Thread: Date type: DATE
Hi everybody. I've got a form where I ask for the date.I store the date in 3 variables: for the day i have $day (dd), for the month i have $month (mm) and for the year $year (yyyy). At the same time i have a table which is as it follows: Table "s_objetivos_caso" Column | Type | Modifiers -----------------+---------+----------- dni | integer | not null fecha | date | not null cod_s_objetivos | integer | not null How do i have to insert the column date?. $date="$year-$month-$day"; insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); Can anyboy help me for how a i must format the date for introduce into the table??? Thank you. And sorry with my English.
> $date="$year-$month-$day"; > insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); By default, postgresql uses "ISO with US(NonEuropean) conventions" for the date, wich means mm/dd/yyyy. You can check that with "show datestyle". Of course, you can change the convention to a European one like dd/mm/yyyy but I don't know how to make it permanent.... anyway, if we stick to the default US convention you would do something like: $date=$month ."/". $day ."/" $year; // Use the dot (.) to join strings And then do the insert. Adrian Tineo
Try putting single quotes around the date, this works for me.. insert into s_objetivos_caso values ($dni,'$date',$cod_s_objetivos,$conn); ----- Original Message ----- From: "Jesus Rios" <galiza-vermelha@wanadoo.es> To: "POSTGRESLQ-PHP" <pgsql-php@postgresql.org> Sent: Thursday, January 23, 2003 5:59 AM Subject: [PHP] Date type: DATE > > Hi everybody. > > I've got a form where I ask for the date.I store the date in 3 variables: for > the day i have $day (dd), for the month i have $month (mm) and for the year > $year (yyyy). > At the same time i have a table which is as it follows: > Table "s_objetivos_caso" > Column | Type | Modifiers > -----------------+---------+----------- > dni | integer | not null > fecha | date | not null > cod_s_objetivos | integer | not null > > How do i have to insert the column date?. > > $date="$year-$month-$day"; > insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); > > Can anyboy help me for how a i must format the date for introduce into the > table??? > > > Thank you. And sorry with my English. > > ---------------------------(end of broadcast)--------------------------- > TIP 3: if posting/reading through Usenet, please send an appropriate > subscribe-nomail command to majordomo@postgresql.org so that your > message can get through to the mailing list cleanly >
Please correct me if I am wrong but it looks like you are asking how to assemble the values in your PHP script into a string that you can insert into your database... if that is the case you need to use the concatenate operator in PHP. You can do this in a few ways... $date = $year . "-" . $month . "-" . $day; insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); or right in the the query itself... insert into s_objetivos_caso values ($dni,$year . "-" . $month . "-" . $day,$cod_s_objetivos,$conn); you would need to check that second way to be sure, but the first one will work for sure. Alan ----- Original Message ----- From: "Mathew Dredge" <mathew@idealsoftwaresolutions.com.au> To: "Jesus Rios" <galiza-vermelha@wanadoo.es>; "POSTGRESLQ-PHP" <pgsql-php@postgresql.org> Sent: Thursday, January 23, 2003 10:05 PM Subject: Re: [PHP] Date type: DATE > Try putting single quotes around the date, this works for me.. > > insert into s_objetivos_caso values ($dni,'$date',$cod_s_objetivos,$conn); > > > ----- Original Message ----- > From: "Jesus Rios" <galiza-vermelha@wanadoo.es> > To: "POSTGRESLQ-PHP" <pgsql-php@postgresql.org> > Sent: Thursday, January 23, 2003 5:59 AM > Subject: [PHP] Date type: DATE > > > > > > Hi everybody. > > > > I've got a form where I ask for the date.I store the date in 3 variables: > for > > the day i have $day (dd), for the month i have $month (mm) and for the > year > > $year (yyyy). > > At the same time i have a table which is as it follows: > > Table "s_objetivos_caso" > > Column | Type | Modifiers > > -----------------+---------+----------- > > dni | integer | not null > > fecha | date | not null > > cod_s_objetivos | integer | not null > > > > How do i have to insert the column date?. > > > > $date="$year-$month-$day"; > > insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); > > > > Can anyboy help me for how a i must format the date for introduce into > the > > table??? > > > > > > Thank you. And sorry with my English. > > > > ---------------------------(end of broadcast)--------------------------- > > TIP 3: if posting/reading through Usenet, please send an appropriate > > subscribe-nomail command to majordomo@postgresql.org so that your > > message can get through to the mailing list cleanly > > > > > ---------------------------(end of broadcast)--------------------------- > TIP 5: Have you checked our extensive FAQ? > > http://www.postgresql.org/users-lounge/docs/faq.html >
At 01:59 PM 1/22/03, Jesus Rios wrote: >I've got a form where I ask for the date.I store the date in 3 variables: for >the day i have $day (dd), for the month i have $month (mm) and for the year >$year (yyyy). >At the same time i have a table which is as it follows: > Table "s_objetivos_caso" > Column | Type | Modifiers > -----------------+---------+----------- > dni | integer | not null > fecha | date | not null > cod_s_objetivos | integer | not null > >How do i have to insert the column date?. > > $date="$year-$month-$day"; > insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); > >Can anyboy help me for how a i must format the date for introduce into the >table??? Perhaps you could tell us what goes wrong with the above coding. Others have mentioned issues with $date variable. Does the code look "exactly" as above? Surely you are coding the insert using pg functions? pg_exec ($conn, "insert into s_objetivos_caso values ($dni,'$date',$cod_s_objetivos)"; Frank
on 1/24/03 12:26 AM, amiller@hollywood101.com purportedly said: > $date = $year . "-" . $month . "-" . $day; > insert into s_objetivos_caso values ($dni,$date,$cod_s_objetivos,$conn); > > or right in the the query itself... > > insert into s_objetivos_caso values ($dni,$year . "-" . $month . "-" . > $day,$cod_s_objetivos,$conn); > > you would need to check that second way to be sure, but the first one will > work for sure. Neither of these will work, nor any previous suggestion, as date formats are ambiguous and *must* be quoted: $date = $year . "-" . $month . "-" . $day; insert into s_objetivos_caso values ($dni,"'$date'", $cod_s_objetivos,$conn ); This *will* work, at least as far as the date is concerned. Keary Suska Esoteritech, Inc. "Leveraging Open Source for a better Internet"