Database Help
 
Forums: » Register « |  User CP |  Games |  Calendar |  Members |  FAQs |  Sitemap |  Support | 
User Name:
Password:
Remember me
Go Back   Codewalkers Forums > Other Technologies > Database Help
Reload this Page

Problem With mysql!

Discuss Problem With mysql! in the Database Help forum on Codewalkers.
Problem With mysql! General SQL help. Help with MySQL, MSSQL, Oracle, mSQL, PostgreSQL, etc.

Reply
Add This Thread To:
  Del.icio.us   Digg   Google   Spurl   Blink   Furl   Simpy   Y! MyWeb 
« Previous Thread | Next Thread »  
Thread Tools Search this Thread Rate Thread Display Modes
 
Unread Codewalkers Forums Sponsor:
  #1  
Old September 27th, 2003, 09:49 AM
mmlug mmlug is offline
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 4 mmlug User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 0
Problem With mysql!
Dear all,
I have a big problem with my PHP news scripts.
I got the following error...
Please help me...

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
My PHP versionHP Version 4.3.3RC2-dev
Mysql : MySQL 4.0.14
php Code:
Original - php Code 
  1.  
  2. <?php
  3. include"config.php";
  4. $db = mysql_connect($db_host,$db_user,$db_pass);
  6.  
  7. //$id=1; //$_GET[id];
  8. //print_r($_GET);
  9.  
  10. $id=$_GET[id];
  11.  
  12. $query="SELECT title,news FROM news WHERE id=$id";
  13. //echo "<br>$query";
  14.  
  15. $result=mysql_query($query)or die(mysql_error());
  16.        while($rows= mysql_fetch_array($result))
  17.         {
  18.             $title=$rows["title"];
  19.             $news=$rows["new"];
  20.   echo "<form name="edit_process.php" method="post"> action="edit_save.php?id=$id">
  21.   
  22.   <p>Title :
  23.     <input type="text" name="title" value="$title">
  24.   </p>
  25.   <p>News :</p>
  26.   <p>
  27.     <textarea name="news" cols="40" rows="6">$news</textarea>
  28.   </p>
  29.   <p>
  30.     <input type="submit" name="Submit" value="Save">
  31.   </p>
  32. </form>";
  33. }
  34.  
  36. ?>


Best regards,
mmlug
Reply With Quote
  #2  
Old September 27th, 2003, 12:02 PM
Blindeddie Blindeddie is offline
Codewalkers Regular (2000 - 2499 posts)
  
Join Date: Apr 2007
Location: NJ - USA
Posts: 2,152 Blindeddie User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 4
RE: Problem With mysql!
if you uncomment the line that says

echo "<br>$query";

and run the script, what is the the query that is output?
Reply With Quote
  #3  
Old September 27th, 2003, 12:08 PM
CodeKadiya CodeKadiya is offline
Codewalkers Regular (2000 - 2499 posts)
  
Join Date: Apr 2007
Location: Colombo,Sri Lanka
Posts: 2,313 CodeKadiya User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 4
Send a message via Yahoo to CodeKadiya
RE: Problem With mysql!
Try and use your SQL query like this.

$query="SELECT title,news FROM news WHERE id='$id'";

Is this working????
Reply With Quote
  #4  
Old September 28th, 2003, 02:29 AM
mdhall mdhall is offline
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 158 mdhall User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 2
RE: Problem With mysql!
Try this on the second line...

replace
include"config.php";
with this
include("config.php");
Reply With Quote
  #5  
Old September 28th, 2003, 07:51 AM
mmlug mmlug is offline
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 4 mmlug User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 0
RE: RE: Problem With mysql!
Quote:
if you uncomment the line that says

php Code:
Original - php Code 
  1. echo "<br>$query";

and run the script, what is the the query that is output?


Yes, I got this error message ..
Code:
SELECT title,news FROM news WHERE id= You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

best reagards,
MMlug
Reply With Quote
  #6  
Old September 28th, 2003, 07:57 AM
Ashkhan Ashkhan is offline
Contributing User
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 372 Ashkhan User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: 1 h 24 m 57 sec
Reputation Power: 2
RE: Problem With mysql!
It seems that value of $_GET['id'] is not set. Do you get it from a form or from url?
Reply With Quote
  #7  
Old September 28th, 2003, 08:13 AM
mmlug mmlug is offline
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 4 mmlug User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: < 1 sec
Reputation Power: 0
RE: Problem With mysql!
Yes I hvae one form to post my news to myssql database. Could be here something wrong?
php Code:
Original - php Code 
  1.  
  2. <form name="post news" method="post" action="postnews.php">
  3.   <p>Title : <input type="text" name="title">
  4.   <br>
  5.   Author : <input type="text" name="author">
  6.     <br>
  7.     News :
  8.     <br>
  9.     <textarea name="news" cols="40" rows="6"></textarea></p>
  10.   <p><input type="submit" name="Submit" value="POST !"></p>
  11.   </form>

postnews.php is like here : :
php Code:
Original - php Code 
  1.  
  2. <?php
  3. include "config.php";
  4.  
  5. $db = mysql_connect($db_host,$db_user,$db_pass);
  6. mysql_select_db("$db_name") or  die(mysql_error());
  7. $title=$_POST[title];
  8. $news=$_POST[news];
  9. $author=$_POST[author];
  10. $query = "INSERT INTO news (title,news,author,date)VALUES('$title','$news','$  author',now())";
  11. mysql_query($query);
  12. echo "News item entered.";
  14. ?>


My view.php is like here
php Code:
Original - php Code 
  1.  
  3.  
  4. $query ="SELECT title, news, author, date FROM news ORDER by id DESC LIMIT 5";
  5.     $result = mysql_query($query);
  6.     while($rows = mysql_fetch_array($result))
  7.         {
  8.             $title = $rows["title"];
  9.             $news =$rows["news"];
  10.             $author=$rows["author"];
  11.             $date =$rows["date"];
  12. echo "<TABLE><TR><TD>$title</TD></TR>
  13. <TR><TD>$news</TD></TR><TR><TD>$author</TD></TR>
  14. <TR><TD>Posted by $author on $date</TD></TR>    
  15. </TABLE>";
  16.             
  17. }
  19. ?>

best reagrds,
mmlug
Reply With Quote
  #8  
Old September 28th, 2003, 08:29 AM
Ashkhan Ashkhan is offline
Contributing User
Codewalkers Newbie (0 - 499 posts)
  
Join Date: Apr 2007
Posts: 372 Ashkhan User rank is Just a Lowly Private (1 - 20 Reputation Level) 
Time spent in forums: 1 h 24 m 57 sec
Reputation Power: 2
RE: Problem With mysql!
Your problem lies in your form method.
The method is post, so that your $id variable will be stored in $_POST['id'] and not in $_GET['id'].

btw: I don't see any element with id name in your form. You are not setting it at all!
Reply With Quote
Reply

Viewing: Codewalkers Forums > Other Technologies > Database Help > Problem With mysql!

« Previous Thread | Next Thread »

Thread Tools  Search this Thread 
Search this Thread:

Advanced Search
Display Modes  Rate This Thread 
Linear Mode Linear Mode
Rate This Thread:


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
View Your Warnings | New Posts | Latest News | Latest Threads | Shoutbox
Forum Jump


Forums: » Register « |  User CP |  Games |  Calendar |  Members |  FAQs |  Sitemap |  Support |