[PHP-DB] [4.3.3] How to buildi with SQLite support ?
I looked the the configure --help but didn't see sqlite? Or did I just miss it? How can I build the module for sqlite. David -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Generating readio buttons
Howdy --
How do I dynamically generate the value for radio buttons off the DB
(MySQL) backend? Here's some code that almost works
TIA,
David
- Almost works
?php
$header = mysql_query (SELECT * FROM chart
ORDER BY acct );
if ($row = mysql_fetch_array($header)) {
do {
print(tr bgcolor=\white\ );
print 'td width=5input type=radio name=gl_acct
value=acct[]/td';
print(td width=\12\);
print $row[acct];
print /tdtd width=\12\;
print $row[cat];
print /tdtd ;
print $row[descript];
print(/td/tr\n);
} while($row = mysql_fetch_array($header));
} else {print Sorry, no records were found!;}
?
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[PHP-DB] Re: LIKE statement
Chris Barnes wrote:
Hi,
I've got a dilly of a problem. I'm probably doing something wrong but I
don't know what. I'm trying to use the LIKE statement in a query where
more than one word is used in with LIKE..e.g.
select count(distinct itemid) from business where name or description
like 'word1 word2 word3%'
Chris --
The answer depends on what database you using? MySQL has a IN operator,
so you might try:
SELECT * from table
WHERE name IN ('value1','value2','value3)
OR
WHERE description IN ('value1','value2','value3)
Your might want to cross post to MySQL list.
HTH,
David
The problem I'm having is probably obvious to you but I don't know why
this returns no matches but if i specify only 1 word in the LIKE
statement then it returns a match.
Am i not able to specify more than 1 word with LIKE or am I just doing
it wrong?
It has been designed to take input from a web form by the variable
$search_string and then the query string is constructed from that e.g.
$query = select count(distinct itemid) from business where name or
description like' . $search_string . ';
Any help or suggestions greatly appreciated.
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Re: [PHP-DB] Generating readio buttons
Peter Beckman wrote:
Peter,
Thanks for your prompt reply.
Gonna need some additional information, such as:
1. The first three rows your query returns
2. The expected output
3. The actual problem and/or error.
What I'm trying todo is build a radio box form
for the selection of ledger accounts for a account app.
( http://mustardandrelish.com/ledger)
Query returns (from HTML table):
input type=radio name=gl_acct value=acct[]
input type=radio name=gl_acct value=acct[]
input type=radio name=gl_acct value=acct[]
Expected results might be:
input type=radio name=gl_acct value=1000
input type=radio name=gl_acct value=2000
input type=radio name=gl_acct value=3000
And then gl_acct will be passed to the G/L entry screen
(sticky forms?)
TIA,
David
Peter
On Sun, 3 Nov 2002, David Jackson wrote:
Howdy --
How do I dynamically generate the value for radio buttons off the DB
(MySQL) backend? Here's some code that almost works
TIA,
David
- Almost works
?php
$header = mysql_query (SELECT * FROM chart
ORDER BY acct );
if ($row = mysql_fetch_array($header)) {
do {
print(tr bgcolor=\white\ );
print 'td width=5input type=radio name=gl_acct
value=acct[]/td';
print(td width=\12\);
print $row[acct];
print /tdtd width=\12\;
print $row[cat];
print /tdtd ;
print $row[descript];
print(/td/tr\n);
} while($row = mysql_fetch_array($header));
} else {print Sorry, no records were found!;}
?
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---
Peter BeckmanSystems Engineer, Fairfax Cable Access Corporation
[EMAIL PROTECTED] http://www.purplecow.com/
---
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Re: [PHP-DB] LIKE statement or IN statement?
Chris Barnes wrote:
Yeah I really need to search for multiple words. Can anyone confirm if
the IN statement will work for me in this situation?
Chris --
Why not just try it you self and let's us know.
Also check to MySQL doc at http://mysql.org
David
On Mon, 2002-11-04 at 09:31, [EMAIL PROTECTED] wrote:
if you want to search for multiple words, u have to use multiple like
operators:
select count(distinct itemid) from business where name like 'word1' or
name like 'word2' or name like 'word3';
or the IN statement with wildcards:
select count(distinct itemid) from business where name IN
('%word1%','%word2%','%word3%'); -- im not too sure of this, would the
experts please shed some more light on this one if its correct?
Amit
On 4 Nov 2002, Chris Barnes wrote:
Hi,
I've got a dilly of a problem. I'm probably doing something wrong but
I
don't know what. I'm trying to use the LIKE statement in a query where
more than one word is used in with LIKE..e.g.
select count(distinct itemid) from business where name or description
like 'word1 word2 word3%'
The problem I'm having is probably obvious to you but I don't know why
this returns no matches but if i specify only 1 word in the LIKE
statement then it returns a match.
Am i not able to specify more than 1 word with LIKE or am I just doing
it wrong?
It has been designed to take input from a web form by the variable
$search_string and then the query string is constructed from that e.g.
$query = select count(distinct itemid) from business where name or
description like' . $search_string . ';
Any help or suggestions greatly appreciated.
---
Peter BeckmanSystems Engineer, Fairfax Cable Access
Corporation
[EMAIL PROTECTED]
http://www.purplecow.com/
---
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Re: [PHP-DB] Generating readio buttons
Peter Beckman wrote:
Here's your problem:
acct[] doesn't equal anything.
$acct[4] might.
So:
echo input type='radio' name='gl_acct' value='{$acct[4]}';
Peter -- Thanks again for you help.
I wonder if it's not a quoting issue on my part?
How would I use the $row[column_name], in the above statement? Since
it not a $_POST[] filed?
Thanks in advance,
David
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[PHP-DB] Re: Generating readio buttons
OK, this works but there has to be a pretty way?
David
- Works -
html
headtitleOperation Sticky Bun/title/head
body
h3 align=centerOperation Sticky Bun /h3
?php require('connect.php'); ?
?php
print 'form action=hello.php method=post';
$header = mysql_query (SELECT * FROM chart
ORDER BY acct );
if ($row = mysql_fetch_array($header)) {
do {
print 'input type=radio name=ledger_acct value=';
print $row[acct];print '';
print $row[descript];print 'br';
} while($row = mysql_fetch_array($header));
} else {print Sorry, no records were found!;}
print '/form';
?
/body
/html
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[PHP-DB] Resolved: Inserting current date(MySQL)?
John, Jeffery, Alex and Jason -- Thanks for you response. As you all suggested it was as simple as CURDATE(). without any quotes or backticks. Thanks again, David -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
