[PHP-DB] Image Bank with php mysql!! ??!!
Hello everyone... I wan't your opinion regarding a program i am constructing. It is about an image bank for graphic studios or anyone else who needs a program that can insert his pictures or descriptions of it and be able to make a fast and reliable search. The Question is: What would be best.? Inserting the pictures in the DB or in the file system ? Also notice that it is a reference db and not an actual bank. I mean that there will be only photos of low quality and not the full picture which could be very large. Expecting your justified opinions that would be a lot of help.. Thanks ___ PGP KEY ID: 0xA86600E9 ___
Re: [PHP-DB] Finding NoMatches in MySQL
Hi, I think this query should do the job SELECT jobs.uid FROM jobs LEFT JOIN employers ON jobs.id1 = employers.uid WHERE employers.uid is Null Dobromir Velev -Original Message- From: Dave Watkinson [EMAIL PROTECTED] To: PHP-MySQL List [EMAIL PROTECTED] Date: Monday, July 30, 2001 12:56 AM Subject: [PHP-DB] Finding NoMatches in MySQL Hi all During the process of importing a lot of Oracle data into MySQL, I have two tables that are giving me a bit of a headache. One is a list of jobs and the other is a list of employers. There's now a unique id for each table, and the job table also has a column called id1, which *should* correspond to employers.uid. With me so far? Well, the problem is that there were loads of duplications in the employers table, and after deleting the duplicates some of the jobs.id1 records now point to non-existing records. How do I do a query that does something like this... select jobs.uid from jobs where jobs.id1 not in employers.uid ??? many thanks in advance! Dave -- 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] -- 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]
RE: [PHP-DB] MySQL newbie: inserting new entry to table?
I'm a learner too. But as I understand it, You make a connection to the database (or open it?). Once it's open or connected to, you can access the different tables within it. I think the difference between mysql_db_query mysql_query is that later assumes that you all ready connected to the database where the first connects at the same time. Anyway glad you fixed your problem. It wasn't a holiday in Australia (just a Sunday night :-)) Cheers Howard -Original Message- From: sg [mailto:[EMAIL PROTECTED]] Sent: Sunday, 29 July 2001 9:40 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] MySQL newbie: inserting new entry to table? Thank You Howard! You sure are quick! I found a way to make it work. I used to have both scripts (the one that displays and the one that inserts new entries) on the same page... Once I separated them I had a link on a third page. I first clicked on the insert entry link then back to the display page... My new entry was there... Now for the solution you gave me. I haven't tried it yet but I'd like to know the difference. I see you don't provide the database name in the mysql_query call. How does MySQL know what base to work with? I've seen both 'mysql_query' and 'mysql_db_query' in several tutorials but haven't yet figured the difference between them (told you a was a newbie...) Thanks again for your interest, I thought everybody would be on holidays... Sébastien -- 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] -- 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]
[PHP-DB] double mysql calls?
Hello all...
Im having a bit of trouble with a mysql update feature. What I am doing is
having a script email a list of people in a database (one email address per
db record) and have the script update the database that it has sent an
email to that particular person. Currently I have:
while ($Row = mysql_fetch_array ($Result)) {
$UserID=$Row-user_id;
if ($Row[EmailSent] == 0) {
if ($Row[email]) {
$Msg0 = Dear $Row[RealName] \n\n;
$To = $Row[email];
$TotalMsg = $Msg0$Msg1$Msg2$Msg3$Msg4$Msg5;
if (mail($To, $Subject, $TotalMsg, $AddHeaders)) {
++$Good;
$Query1= UPDATE $connectdb1 SET
EmailSent=\1\ WHERE user_id=\$UserID\ LIMIT 1;
mysql_query($Query1, $Link);
} else {
++$Bad;
}
} else {
++$CountNoEmail;
}
} else {
++$CountAlreadySent;
}
++$CountTotal;
}
What am I doing wrong? can I have multiple incarnations of mysql queries
like this? How could I get a UserID out of a while statement for later
updating? Thanks for any help.
--Photocon
Conrad Hunziker III
www.nightskyent.com
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[PHP-DB] Re: Image Bank with php mysql!! ??!!
Koutsogiannopoulos Karolos [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I wan't your opinion regarding a program i am constructing. It is about an image bank for graphic studios or anyone else who needs a program that can insert his pictures or descriptions of it and be able to make a fast and reliable search. http://www.hotscripts.com/PHP/Scripts_and_Programs/Image_Galleries/more2.htm l What would be best.? Inserting the pictures in the DB or in the file system I would keep the image name in the database; save the original image in a dedicated directory with a numeric name like 4920185.jpg and then the thumbnail either with a suffix (like 4920185b.jpg) or with the same name in a thumbnails directory; store the number as the picture ID. The numbers should be large and pseudorandom, ie you don't want to make it easy for someone to write a script to leech all your pictures (it'll still be possible, but harder). Maybe use the database ID to seed a random number generator, and use the Nth generated number? Have a text field for keywords, a text field for description, a price field?, the author's name, email, and homepage. If you put the image directory below the web root directory, then the images can only be accessed via a script, and you can restore the original name at download (this solves the problem of name collisions). If you're going to sell the pictures, you should allow a buyer more than one download, in case they screw up or accidentally delete it on their system; I would have a table for 'current purchases' with a buyer ID and picture ID, and allow them something like access for a week, up to 20 downloads of that image. ? Also notice that it is a reference db and not an actual bank. I mean that there will be only photos of low quality and not the full picture which could be very large. (shrug) as above, keep the images in the filesystem; then it doesn't matter how large it is. -- 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]
[PHP-DB] Re: Storing last access
Olinux O [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hi all, I would like to display messages that are entered into a database since the last time i checked it. [Much like web based email marks NEW messages.] What is the best way to do this? By adding an additional column to the Messages Table OR creating a new table with the field Table_Name and Last_Checked Adding a field to the table is a single-user solution; if you want this useable by more than one person, a second table is needed. There are two ways of going about this; if you want to know whether individual messages are read or not, you will have to keep individual user/message records; if all you want is messages since a given date, you can just add a 'last-read' timestamp to your users table. -- 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]
[PHP-DB] A Join Question
I'm trying to create a join statement that pulls out a CompanyName based on a given CompanyID that is tied to a specified ContactID. For example, if ContactID=1 then the corresponding CompanyName might be Smith, Inc. depending on what was entered. Here are the tables. table = Contacts FIELD | TYPE | ATTRIBUTES | NULL | DEFAULT | EXTRA | ContactID | INT(4) | UNSIGNEED | NO | AUTO_INCREMENT | Primary Key/Unique CompanyID | INT(4) | // more fields table = Company FIELD | TYPE | ATTRIBUTES | NULL | DEFAULT | EXTRA | CompanyID | INT(4) | UNSIGNEED | NO | AUTO_INCREMENT | Primary Key/Unique CompanyName | INT(4) | //more fields - What I have tried so far is: SELECT * FROM contacts,company WHERE contacts.CompanyID=$ContactID and companyID= $ContactID I've tried slightly other variations including: SELECT CompanyName,WebSite FROM $table_name1,$table_name2 WHERE contacts.CompanyID='$CompanyID' also, SELECT * FROM Contacts,Company WHERE Contacts.CompanyID=Company.CompanyID AND Contacts.CompanyID='$ContactID' With this I get an empty set. None of these seem to work. I'm obviously not doing the join right, but I'm not sure what I'm leaving out. Thanks. Steve -- 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]
[PHP-DB] Re: Query displays one
Mike Gifford [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... The problem is that I am only getting one response from this query rather than 5. I can't figure this out. What did I do to only get one response? Try removing the ORDER BY and LIMIT and run the query by hand - how many responses do you get? The search may only return one article. -- 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]
[PHP-DB] Re: A Join Question
Steve Fitzgerald [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I'm trying to create a join statement that pulls out a CompanyName based on a given CompanyID that is tied to a specified ContactID. For example, if ContactID=1 then the corresponding CompanyName might be Smith, Inc. depending on what was entered. SELECT CompanyName FROM Company INNER JOIN Contacts ON Company.CompanyID = Contacts.CompanyID WHERE ContactID=$ContactID -- 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]
RE: [PHP-DB] ODBC and PHP
Ariadne, What is the exact error you get? Among other debugging tricks, you can get an ODBC trace from the ODBC Administrator control panel (if using Win32 on client side) and look at where the API calls are failing. Best regards, Andrew Hill Director of Technology Evangelism OpenLink Software http://www.openlinksw.com Universal Data Access Data Integration Technology Providers -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] Sent: Friday, July 27, 2001 6:30 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] ODBC and PHP I´m using PHP 4 with ODBC and sometimes I get a response that was unable to connect to the MSSQL database. Then I go to the ODBC configuration to test the connection and I get a response that everything is working fine, but my pages still can´t connect. Is that normal? What can I do? I´m using NT Server 4 with apache 1.3. []s Ariadne -- 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] -- 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]
Re: [PHP-DB] Re: Image Bank with php mysql!! ??!!
To take it a step further, I would suggest creating a php script downloadimage.php as follows: ?php //downloadimage.php //some code to verify requested download against customers details, remembering not to send any output to the browser (would upset the header() function) // header(Content-type: image/jpeg); $image_name=urldecode($requiredimage); header( Content-Description: JPEG Image from Freds Library ($image_name) ); header( Content-Disposition: attachment; filename=$image_name ); $the_image=/imagelibrary/highresolution/$image_name; readfile($the_image); ? Then link to that with a href=downloadimage.php?imagename=10293492.jpgcustomerid=11217Download Image/a Make sure the images are outside of your webservers document root. Then you can programatically decide whether to serve the image to the user. Removes the possibility of leeching the entire library.. Tony -- Tony McCrory IT, Trinity Mirror group (Ireland) (028) 9068 0168 [EMAIL PROTECTED] Hugh Bothwell hugh_bothwell@hoTo: [EMAIL PROTECTED] tmail.com cc: Subject: [PHP-DB] Re: Image Bank with php mysql!! ??!! 07/30/2001 02:14 PM Koutsogiannopoulos Karolos [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I wan't your opinion regarding a program i am constructing. It is about an image bank for graphic studios or anyone else who needs a program that can insert his pictures or descriptions of it and be able to make a fast and reliable search. http://www.hotscripts.com/PHP/Scripts_and_Programs/Image_Galleries/more2.htm l What would be best.? Inserting the pictures in the DB or in the file system I would keep the image name in the database; save the original image in a dedicated directory with a numeric name like 4920185.jpg and then the thumbnail either with a suffix (like 4920185b.jpg) or with the same name in a thumbnails directory; store the number as the picture ID. The numbers should be large and pseudorandom, ie you don't want to make it easy for someone to write a script to leech all your pictures (it'll still be possible, but harder). Maybe use the database ID to seed a random number generator, and use the Nth generated number? Have a text field for keywords, a text field for description, a price field?, the author's name, email, and homepage. If you put the image directory below the web root directory, then the images can only be accessed via a script, and you can restore the original name at download (this solves the problem of name collisions). If you're going to sell the pictures, you should allow a buyer more than one download, in case they screw up or accidentally delete it on their system; I would have a table for 'current purchases' with a buyer ID and picture ID, and allow them something like access for a week, up to 20 downloads of that image. ? Also notice that it is a reference db and not an actual bank. I mean that there will be only photos of low quality and not the full picture which could be very large. (shrug) as above, keep the images in the filesystem; then it doesn't matter how large it is. -- 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] IMPORTANT NOTICE The information in this e-mail is confidential and should only be read by those persons to whom it is addressed and is not intended to be relied upon by any person without subsequent written confirmation of its contents. Furthermore, the content of this e-mail is the personal view of the sender and does not represent the advice, views or opinion of our company. Accordingly, our company disclaim all responsibility and accept no liability (including in negligence) for the consequences of any person acting, or refraining from acting, on such information prior to the receipt by those persons of subsequent written confirmation. In particular (but not by way of limitation) our company disclaims all responsibility and accepts no
[PHP-DB] RE: Tree Display
Have your table set up with the following two fields (among others!): -Category -Parent Result1 parent = 0 (it has no parent). Item1 parent is Result1 (ie the parent field of item1 is result1's unique identifier). Item2 parent is Result1. Result2 parent = 0. Result2's children have Result2 as their parent. Now you have to write the PHP code so that the corrent items are display. If no results are clicked then display all results in a column. Result1's URL would be something like samepage.php3?result=result1. Have PHP code that checks if you have a value for Result then displays that result's items . . . and if you have a value for result1 and item1 then display the items for result1 and the content for item1. I use a tree display for website menus and submenus. And I've created an entire administrative interface that makes it really easy to move sections of the trees around. It makes it really easy when people come to me at the last minute and want me to rearrange some part of the main menu, or add a whole section. You should be able to use this from other tree structures like product lists, etc etc. Hope this makes sense . . . Rita. -Original Message- From: Sharif Islam [mailto:[EMAIL PROTECTED]] Sent: Friday, July 27, 2001 2:13 PM To: [EMAIL PROTECTED] Subject: Tree Display I am still trying to figure out how to do it. I want to display my result from the data base in tree/menu like system. So the result page will look like this: +Result1 (you click this , it will expand) -Item1 (you click this and get more info) -Item2 -Item3 +Result2 -Item1 -Item2 All the records will be pulled from the same table. Any hint? -- 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]
[PHP-DB] new lines in text field
Greetings. I'm trying to concatenate several values entered in my form, then write them to a TEXT field in my mySQL database. All is well, except that I want to each value to appear on a different line. I've tried \n and that didn't work. What am I doing wrong? Thanks in advance for the help. Kenn -- 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]
[PHP-DB] PHP on Linux and MS SQL
Anybody know of a good installation description for connection PHP on linux to MS SQL on a 2000 machine? I've been through about 3 so far and none of them have worked. Adam -- 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]
[PHP-DB] Re: Query displays one
Hi Hugh,
Hugh Bothwell wrote:
Mike Gifford [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
The problem is that I am only getting one response from this query rather
than
5. I can't figure this out. What did I do to only get one response?
Try removing the ORDER BY and LIMIT and run the query by
hand - how many responses do you get? The search may only
return one article.
I tried removing these and I got a similarly disappoinging response (one reply),
so I decided to look again for code that is actually working and copy it from
scratch. I hit upon this and it is working now.
Thanks for your help.
function latest_articles() {
global $id,$main_file,$front_end_url,$alt, $articlestable, $categorytable;
$alt_var=;
$contentquery4 = mysql_query(SELECT a.*,s.URLname FROM $articlestable
a,
$categorytable s WHERE a.articleSectionID = s.articleSectionID ORDER BY a.date
DESC) or mysql_die();
if ($result = mysql_num_rows($contentquery4)) {
while ($contentarray4 = mysql_fetch_array($contentquery4)) {
$category = $contentarray4[URLname];
if ($alt) {
if ($contentarray4[alt_title]) {
$link_name =
$contentarray4[alt_title];
$alt_var = ?alt=french;
} else {
$link_name = $contentarray4[title];
$alt_var = ?alt=french;
}
} else {
$link_name = $contentarray4[title];
$alt_var = ;
}
$content .= \nlia
href=\.$front_end_url./.$main_file./.urlencode($category)./.$contentarray4[articleID]./
. $alt_var .\ class=\articleslink\.stripslashes($link_name)./a;
$i++;
if ($i==5) {
break;
}
}
}
return $content;
}
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[PHP-DB] Re: new lines in text field
Kenn Murrah [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I'm trying to concatenate several values entered in my form, then write them to a TEXT field in my mySQL database. All is well, except that I want to each value to appear on a different line. I've tried \n and that didn't work. What am I doing wrong? You skipped a step here, between write it to my database and appear on a different line. I assume that you insert to the database, then pull it out and display it as HTML. '\n' doesn't cause a line-break in HTML. You can either use br instead of \n before putting it in the database (less portable), or look at the nl2br() function (I would prefer this option, personally). -- 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]
[PHP-DB] Re: mysql_fetch_array
try just changing mysql_fetch_array to mysql_fetch_assoc
Bryan [EMAIL PROTECTED] wrote in message
006801c1192c$649f6e30$272478cc@bryanspc">news:006801c1192c$649f6e30$272478cc@bryanspc...
mysql_fetch_array places 2 records for each one returned (example)
Array
(
[0] = 56
[location_id] = 56
)
Is there any way to delimate that and have the key present only once?
here is my code :
/*
*
*
* Start code
*
*
*/
while ($aRow = mysql_fetch_array($sResult) ) {
print_r($aRow);
foreach ($aRow as $key = $sValue) {
$sData[$key] = SELECT * FROM location WHERE region='$sValue' AND region
!='yes';
$sDataResult[$key] = mysql_query($sData[$key], $sDbcnx) or
die(mysql_error());
/* print Main heading linking to no where */
print OPTION VALUE=\#\.$sValue./OPTION;
while($aData[$key] = mysql_fetch_array($sDataResult[$key]) ) {
/* grab the data from the main heading and print all the info for that
record */
print
OPTIONVALUE=\indexes/index.php?iLoc_id=.trim($aData[$key]['location_
id']).\nbsp;nbsp;nbsp;.trim($aData[$key]['location'])./OPTION\
n;
} /* end while */
} /* end foreach */
} /* end main while */
/*
*
*
* End Code
*
*
*/
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[PHP-DB] INSERT with a JOIN
I'm trying to create an INSERT statement that will change a field in one
table based upon the id defined in another table.
Here is what I tried:
$sql1 = INSERT INTO $table_name1 (FirstName,LastName,WorkPhone,
HomePhone,EmailName,Birthday)
VALUES
('[$FirstName]','[$LastName]','[$WorkPhone]','[$HomePhone]','[$EmailName]','
[Birthday]')
;
$sql2 = INSERT INTO $table_name2 (CompanyName,WebSite)
VALUES
('[$CompanyName]','[$WebSite]')
WHERE contacts ON company.CompanyID =
contacts.CompanyID WHERE ContactID='$ContactID'
;
$result_1 = @mysql_query($sql1,$connection) or die(Couldn't execute
query.);
$result_2 = @mysql_query($sql2,$connection) or die(Couldn't execute
query.);
?
The goal is based upon a unique field call ContactID I want to be able to
update the CompanyName associated with the ContactID.
For example: ContactID=1 might have an associated CompanyName= Smith, Inc.
What I want to do is UPDATE/INSERT a new CompanyName and have that reflected
by changing the CompanyID associated with the ContactID.
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[PHP-DB] Application dev w/ PHP MySql
Hello I am having to learn php and mysql, at the sametime and with a time constraint. So I am hoping I can have a lending hand from all the coding gurus out there. Let my give you a little back ground. I the owner of an ISP tech support outsourcing company and am trying to develop a call note system where the techs input to mysql notes on the call the teched and then have our clients (isps) use and authenicated site to do a search view the browser on a particuliar user's call note history. At the present time I am working on the Insertion part of the php script. I have created a data base in mysql and have been able to populate the data base.When the tech hits the submit button and the data is inserted into the customer table, with cust_key, tech_key, cust_fname, cust_lname, cust_username, cust_timestamp columns and into table issues with issue_key, issue_type, issue_entry, issue_date columns. I have this part working great. There is one more table in need to populate when the tech submits a call note and this is labled cust_issue_association. The cust_issue_association table has the following columns cust_key and issue_key. This table is being use to tie in the tables for a search. My problem is I have no idea on how to populate this table. When a call note is submitted into the customer table the cust_key is auto_incremented and in the issues table the issue_key is auto incremented. So the million dollar question that is driving me crazy with is how do I get these two keys to be inserted into cust_issue_association table when the call note is submitted. I have attached what I have done so far (please excuse the readabiltiy and basicness of the code this is my first time doing this) I hope I have described this clearly enough. Of course, I will give created in the code and on the appropriate html page, where create is do for the people who help me. Thank you ahead of time for all your inputs. Bob -- 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]
[PHP-DB] INSERTING DATA into a MySQL DB
Hi All,
I am having a problem with Inserting Data from a generated form.
To explain this clearly the code is listed below. There are three pages and
a
table in the db.
When the sql statement is run, it ends at the INSERT clause.
Any help would be appreciated ...
---
TABLE
---
# phpMyAdmin MySQL-Dump
# version 2.2.0rc3
# http://phpwizard.net/phpMyAdmin/
# http://phpmyadmin.sourceforge.net/ (download page)
#
# Host: ww2
# Generation Time: July 31, 2001, 9:39 am
# Server version: 3.23.25
# PHP Version: 4.0.2
# Database : test
#
#
# Table structure for table 'server'
#
CREATE TABLE server (
server_type varchar(4) NOT NULL default '0',
server_name varchar(10) NOT NULL default '0',
server_intip varchar(15) NOT NULL default '0',
server_intport int(5) NOT NULL default '0',
server_protocol char(3) NOT NULL default ' '
) TYPE=MyISAM;
---
Page One
---
HTMLHEADTITLESelect the Number of Onsite Servers you need: Step
1/TITLE/HEAD
BODY
FORM method=POST action=create.servers.php3
PPlease Enter the Number of Onsite Servers your school needs./P
PNumber of Servers: INPUT type=text name=num_servers size=5/p
INPUT type=submit value=Display Servers
/FORM/BODY/HTML
---
Page Two
---
HTML
HEADTITLECreate a Database Table: Step 2/TITLE/HEAD
BODY
FORM method=POST action=create.servers.process.php
table cellspacing=5 cellpadding=5
tr
thServer Type/th
thServer Name/th
thInternal Server IP/th
thInternal Server Port/th
thUDP/TCP/th
/tr
?
for ($i = 0 ; $i $num_servers; $i++) {
echo
tr
td align=center
select name=\server_type[]\
option value=\float\web/option
option value=\float\mail/option
option value=\float\other/option
/select
/td
td align=center
input type=\text\ name=\server_name[]\ size=\20\
/td
td align=center
input type=\text\ name=\iserver_intip[]\ size=\20\
/td
td align=center
input type=\text\ name=\server_intport[]\ size=\5\
/td
td align=center
select name=\server_protocol[]\
option value=\float\TCP/option
option value=\float\UDP/option
/select
/td
/tr;
}
?
trtd align=center colspan=3INPUT type=submit value=Create
Table/td/tr
/table
/FORM
/BODY
/HTML
---
Page Three
---
HTML
HEAD
TITLEInserting Data/TITLE/HEAD
BODY
h1Inserting Data/h1
?
$sql = INSERT INTO SERVER VALUES (; for ($i = 0; $i
count($server_type); $i++) {
$sql .= $server_type[$i] $server_name[$i] $server_intip[$i]
$server_intport[$i];
if ($server_protocol[$i] != ) {
$sql .= ($server_protocol[$i]),;
} else {
$sql .= ,;
}
}
$sql = substr($sql, 0, -1);
$sql .= );
$connection = mysql_connect(localhost,username,password)
or die(Couldn't connect to server.);
$db = mysql_select_db(test, $connection)
or die(Couldn't select database.);
$sql_result = mysql_query($sql,$connection)
or die(Couldn't execute query.);
if (!$sql_result) {
echo PCouldn't insert data!;
} else {
echo Pdata inserted!;
}
?
/BODY
/HTML
_
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Re: [PHP-DB] INSERTING DATA into a MySQL DB
My last post may have been a little unclear, I have attached the pages. Thanks ... From: Christopher Trewin [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [PHP-DB] INSERTING DATA into a MySQL DB Date: Tue, 31 Jul 2001 09:48:23 +1000 Hi All, I am having a problem with Inserting Data from a generated form. To explain this clearly the code is listed below. There are three pages and a table in the db. When the sql statement is run, it ends at the INSERT clause. Any help would be appreciated ... _ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp # phpMyAdmin MySQL-Dump # version 2.2.0rc3 # http://phpwizard.net/phpMyAdmin/ # http://phpmyadmin.sourceforge.net/ (download page) # # Host: ww2 # Generation Time: July 31, 2001, 9:39 am # Server version: 3.23.25 # PHP Version: 4.0.2 # Database : test # # # Table structure for table 'server' # CREATE TABLE server ( server_type varchar(4) NOT NULL default '0', server_name varchar(10) NOT NULL default '0', server_intip varchar(15) NOT NULL default '0', server_intport int(5) NOT NULL default '0', server_protocol char(3) NOT NULL default ' ' ) TYPE=MyISAM; -- 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]
[PHP-DB] Can't connect to MS SQL
I have PHP running on Apache for win32. The mssql module is enabled. I am trying to connect to a database with the following code. ?php $hostname = 192.168.1.1:1433; $username = test; $password = smiley; $dbName = meyedev; $dbc = mssql_connect($hostname,$username,$password) or DIE(DATABASE FAILED TO RESPOND.); mssql_select_db($dbName) or DIE(Table unavailable); mssql_close($dbName); ? However I get the following error. Warning: MS SQL: Unable to connect to server: 192.168.1.1:1433 in c:\program files\apache group\apache\htdocs\meyedev\connect.php4 on line 11 DATABASE FAILED TO RESPOND. What am I doing wrong? Adam -- 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]
[PHP-DB] Linux, Oracle 8.1.5, configure: libclntsh.so: Don't use llseek; slpmprodstab undefined
I'm at wit's end. .../php-4.0.6/configure gives an error, and debug.log says libclntsh.so: don't use llseek, use lseek64, and slpmprodstab is undefined. I've looked everywhere. All of php.org. All of this newsgroup. The linux forum at Oracle TechNet. The one lead I found was a reference to an Oracle note 68151.1, which apparently no longer exists on their site. I see a gensyslib in $ORACLE_HOME/bin which apparently creates $ORACLE_HOME/lib/sysliblist. This latter contains one line: -ldl -lm -lpthread. I note that -plthread isn't in the command line producing the error in debug.log. Should it be? How do I get it there? TIA for any advice. (I'm not a C programmer, just a DBA trying to get something working) My debug.log is below. Rick -- Rick HickersonT(978)659-4901 _\|/_ Agilent Technologies IT Consulting Engineer (646)452-7668 /|\ Innovating the HP Way 3000 Minuteman Rd, MS 596 (877)868-9829 Andover, MA 01810IT | Information Technology CONFIGURE: './configure' '--with-mysql' '--with-apxs' '--with-pgsql' '--with-oci8' '--prefix=/usr' CC: gcc CFLAGS: -g -O2 CPPFLAGS:-DLINUX=22 -DUSE_HSREGEX -DUSE_EXPAT -DNO_DL_NEEDED -DSUPPORT_UTF8 CXX: CXXFLAGS: INCLUDES:-I/opt/php/php-4.0.6/@includedir@ -I$(top_builddir)/Zend -I/opt/php/php-4.0.6/ext/mysql/libmysql -I/opt/oracle/app/oracle/product/8.1.5/rdbms/demo -I/opt/oracle/app/oracle/product/8.1.5/network/public -I/opt/oracle/app/oracle/product/8.1.5/plsql/public LDFLAGS: -Wl,-rpath,/opt/oracle/app/oracle/product/8.1.5/lib -L/opt/oracle/app/oracle/product/8.1.5/lib -Wl,-rpath,/usr/lib/pgsql -L/usr/lib/pgsql LIBS: -lpq -lm -ldl -lcrypt -lresolv -lm -ldl -lnsl -lresolv -lclntsh DLIBS: SAPI: apache PHP_RPATHS: /opt/oracle/app/oracle/product/8.1.5/lib /usr/lib/pgsql uname -a: Linux tackroom.and.agilent.com 2.4.3-20mdk #1 Sun Apr 15 23:03:10 CEST 2001 i686 unknown gcc -o conftest -g -O2 -DLINUX=22 -DUSE_HSREGEX -DUSE_EXPAT -DNO_DL_NEEDED -DSUPPORT_UTF8 -Wl,-rpath,/opt/oracle/app/oracle/product/8.1.5/lib -L/opt/oracle/app/oracle/product/8.1.5/lib -Wl,-rpath,/usr/lib/pgsql -L/usr/lib/pgsql conftest.c -lpq -lm -ldl -lcrypt -lresolv -lm -ldl -lnsl -lresolv -lclntsh 15 /usr/lib/gcc-lib/i586-mandrake-linux-gnu/2.96/../../../libclntsh.so: the `llseek' function may be dangerous; use `lseek64' instead. /usr/lib/gcc-lib/i586-mandrake-linux-gnu/2.96/../../../libclntsh.so: undefined reference to `slpmprodstab'collect2: ld returned 1 exit status -- 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]
[PHP-DB] Can I UPDATE 2 tables with 1 SQL statement?
Heres what I'm trying to do: $data=addslashes(fread(fopen($form_data, rb), filesize($form_data))); //some lines of code edited out $result = mysql_query ( UPDATE bands, binary_data SET genreid='$genre', bandname='$name', bandesc='$description', bandurl='$url', bandemail='$email', bin_data='$data', filename='$form_data_name', filesize= '$form_data_size', filetype='$form_data_type' WHERE bandname='$name' AND bands.bin_id=binary_data.bin_id ); Thanks all you smart people. = Mark [EMAIL PROTECTED] __ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/ -- 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]
Re: [PHP-DB] Can't connect to MS SQL
Hi Adam, Host name should not include the port number. If you need to use a non standard port number you should create a client configuration (using client network utility) and then use the name of the configuration as the host name. - Frank I have PHP running on Apache for win32. The mssql module is enabled. I am trying to connect to a database with the following code. ?php $hostname = 192.168.1.1:1433; $username = test; $password = smiley; $dbName = meyedev; $dbc = mssql_connect($hostname,$username,$password) or DIE(DATABASE FAILED TO RESPOND.); mssql_select_db($dbName) or DIE(Table unavailable); mssql_close($dbName); ? However I get the following error. Warning: MS SQL: Unable to connect to server: 192.168.1.1:1433 in c:\program files\apache group\apache\htdocs\meyedev\connect.php4 on line 11 DATABASE FAILED TO RESPOND. What am I doing wrong? Adam -- 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] -- 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]
