NULL in the database > > My Code: > > if(empty($rs->Fields(22))){ > $q4 = ""; > }else{ > $q4 = $rs->Fields(22); > } > > Produces this error: > Fatal error: Can't use method return value in write context in > D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32 > > Line 32 is the "if" line... > > If I switch the code to (using is_null): > if(is_null($rs->Fields(22))){ > $q4 = ""; > }else{ > $q4 = $rs->Fields(22); > } > > It produces this error: > Catchable fatal error: Object of class variant could not be converted to > string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196 > > Line 196 is: <?php echo $q4;?> > > What am I doing wrong? > > Thanks! Just a thought... do you really mean $rs->Fields(22) or do you mean $rs->Fields[22]? The former is a function call and the latter is an array variable. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
From: hack988 hack988 on 26 Aug 2009 07:39 use is_null() check it 2009/8/26 David Stoltz <Dstoltz(a)shh.org>: > Paul, > > This all started because when I try this: > > <?php echo $rs->Fields(22);?> > > It work fine, as long as there is a non-null value there, otherwise it > produces an error. > > Also, I'm working with a Microsoft SQL 2000 database, not MySQL....not > sure if that matters.... > > But "echo $rs->Fields(22)" works perfectly for dumping values out of my > $rs recordset...that is, unless the value is NULL is the database - then > I get: > > Catchable fatal error: Object of class variant could not be converted to > string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176 > > > -----Original Message----- > From: Paul M Foster [mailto:paulf(a)quillandmouse.com] > Sent: Tuesday, August 25, 2009 4:39 PM > To: php-general(a)lists.php.net > Subject: Re: [PHP] How to output a NULL field? > > On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote: > >> $rs->Fields(22) equals a NULL in the database >> >> My Code: >> >> if(empty($rs->Fields(22))){ >> $q4 = ""; >> }else{ >> $q4 = $rs->Fields(22); >> } >> >> Produces this error: >> Fatal error: Can't use method return value in write context in >> D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32 >> >> Line 32 is the "if" line... >> >> If I switch the code to (using is_null): >> if(is_null($rs->Fields(22))){ >> $q4 = ""; >> }else{ >> $q4 = $rs->Fields(22); >> } >> >> It produces this error: >> Catchable fatal error: Object of class variant could not be converted > to >> string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196 >> >> Line 196 is: <?php echo $q4;?> >> >> What am I doing wrong? >> >> Thanks! > > Just a thought... do you really mean $rs->Fields(22) or do you mean > $rs->Fields[22]? The former is a function call and the latter is an > array variable. > > Paul > > -- > Paul M. Foster > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >
From: "David Stoltz" on 26 Aug 2009 07:51
I tried that -it's in the first part of my message -----Original Message----- From: hack988 hack988 [mailto:hack988(a)dev.htwap.com] Sent: Wednesday, August 26, 2009 7:39 AM To: David Stoltz Cc: Paul M Foster; php-general(a)lists.php.net Subject: Re: [PHP] How to output a NULL field? use is_null() check it 2009/8/26 David Stoltz <Dstoltz(a)shh.org>: > Paul, > > This all started because when I try this: > > <?php echo $rs->Fields(22);?> > > It work fine, as long as there is a non-null value there, otherwise it > produces an error. > > Also, I'm working with a Microsoft SQL 2000 database, not MySQL....not > sure if that matters.... > > But "echo $rs->Fields(22)" works perfectly for dumping values out of my > $rs recordset...that is, unless the value is NULL is the database - then > I get: > > Catchable fatal error: Object of class variant could not be converted to > string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176 > > > -----Original Message----- > From: Paul M Foster [mailto:paulf(a)quillandmouse.com] > Sent: Tuesday, August 25, 2009 4:39 PM > To: php-general(a)lists.php.net > Subject: Re: [PHP] How to output a NULL field? > > On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote: > >> $rs->Fields(22) equals a NULL in the database >> >> My Code: >> >> if(empty($rs->Fields(22))){ >> $q4 = ""; >> }else{ >> $q4 = $rs->Fields(22); >> } >> >> Produces this error: >> Fatal error: Can't use method return value in write context in >> D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32 >> >> Line 32 is the "if" line... >> >> If I switch the code to (using is_null): >> if(is_null($rs->Fields(22))){ >> $q4 = ""; >> }else{ >> $q4 = $rs->Fields(22); >> } >> >> It produces this error: >> Catchable fatal error: Object of class variant could not be converted > to >> string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196 >> >> Line 196 is: <?php echo $q4;?> >> >> What am I doing wrong? >> >> Thanks! > > Just a thought... do you really mean $rs->Fields(22) or do you mean > $rs->Fields[22]? The former is a function call and the latter is an > array variable. > > Paul > > -- > Paul M. Foster > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > |