Small insert problem

[emediastudios]

Can someone pick whats wrong with my code please

 

if($_SESSION[passed] == "y"){

$id=$_POST['id'];
$title=$_POST['title'];
$date=$_POST['date'];
$text=$_POST['text'];


$query = "INSERT INTO news id='NULL', title='$title', date='$date', text='$text'";
$result = mysql_query($query) or die(mysql_error());

header("Location: admin.php"); 
}
  else{ 
  echo "You Must Be Logged In To Do That";
  }

[kenrbnsn]

Can you give us a hint? What's it doing? What's it supposed to do?

 

A quick glance shows me that

<?php
query = "INSERT INTO news id='NULL', title='$title', date='$date', text='$text'";
?>

is incorrect, try

<?php
query = "INSERT INTO news SET id='NULL', title='$title', date='$date', text='$text'";
?>

 

 

Ken

[emediastudios]

Parse error: syntax error, unexpected '=' in C:\Program Files\Apache Group\Apache2\htdocs\wg\wgp\news_edit.php on line 24

line 24 ---------- query = "INSERT INTO news SET id='NULL', title='$title', date='$date', text='$text'";

[kenrbnsn]

And which is line 24? The code snippet above doesn't have 24 lines in it.

 

Ken

[emediastudios]

line 24 ---------- query = "INSERT INTO news SET id='NULL', title='$title', date='$date', text='$text'";

[emediastudios]

anyone?

Please!!

[kenrbnsn]

The snippet of code you posted has no syntax errors, please post more of your code.

 

Don't bump too often.

 

Ken

[emediastudios]

Stupid me, no $ in query

But now i get this problem

Out of range value adjusted for column 'id' at row 1

[kenrbnsn]

That's what happens when I post when I should be asleep.

 

Ken

[emediastudios]

<?php
include_once('includes/include.php');
$password = "*******";


if($_POST[password] == "" && $_SESSION[passed] != "y"){
  $content .= "<form method='post' action='admin.php'><input type='password' name='password' value='password'><input type=submit></form>";
}
else if($_POST[password] != ""){
  if($password == $_POST[password]){
   $_SESSION[passed] = "y";
  }
  else{
    echo "<script>window.location = 'admin.php';</script>";
  }
}
if($_SESSION[passed] == "y"){


$title=$_POST['title'];
$date=$_POST['date'];
$text=$_POST['text'];


$query = "INSERT INTO news SET id='', title='$title', date='$date', text='$text'";
$result = mysql_query($query) or die(mysql_error());

header("Location: admin.php"); 
}
  else{ 
  echo "You Must Be Logged In To Do That";
  }

?>

 

My form looks like this

 

<?php
$content .= '<form action="news_edit.php" method="post"><table width="500">
  <tr>
    <td colspan="2"><span class="bolder">Manage News</span></td>
    </tr>
  <tr>
    <td><span class="bolder">Title</span></td>
    <td>
      <input type="text" name="title" id="title" />
</td>
  </tr>
  <tr>
    <td><span class="bolder">Date</span></td>
    <td><input type="text" name="date" id="date" /></td>
  </tr>
  <tr>
    <td valign="top"><span class="bolder">News</span></td>
    <td>
      <textarea name="text" id="text" cols="45" rows="5"></textarea></td>
  </tr>
  <tr>
    <td><input name="id" type="hidden" value="id" /></td>
    <td><input type="reset" name="reset" id="reset" value="Reset" />
      <input type="submit" name="submit" id="submit" value="Submit" /></td>
  </tr>
</table>
</form>';
}
?>

 

 

(edited by kenrbnsn to add

tags)

[kenrbnsn]

If the field "id" in your table is of type "int" and set to "autoincrement", then you can just leave it out of the mysql insert statement:

<?php
$query = "INSERT INTO news SET title='$title', date='$date', text='$text'";
$result = mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());
?>

 

I also find it very helpful to include the query in the die() statement.

 

Ken