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I created database(my_db) and table(persons) in which i used three fields FirstName,LastName and Age. I run below php script it shows error like:

( ! ) Notice: Undefined index: FirstName in C:\wamp\www\insert.php on line 5 Call Stack

Time Memory Function Location

1 0.0007 369800 {main}( ) ..\insert.php:0

( ! ) Notice: Undefined index: LastName in C:\wamp\www\insert.php on line 6 Call Stack

Time Memory Function Location

1 0.0007 369800 {main}( ) ..\insert.php:0

( ! ) Notice: Undefined index: Age in C:\wamp\www\insert.php on line 7 Call Stack

Time Memory Function Location

1 0.0007 369800 {main}( ) ..\insert.php:0 Records added to the database

BELOW IS THE PHP CODE: PLS HELP ME AND THANKS IN ADVANCE

  <html>
 <body>

   <?php
  $first=$_POST ['FirstName']; 
  $last=$_POST ['LastName'];
  $a=$_POST ['Age'];

  $user_name = "root";
  $password = "";
  $database = "my_db";
  $server = "127.0.0.1";
  $db_handle = mysql_connect($server, $user_name, $password);

  $db_found = mysql_select_db($database, $db_handle);

  if ($db_found) {

  $SQL = "INSERT INTO persons(FirstName, LastName, Age) VALUES     ('$first', '$last', '$a')";

     $result = mysql_query($SQL);

     mysql_close($db_handle);

     print "Records added to the database";
    }
    else 
      {
        print "Database NOT Found ";
     mysql_close($db_handle);
     }

     ?>


    AND HTML FORM IS BELOW:

     <html>
       <body>
        <form action="insert.php" mehtod="POST">
          First Name: <input type="text" name="FirstName">
          <br />
          Last Name:<input type="text" name="LastName">
          <br />
          Age:<input type="text" name="Age">
          <br />
         <input type="submit" value="ADD" />
       </form>  
      </body>
     </html>
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1
  • is that in the same page as the form? Commented Jan 30, 2012 at 16:57

5 Answers 5

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2

It is most likely your $_POST-variables are not set.

Try a...

var_dump($_POST);

...instead of...

$first=$_POST['FirstName']; 
$last=$_POST['LastName'];
$a=$_POST['Age'];

...to debug, if the values you're looking for are stored in there (get rid of the space between the $_POST and the identifier, too, please!).

Not your question, but if you're INSERTing those variables without sanitizing them first, that opens any possibilities for SQL injections. Sanitize them properly to counter this risk.

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Comments

1

You have used mehtod in

<Form mehtod="post">

instead of 

<Form method="post">
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2 Comments

@enb081, in this question user had wrote "mehtod" instead of "method"
I know, I am sorry. Good job suggested edit reviewers who accepted my first wrong edit and declined my attempt to fix it! @OldPro
0

Are you sure you are submitting your form with POST instead of GET??

Try to add dumps to see what you get

<?php
  var_dump($_POST);
  var_dump($_GET);

  //Anyway you can use $_REQUEST instead of $_POST or $_GET
  $first= $_REQUEST['FirstName']; 
  $last= $_REQUEST['LastName'];
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2 Comments

Always be careful with $_REQUEST as it is easy to overwrite the data with the COOKIE side of it. If you are going to use request in your script ensure that you do something like this: $_REQUEST = array_merge($_POST, $_GET);
+1 @DarkMantis. I agree, it was just to test if the variables were set in any of those arrays.
0

The "errors" your seeing are actually notices. They occur because you are trying to access array indices that don't exist.

When you try to get the values out of $_POST, they do not exist, and the notice is generated.

It is possible to turn these notices off, but you shouldn't. Instead, you should check to make sure the input is valid (AND SANITIZE IT FOR GOODNESS' SAKE!) using isset or empty.

The reason why the errors are being displayed is hard to guess, since you haven't included the HTML that is used to post the data. But most likely, you misnamed the variables or sent them via GET instead of POST.

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Comments

0

Try this: 

    if ($db_found) {

    $SQL = sprintf("
        INSERT INTO persons 
            (FirstName, LastName, Age) 
        VALUES ('%s', '%s', '%s')",
        mysql_real_escape_string($first), 
        mysql_real_escape_string($last), 
        mysql_real_escape_string($a));

    $result = mysql_query($SQL);
    if( $result == true ){
        print "Records added to the database";
    }

    mysql_close($db_handle);
}

The reason why it wasn't working is that you weren't calling your $result anywhere, you just declared it.

Also, I put your query into a sprintf() function which means that it is easier to see what data you are passing it.

Another thing, is that you want to check if the affected rows are > 0 as that will mean that something has been inserted.

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4 Comments

Thanks for your replay, i tried your code but it shows error like: Warning: mysql_affected_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\insert.php
Okay try the edited code, it will check whether mysql_query($SQL) was successful (true/false)
within if ($result == true) you either remove == true or you compare using ===. The way it is now makes no sense.
No it makes sense now? it will compare the two; If mysql_query() returns true/1/'true' it will execute the subsequent code. Whether it returns (boolean) true is neither here nor there as long as it doesn't return false/0/'false'

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