Re: [PHP-DB] viewing search result
Hi,
Iam new in this , can you tell me how to make the pager.php works? I can't
use it...
Best Regards,
Hadi Tjen
Bossek [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
you can use PEAR DB::Pager
http://pear.php.net/package/DB_Pager
Fedde Van Feggelen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
str_replace(' ','%',$searchwords);
$query=select * from table where colom like '%$searchwords%' order by
id
desc ;
$result=mysql_query($query);
$numresult=mysql_num_rows($result);
Use limit in your select statement.
laters,
Fedde
~=There's a fine line between genius and insanity. I have erased this
line=~
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[PHP-DB] viewing search result
Hi Everybody,
Please help me in this,
I have a search page, when people search they keywords and the result is 50
rows from mysql , how to display them page by page which each page contain
20 results , thanks in advance :
str_replace(' ','%',$searchwords);
$query=select * from table where colom like '%$searchwords%' order by id
desc ;
$result=mysql_query($query);
$numresult=mysql_num_rows($result);
if (mysql_num_rows($hasil) 0 ) {
echo (bFound :/b $searchwords brbr);
//page display
I have tried my own recipe but didn't work
while ($data=mysql_fetch_array($result)) {
echo (
tr
td width=\330px\font size =\3\br$data[colom1]brbr/td
td width=\50px\p align =\center\font size
=\2\$data[colom2]br/td
td width=\50px\p align =\center\font size
=\2\$data[colom3]br/td
/tr
);
Best Regards,
Hadi Tjen
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[PHP-DB] Re: Problem with INSERT Query
[EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi all:
Well, when i bring out the page with the drop down list it was able to
display
all tutors' names from tutor_name column. Anyway here's a review of my
code
(snip) again before i continue:
--
-
snip
$sql = INSERT INTO class (class_code, tutor_name, edu_level,
timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
);
?//retrieve data from DB display in dynamic drop down ?
SELECT class=textarea name=tutor_name /
?
$sqltutor = mysql_query(SELECT DISTINCT tutor_name FROM tutor );
while ($row = mysql_fetch_array($sqltutor))
{
print OPTION VALUE=\$row [tutor_name]. \ SELECTED .$row
[tutor_name]. /option ;
}
$result = $db- query($sql);
?
/select
?
while($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
How about
if ($tutor_name ){
echo $_POST[tutor_name];
$sql = INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
);}
is it working?
Hadi
?
/snip
--
-
so when i submit the form, i am suppose to echo the values i have entered
into
the field and then INSERT the values into DB (Queries stated above).
However i
was able to echo all other values eg. class_code, edu_level, etc...but
not tutor_namesame thing happen when i do an INSERT, all other
values
are inserted into DB but not $tutor_namewhy is this so???Really need
some
help here...Anyway i have already specify a name to be reference :
SELECT class=textarea name=tutor_name
and then I also did an echo of tutor_name being selected:
while($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
All help are greatly appreciated =)
Irin.
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[PHP-DB] password---simple questions but...
Hi, all
please help me in this, Iam using MYSQL , my colomn is password varchar(12)
insert into tablepassword values ('',password('def'));
it is executed, (select * from tablepassword) showing the password is
hashed, and then to view the record using 'where' :
select * from tablepassword where password=password('def');
display empty set, how to solve this?
Thanks in advance.
Merry Christmas and Happy New Year
Hadi Tjen
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[PHP-DB] anyone trickier than 'LIKE' ?
Hi all, I have this data (eg.) ..big fat buddy... on the database . If a person enter big buddy in the search form , the ...big fat buddy. data will not be found . Any suggestion ? Thanks in advance. select * from table where data like %keywords%; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] help newbie
Hi , please help this newbie , how to make my database view from
[1] [2] [3] [4] [5] [6] [7] [8]
to
1 2 34
Thanks very much
Hadi
require(cobaconfig.php);
$numresult = mysql_query(select * from dataiklanbaris order by id desc);
$jumlah = mysql_num_rows($numresult);
$limit = 20;
if (empty($offset)) {
$offset = 0;
}
$query = select * from dataiklanbaris order by id desc limit $offset,
$limit;
//pembagian halaman dimulai
$result = mysql_query($query);
$halaman = intval($jumlah/$limit);
if ($jumlah%$limit) {
$halaman++;
}
for ($i = 1; $i=$halaman; $i++) {
$newoffset=$limit*($i-1);
if ($offset != $newoffset) {
echo [a href=\lihatiklan.php?offset=$newoffset\$i/a];
} else {
echo [$i];
}
}
//pembagian halaman selesai
while ($row = mysql_fetch_array($result)) {
echo (
tr
td width=\330px\font size =\3\$row[isiiklan]br/td
td width=\50px\p align =\center\font size
=\2\$row[lt]br/td
td width=\50px\p align =\center\font size
=\2\$row[lb]br/td
td width=\70px\p align =\center\font size
=\2\$row[harga]br/td
td width=\100px\p align =\center\font size
=\2\$row[kecamatan]br/td
td width=\100px\p align =\center\font size
=\2\$row[kota]br/td
td width=\100px\a href=\detail.php?id=$row[id]\
target=\blank\p
align =\center\font size =\2\$row[gambar]br/a/td
/tr
);
}
?
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[PHP-DB] Hi all, how the sript to view like...
I have my database page look like this:
[1][2] [3] [4] [5] [6]
the problem is when I have my database growing it will be not nice to look
at because the number will keep growing,
how to make it like this :
1234
Thanks for the help
require(cobaconfig.php);
$numresult = mysql_query(select * from dataiklanbaris order by id desc);
$total = mysql_num_rows($numresult);
$limit = 20;
if (empty($offset)) {
$offset = 0;
}
$query = select * from table order by id desc limit $offset, $limit;
$result = mysql_query($query);
$halaman = intval($total/$limit);
if ($jumlah%$limit) {
$page++;
}
for ($i = 1; $i=$page; $i++) {
$newoffset=$limit*($i-1);
if ($offset != $newoffset) {
echo [a href=\view.php?offset=$newoffset\$i/a];
} else {
echo [$i];
}
}
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