mysql insert query not working

Question

I have a table disease, which has the columns diid, diseasename and descrption, in a database called hcp.

I want to simply input from a HTML form; here is my code so far.

<html><head><title>Disease Inssert</title></head>
<body>
<form action="diseaselist.php" method="post">
Disease Name : <input type="text" name="txtDiseaseName" id="textbox" /><br />
Description : <input type="text" name="txtDiseaseDescription" id="textbox" /><br />
<input type="submit" name="btnDiseaseSubmit" id="button"/>

</form>
</body>
</html>

<?php
mysql_connect("localhost","root","");
mysql_select_db("hcp");

if($_REQUEST['btnDiseaseSubmit'])
{
$queryDiseaseInsert="INSERT INTO  disease (`diseasename` ,`description`) VALUES ('".$_REQUEST['txtDiseaseName']."', '".$_REQUEST['txtDiseaseDescription']."');";

$resultDI=mysql_query($queryDiseaseInsert) or die(mysql_error());
}
?>

After submitting, page is redirecting to the diseaselist.php, but the data is not saving in the database.

Can anyone see anything wrong with my query?

yes there is at least one thing wrong: your input data is not properly sanitized, see filter extension for more on php.net/filter — Hajo
– Hajo, Commented Apr 24, 2012 at 6:20

Stelian Matei · Accepted Answer · 2012-04-24 06:27:25Z

2

The form is submitting directly to the list <form action="diseaselist.php" method="post">. The code saving to the database is never executed.

You could create an intermediate page, let's say process.php containing the code for adding to the database:

<?php
mysql_connect("localhost","root","");
mysql_select_db("hcp");

if($_REQUEST['btnDiseaseSubmit'])
{
    $queryDiseaseInsert="INSERT INTO  disease (`diseasename` ,`description`) VALUES ('". mysql_real_escape_string($_REQUEST['txtDiseaseName'])."', '".mysql_real_escape_string($_REQUEST['txtDiseaseDescription'])."');";

   $resultDI=mysql_query($queryDiseaseInsert) or die(mysql_error());
   // insert is complete, redirect to list
   header("location: diseaselist.php");
} else {
   // the form was not submitted
   // redirect back to the form or show error message
}

?>

Changes:

added mysql_real_escape_string
after insert, send the header instruction to redirect the browser to the list

Now, in your current file, leave only the html code and update the action attribute to submit the form to the process.php script:

<html><head><title>Disease Inssert</title></head>
<body>
<form action="process.php" method="post">
Disease Name : <input type="text" name="txtDiseaseName" id="textbox" /><br />
Description : <input type="text" name="txtDiseaseDescription" id="textbox" /><br />
<input type="submit" name="btnDiseaseSubmit" id="button"/>

</form>
</body>
</html>

Changes:

updated the form action attribute to process.php

answered Apr 24, 2012 at 6:27

Stelian Matei

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2 Comments

Abdur Rahim Over a year ago

i have made the changes. but one new problem arised: " Cannot modify header information - headers already sent by (output started at C:\wamp\www\HCP\insertdisease.php:3) in C:\wamp\www\HCP\insertdisease.php on line 13"

Stelian Matei Over a year ago

The error means that there was something already printed (echoed) when the header function was called. Make sure the PHP code is right in the begining of the file and that you don't output anything before calling the header function

Shane · Accepted Answer · 2012-04-24 06:33:59Z

There are a few problems with your script. First of all is you are not checking for data before trying to insert it. The typical flow of an HTML form is like so:

Display form. User enters data and submits.
The data is sent to the server where it (should) check for all required data. If it is not displayed, the user (usually) receives the form with an error.
This goes on until the user gives in or submits a fully valid form.

Now having said the above, here in your form:

<?php

  $mysql = mysql_connect("localhost","root","");   // assign to variable
  mysql_select_db("hcp", $mysql);

  // Get into the habit of using $_POST and $_GET instead of $_REQUEST
  if(isset($_POST['btnDiseaseSubmit']) && !empty($_POST['btnDiseaseSubmit'])) 
  {  // Form was completely submitted
    $sql = "INSERT INTO  disease (`diseasename` ,`description`) VALUES ('".$_POST['txtDiseaseName']."', '".$_POST['txtDiseaseDescription']."');";  // No need for massive variable names

    $resultDI = mysql_query($sql, $mysql) or die(mysql_error());

    mysql_close($mysql);    // Close database connection - cleans things up
    echo "Successfolly ran database query!";
  } else {  // Form was not completely submitted, if at all - display form.
?>
<html>
  <head><title>Disease Inssert</title></head>
  <body>
    <!-- If action is not present, the form will submit to the current page. ie. If current page is a.php and we leave out action, the form will submit to a.php! -->
    <form method="post">
      Disease Name : <input type="text" name="txtDiseaseName" id="textbox" /><br />
      Description : <input type="text" name="txtDiseaseDescription" id="textbox" /><br />
      <input type="submit" name="btnDiseaseSubmit" id="button"/>
    </form>
  </body>
</html>
<?php
  }
?>

That should work fine for you. Please note the comments in the code so you know what I have done and why I have done it. Google around for more examples. It will all become clear to you soon.

Poonam · Accepted Answer · 2012-04-24 06:28:35Z

0

on diseaselist.php page try with following code

<?php

   mysql_connect("localhost", "root", "");
mysql_select_db("hcp");

if (isset ($_REQUEST['btnDiseaseSubmit'])) {
    $queryDiseaseInsert = "INSERT INTO disease ('diseasename' , 'description') VALUES('" . $_REQUEST['txtDiseaseName'] . "','" . $_REQUEST['txtDiseaseDescription'] . "')";

    $resultDI = mysql_query($queryDiseaseInsert) or die(mysql_error());
}
?>

answered Apr 24, 2012 at 6:28

Poonam

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3 Comments

Shane Over a year ago

Who is to say the form isnt also on diseaselist.php?

Poonam Over a year ago

decoyer didn't mentioned that the page is one and the same i.e. redirecting to same page diseaselist.php.If is it then also isset() code will handle the rest of the thing @Shane

Shane Over a year ago

Fair enough. In my comment I have assumed the code is in the same file, as that is what the OP seems to be doing.

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