[PHP-DB] Two sql queries
Hi there,
Wonder if someone could help me, I have not touched PHP in a while now so a little
rusty.
What i have is a news page now the news story is disaplayed when someone clicks the
item from the home page, which then loads a news page, but the problem I am having is
when the news page loads it has the links down the left with the other news stories,
all with the correct id numbers appended to the link, but the news item on the page is
loading as the last one, and when you click the links it does not load the correct
item.
Now I know this is something to do with having two sql queries on the same page, and
the id number being fetched.
Here is the code for the two sql statements on the same page, if someone could help it
would be most appreciated.
?
//start query code
$sql= SELECT * FROM news WHERE 'Corporate' = division ;
$sql_result = mysql_query($sql);
if (mysql_num_rows($sql_result) ==0)
{
echo ( \n);
}
else {
while ($row = mysql_fetch_array($sql_result)){
$title = $row[title];
$id = $row[id];
echo (tr \n);
echo (td width=\13%\ class=\body\p /p/td \n);
echo (td width=\87%\pa href=\news.php?id=$id\$title/a/p/td
\n);
echo (/tr \n);
}}
?
?
//start query code
$sql= SELECT * FROM news WHERE ('$id' = id) ;
$sql_result = mysql_query($sql);
if (mysql_num_rows($sql_result) ==0)
{
echo ( \n);
}
else {
while ($row = mysql_fetch_array($sql_result)){
$division = $row[division];
$title = $row[title];
$date = $row[date];
$content = $row[content];
$link = $row[link];
echo (h3$title/h3 \n);
echo (p . date ('F jS Y',strtotime($date)) . /p \n);
echo (p$content/p \n);
echo (p /p \n);
}}
?
There is html code inbetween these two statements, and the id number is passed from
the index page with the news item listed.
Hope someone can help me in the right direction.
Thanks
Barry Zimmermn
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[PHP-DB] Two sql queries
Hi there,
Wonder if someone could help me, I have not touched PHP in a while now so a little
rusty.
What i have is a news page now the news story is disaplayed when someone clicks the
item from the home page, which then loads a news page, but the problem I am having is
when the news page loads it has the links down the left with the other news stories,
all with the correct id numbers appended to the link, but the news item on the page is
loading as the last one, and when you click the links it does not load the correct
item.
Now I know this is something to do with having two sql queries on the same page, and
the id number being fetched.
Here is the code for the two sql statements on the same page, if someone could help it
would be most appreciated.
?
//start query code
$sql= SELECT * FROM news WHERE 'Corporate' = division ;
$sql_result = mysql_query($sql);
if (mysql_num_rows($sql_result) ==0)
{
echo ( \n);
}
else {
while ($row = mysql_fetch_array($sql_result)){
$title = $row[title];
$id = $row[id];
echo (tr \n);
echo (td width=\13%\ class=\body\p /p/td \n);
echo (td width=\87%\pa href=\news.php?id=$id\$title/a/p/td
\n);
echo (/tr \n);
}}
?
?
//start query code
$sql= SELECT * FROM news WHERE ('$id' = id) ;
$sql_result = mysql_query($sql);
if (mysql_num_rows($sql_result) ==0)
{
echo ( \n);
}
else {
while ($row = mysql_fetch_array($sql_result)){
$division = $row[division];
$title = $row[title];
$date = $row[date];
$content = $row[content];
$link = $row[link];
echo (h3$title/h3 \n);
echo (p . date ('F jS Y',strtotime($date)) . /p \n);
echo (p$content/p \n);
echo (p /p \n);
}}
?
There is html code inbetween these two statements, and the id number is passed from
the index page with the news item listed.
Hope someone can help me in the right direction.
Thanks
Barry Zimmermn
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Updating Many Records
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[PHP-DB] Re: [PHP-DB] Updating Forms Values
Hi there again.
I have tried using the urlencode method but for some reason it is still doing
the same thing and stripping the rest of the name.
I will try to explain more how this works.
I have a site where you login using a name and passord. These values are
stored using cookies and we have the cookie script on each page before the
headers to check to see if the user is allowed at the page there are trying
to access.
This is working fine until we come to editing or adding records.
What happens is the name is tagged onto the standard url links on each page.
So therefore knowing what records to show according to which user is logged
into the system.
Now I have tried using the urlencode but for some reason it just does not
seem to work.
Basically when the user of the system adds or edits a new record, once they
click submit and are taken to the next page which either updates or adds the
new entry, the links which takes them back to the start of the system loses
the end of there name as there is a a space between there first and second
name.
I have pasted the code which updates the record in the database, together
with some of the other code on the page.
The cookie is at the top and if you look at the link to go back to the index
it contains a urlencode there.
I hope someone can help as I am pulling my hair on this one.
Many thanks
Barry
*
?
if ($name == $name) {
$msg = PWelcome to secret page A, authorized user!/p;
} else {
header(Location: http://*/administration/error.php;);
exit;
}
?
!DOCTYPE HTML PUBLIC -//W3C//DTD HTML 4.0 Transitional//EN
HTML
HEAD
TITLEedited/TITLE
META NAME=GENERATOR CONTENT=Freeway 3.1
STYLE TYPE=text/css
!--
style6 { margin-top: 0px; }
style3 { font-size: 10px; }
--
/STYLE
/HEAD
BODY TOPMARGIN=0 LEFTMARGIN=0 MARGINHEIGHT=0 MARGINWIDTH=0
BGCOLOR=#ffCENTER
TABLE BORDER=0 CELLSPACING=0 CELLPADDING=0 WIDTH=693
TR VALIGN=TOP
TD COLSPAN=3/TD
TD HEIGHT=8/TD
/TR
TR VALIGN=TOP
TD ROWSPAN=3
TABLE BORDER=0 CELLSPACING=0 CELLPADDING=0 WIDTH=312
TR VALIGN=TOP
TD ROWSPAN=2IMG SRC=../../Resources/ip20tm.gif BORDER=0
WIDTH=90 HEIGHT=39 ALT=/TD
TD COLSPAN=2/TD
TD HEIGHT=22/TD
/TR
TR VALIGN=TOP
TD/TD
TDIMG SRC=../../Resources/admintitle.gif BORDER=0
WIDTH=209 HEIGHT=17 ALT=/TD
TD HEIGHT=17/TD
/TR
TR
TD WIDTH=90IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=90 HEIGHT=1 ALT=/TD
TD WIDTH=12IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=12 HEIGHT=1 ALT=/TD
TD WIDTH=209IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=209 HEIGHT=1 ALT=/TD
TD WIDTH=1 HEIGHT=1IMG SRC=../../Resources/_clear.gif
BORDER=0 WIDTH=1 HEIGHT=1 ALT=/TD
/TR
/TABLE
/TD
TD COLSPAN=2/TD
TD HEIGHT=26/TD
/TR
TR VALIGN=TOP
TD/TD
TD
P CLASS=style6 ALIGN=RIGHTFONT FACE=Verdana, Arial
SIZE=2SPAN CLASS=style3| /SPAN/FONTA
HREF=../index.php?name=? echo urlencode($name); ?FONT
FACE=Verdana, Arial SIZE=2SPAN CLASS=style3back to my
clients/SPAN/FONT/AFONT FACE=Verdana, Arial SIZE=2SPAN
CLASS=style3 | /SPAN/FONTA HREF=../../index.php
TARGET=_selfFONT FACE=Verdana, Arial SIZE=2SPAN
CLASS=style3log out/SPAN/FONT/AFONT FACE=Verdana, Arial
SIZE=2SPAN CLASS=style3 |/SPAN/FONT/P/TD
TD HEIGHT=13/TD
/TR
TR VALIGN=TOP
TD COLSPAN=2/TD
TD HEIGHT=1/TD
/TR
TR
TD WIDTH=312IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=312 HEIGHT=1 ALT=/TD
TD WIDTH=8IMG SRC=../../Resources/_clear.gif BORDER=0 WIDTH=8
HEIGHT=1 ALT=/TD
TD WIDTH=371IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=371 HEIGHT=1 ALT=/TD
TD WIDTH=2 HEIGHT=1IMG SRC=../../Resources/_clear.gif BORDER=0
WIDTH=2 HEIGHT=1 ALT=/TD
/TR
/TABLE
TABLE BORDER=0 CELLSPACING=0 CELLPADDING=0 WIDTH=693
TR VALIGN=TOP
TD/TD
TD HEIGHT=6/TD
/TR
TR VALIGN=TOP
TD?php
// create connection
$connection = mysql_connect(*,*,*)
or die(Couldn't make connection.);
// select database
$db = mysql_select_db(*, $connection)
or die(Couldn't select database.);
?
?
if ($REQUEST_METHOD==POST) {
# setup SQL statement
$SQL = UPDATE case_study SET;
$SQL = $SQL . clientname = '$clientname', ;
$SQL = $SQL . activity = '$activity', ;
$SQL = $SQL . title = '$title', ;
$SQL = $SQL . brief = '$brief', ;
$SQL = $SQL . solution = '$solution', ;
$SQL = $SQL . result = '$result' ;
$SQL = $SQL . WHERE id = '$id' ;
#execute SQL statement
$result = mysql_db_query( inpress,$SQL,$connection );
# check for error
if (!$result) { echo(ERROR: . mysql_error() . \n$SQL\n);}
echo (P ALIGN=CENTERFONT FACE=\Verdana\SPAN
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HI People, I do not use dreamweaver normally, but have to do a few changes and add some php code to this site. I wonder if someone could quickly look at the attached php page, and look at the navigation i use and see why it is looping within the html. Basicall I need the records to look but not the next and previous section for the navigation. If someone could let me know where I am going wrong, as its urgent. It is the html formatting that I think I cannot quite get to grips with. Thanks Barry -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Drop Down Menus
Hi wonder if anyone knows what I am doing wrong here.
I have a drop down selection menu that is generated from a mysql database. In the
database we have over 15 fields one of them contains text for the catergory that the
entry belongs to. I have used the following code to generate my drop down menu but
when i view it in the browser the drop down menu has not listed the categories instead
we have blank entries for each selection.
? mysql_connect(localhost,user,password);
mysql_select_db(database);
$sql = select distinct category from books ORDER BY category ASC;
$makes_result = mysql_query($sql);
print (select name=\category\\n);
print(option selected value=\\Please select a Category/option\n);
while($row = mysql_fetch_row($makes_result))
{
print(option value=\$row[1]\$row[1]/option\n);
}
print(/select); ?
Thanks for the help.
Barry
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[PHP-DB] Forms carrying values.
Hi list, hope someone can help me here. I have content management system I have designed. I am still learning php so help here would be appreciated. The user logs into the system, and we carry there user name over the pages using a link to each page with the value of their name carried over as follows: ../index.php?name=?echo($name)? This works fine for the whole site, except when a user makes changes to a entry. Once they submit adding a new entry or modifying an existing one, the value of : ../index.php?name=My Name ends up being ./index.php?name=My It deletes the end of the name of the url, so that when the user clicks the link it does not really know who they are. I have used a hidden field to carry the value of the name over the pages, but it does not work on forms. I hope this makes some sort of sense. If anyone knows of a workaround. Thanks Barry -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Database query
Hi group,
I am still getting to grips with PHP and MySQL and have a question that hopefully
someone can easily answer:
I have a search page which consists of two drop down menus, the user has 3 ways of
finding information.
The first is by selecting a location and a discipline which would look in the database
and return any entries with the data in the two seperate fields.
The second is where they can select the location only, and this will return all
entries with that location
The third is where they can select the discipline only, and again this will return all
entries with that discipline.
The problem I have is when the user selects the location and a discipline which should
return no results as it is not an exact match it is returning results as it contains
part of the information.
My sql query is as follows: $sql = SELECT * FROM vacancy WHERE ('$related_discipline'
= related_discipline and '$location' = location) or ('$related_discipline' =
related_discipline) or ('$location' = location) ORDER BY ID LIMIT $limitvalue1,
$limit;
Could someone tell me what I need to change?
Again help is always appreciated.
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[PHP-DB] Assistance Urgent.
Hi I am new to this list and need some help with PHP and MYSQL. We have a database that has lots of fields in the table. What we are trying to do is get specific results, i will explain. Basicall we need to be able to retrieve the total number of records in the database but for individual entries. We have a field named: related_discipline which contains 1 of 12 different disciplines. So for example we have Rail which could be in the field, or Utilities and so on.. what we need to do is know how many of each there are in the database, I am still developing my knowledge of PHP and MySql so help would be appreciated. We need to be able to display the 12 different options that could be in that field in the database and how many entries there are for each option. I would truly appreciate if someone could assist me here. With kindest regards Barry Zimmerman -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
