PHP Newbie, error

[Vultima]

Hi I am in process of doing a PHP script, my other one works fine, I put them both in below, however i get this error

[code]Error: 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 'order (Firstname, surname, telephone, email, title, description) VALUES ('Stef' at line 1

 

The code I am using for this section is

 

 

<?php

$con = mysql_connect("localhost","root","rhysrhys");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

 

mysql_select_db("hagidolls", $con);

 

$sql="INSERT INTO order (Firstname, surname, telephone, email, title, description)

VALUES

('$_POST[name]','$_POST[lname]','$_POST[tele]','$_POST[mail]','$_POST[title]','$_POST[descriptions]')";

 

 

 

if (!mysql_query($sql,$con))

  {

  die('Error: ' . mysql_error());

  }

echo "Order Has been Submitted, Being Redirected in 5 Seconds";

 

mysql_close($con);

?>

[/code]

 

The database tables I am using are idorder, Firstname, surname, telephone, email, title, description

[Vultima]

WOW I posted that terribly, heres what I meant..

 

Hi I am in process of doing a PHP script, my other one works fine, I put them both in below, however i get this error.

 

Error: 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 'order (Firstname, surname, telephone, email, title, description) VALUES ('Stef' at line 1

 

The code I am using for this section is

 

<?php
$con = mysql_connect("localhost","root","rhysrhys");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hagidolls", $con);

$sql="INSERT INTO order (Firstname, surname, telephone, email, title, description)
VALUES
('$_POST[name]','$_POST[lname]','$_POST[tele]','$_POST[mail]','$_POST[title]','$_POST[descriptions]')";



if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Order Has been Submitted, Being Redirected in 5 Seconds";

mysql_close($con);
?>

 

The database tables I am using are idorder, Firstname, surname, telephone, email, title, description

[EmperorJazzy]

Hey Vultima,

 

Try adding the following echo line to your code; it may help determine if the query string is a problem.

 

$sql="INSERT INTO order (Firstname, surname, telephone, email, title, description)
VALUES
('$_POST[name]','$_POST[lname]','$_POST[tele]','$_POST[mail]','$_POST[title]','$_POST[descriptions]')";

echo $sql;

 

I would comment out the execution and Die for the moment.

 

//if (!mysql_query($sql,$con))
// {
  //die('Error: ' . mysql_error());
  //}
//echo "Order Has been Submitted, Being Redirected in 5 Seconds";

 

Post your result.

[Vultima]

Hi I did as you said and got the result of

 

INSERT INTO order (Firstname, surname, telephone, email, title, description) VALUES ([redacted])

 

Edited June 27, 2018 by requinix

[Pikachu2000]

`order` is a MySQL reserved word. Either rename the field, or enclose `order` in `backticks` (not 'single quotes') whereever it is used in a query string.

[Vultima]

Hi that didn't seem to work, first i changed the name to orders, didn't work, then i added the ` to it and same result

 

INSERT INTO `orders` (Firstname, surname, telephone, email, title, description) VALUES ([redacted])

Edited June 27, 2018 by requinix

[Pikachu2000]

Was there an error returned? Post the current revision of the relevant code, please.

[Vultima]

     <?php
$con = mysql_connect("localhost","root","rhysrhys");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hagidolls", $con);

$sql="INSERT INTO `orders` (Firstname, surname, telephone, email, title, description)
VALUES
('$_POST[name]','$_POST[lname]','$_POST[tele]','$_POST[mail]','$_POST[title]','$_POST[descriptions]')";

echo $sql;


//if (!mysql_query($sql,$con))
// {
  //die('Error: ' . mysql_error());
  //}
//echo "Order Has been Submitted, Being Redirected in 5 Seconds";


mysql_close($con);
?>

 

Also this is page where the form is

 

     

<form action="admin/insertorder.php" method="post">
      <p>First Name: <input type="text" name="name" /><br />
      Last Name: <input type="text" name="lname" /><br />
      Telephone:  <input type="text" name="tele" /><br />
      Email:  <input type="text" name="mail" /><br />
      <br/>
      Artwork Title: <input type="text" name="title" /><br />
       <br/>
       
Description of what you would like to be made..<br>
<textarea type="text" onKeyPress="return taLimit(this)" onKeyUp="return taCount(this,'myCounter')" name="descriptions" rows=7 wrap="physical" cols=40>
</textarea>
<br><br>
You have <B><SPAN id=myCounter>2000</SPAN></B> characters remaining 
for your description...</font>
<br/>
<br/>
<input type="submit" />
</form>

[EmperorJazzy]

Ok; strange, but could you try renaming the table to tblOrders or alternatively, using [orders] for the table name.

 

[Pikachu2000]

You didn't uncomment the query execution, but try this and see if it works. If it does, then you can continue by escaping the data to help prevent SQL injection . . .

 

mysql_select_db("hagidolls", $con);

$sql="INSERT INTO `orders` (Firstname, surname, telephone, email, title, description)
VALUES
('$_POST[name]','$_POST[lname]','$_POST[tele]','$_POST[mail]','$_POST[title]','$_POST[descriptions]')";
if( !mysql_query($sql) ) {
// what to do if query fails to execute
echo "<br>Query: $sql<br>Failed with error: " . mysql_error() . '<br>';
} else {
// what to do if query runs . . . 
if( mysql_affected_rows() === 1 ) {
	// only one record is expected to be inserted
	echo "Query inserted one record.";
} else {
	// any number of rows other than one can be considered an error condition
	echo "Error: query ran, but inserted " . mysql_affected_rows() . " records.";
}
}

mysql_close($con);