Problem With mysql! - Codewalkers

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September 27th, 2003, 09:49 AM

mmlug

Join Date: Apr 2007

Posts: 4

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 version

HP Version 4.3.3RC2-dev
Mysql : MySQL 4.0.14

php Code:

Original - php Code

 
<?php
include"config.php";
$db = mysql_connect($db_host,$db_user,$db_pass);
mysql_select_db($db_name) or die(mysql_error());
 
//$id=1; //$_GET[id];
//print_r($_GET);
 
$id=$_GET[id];
 
$query="SELECT title,news FROM news WHERE id=$id";
//echo "<br>$query";
 
$result=mysql_query($query)or die(mysql_error());
       while($rows= mysql_fetch_array($result))
        {
            $title=$rows["title"];
            $news=$rows["new"];
  echo "<form name="edit_process.php" method="post"> action="edit_save.php?id=$id">
  
  <p>Title :
    <input type="text" name="title" value="$title">
  </p>
  <p>News :</p>
  <p>
    <textarea name="news" cols="40" rows="6">$news</textarea>
  </p>
  <p>
    <input type="submit" name="Submit" value="Save">
  </p>
</form>";
}
 
mysql_close($db);
?>

Best regards,
mmlug

September 27th, 2003, 12:02 PM

Blindeddie

Join Date: Apr 2007

Location: NJ - USA

Posts: 2,152

Time spent in forums: < 1 sec
Reputation Power: 5

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?

September 27th, 2003, 12:08 PM

CodeKadiya

Contributing User

Join Date: Apr 2007

Location: Colombo,Sri Lanka

Posts: 2,330

Time spent in forums: 8 h 50 m 36 sec
Reputation Power: 5

RE: Problem With mysql!

Try and use your SQL query like this.

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

Is this working????

September 28th, 2003, 02:29 AM

mdhall

Join Date: Apr 2007

Posts: 158

Time spent in forums: < 1 sec
Reputation Power: 3

RE: Problem With mysql!

Try this on the second line...

replace
include"config.php";
with this
include("config.php");

September 28th, 2003, 07:51 AM

mmlug

Join Date: Apr 2007

Posts: 4

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
echo "<br>$query";

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

September 28th, 2003, 07:57 AM

Ashkhan

Contributing User

Join Date: Apr 2007

Posts: 372

Time spent in forums: 1 h 24 m 57 sec
Reputation Power: 3

RE: Problem With mysql!

It seems that value of $_GET['id'] is not set. Do you get it from a form or from url?

September 28th, 2003, 08:13 AM

mmlug

Join Date: Apr 2007

Posts: 4

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
<form name="post news" method="post" action="postnews.php"> <p>Title : <input type="text" name="title"> <br> Author : <input type="text" name="author"> <br> News : <br> <textarea name="news" cols="40" rows="6"></textarea></p> <p><input type="submit" name="Submit" value="POST !"></p> </form>

 
<form name="post news" method="post" action="postnews.php">
  <p>Title : <input type="text" name="title">
  <br>
  Author : <input type="text" name="author">
    <br>
    News :
    <br>
    <textarea name="news" cols="40" rows="6"></textarea></p>
  <p><input type="submit" name="Submit" value="POST !"></p>
  </form>

postnews.php is like here : :

php Code:

Original - php Code
<?php include "config.php"; $db = mysql_connect($db_host,$db_user,$db_pass); mysql_select_db("$db_name") or die(mysql_error()); $title=$_POST[title]; $news=$_POST[news]; $author=$_POST[author]; $query = "INSERT INTO news (title,news,author,date)VALUES('$title','$news','$ author',now())"; mysql_query($query); echo "News item entered."; mysql_close($db); ?>

 
<?php
include "config.php";
 
$db = mysql_connect($db_host,$db_user,$db_pass);
mysql_select_db("$db_name") or  die(mysql_error());
$title=$_POST[title];
$news=$_POST[news];
$author=$_POST[author];
$query = "INSERT INTO news (title,news,author,date)VALUES('$title','$news','$  author',now())";
mysql_query($query);
echo "News item entered.";
mysql_close($db);
?>

My view.php is like here

php Code:

Original - php Code

 
mysql_select_db($db_name) or die(mysql_error());
 
$query ="SELECT title, news, author, date FROM news ORDER by id DESC LIMIT 5";
    $result = mysql_query($query);
    while($rows = mysql_fetch_array($result))
        {
            $title = $rows["title"];
            $news =$rows["news"];
            $author=$rows["author"];
            $date =$rows["date"];
echo "<TABLE><TR><TD>$title</TD></TR>
<TR><TD>$news</TD></TR><TR><TD>$author</TD></TR>
<TR><TD>Posted by $author on $date</TD></TR>    
</TABLE>";
            
}
mysql_close($db);
?>

best reagrds,
mmlug

September 28th, 2003, 08:29 AM

Ashkhan

Contributing User

Join Date: Apr 2007

Posts: 372

Time spent in forums: 1 h 24 m 57 sec
Reputation Power: 3

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!

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