Form validation in php getting Notice?

Question

When I click the submit button from the form without entering any values, it displays the message Please fill all the fields and error for the query. But it also generates the following notice:

Notice: Undefined variable: sql in C:\XAMPP\htdocs\statistics\savecourseadminscores.php on line 31

Is this normal?

saveITScores.php

 <?php

 define('DB_NAME','some');//false name
 define('DB_USER','some');//false name
 define('DB_PASSWORD','some');//false name
 define('DB_HOST','localhost');

 $connect = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);

 if(!$connect){
      die('Could not connect:'.mysql_error());
 }

 $db_selected=mysql_select_db(DB_NAME,$connect);

 if(!$db_selected){
   die('Can\'t use'.DB_NAME.':'.mysql_error());
 }

$value1=$_POST['s1'];
$value2=$_POST['s2'];
$value3=$_POST['year'];

if(isset($_POST['submit'])){

    if($value1 && $value2 && $value3){
       $sql="INSERT INTO itservice(s1,s2,year)    VALUES('$value1','$value2','$value3')";
    }
    else{
       echo "Please fill all the fields.";
    }
}
  if(!mysql_query($sql)){
     die('Error:'.mysql_error());

  }

  mysql_close();

  ?>

saveForm.php

  <div id="form">

<form action="saveITScores.php" method="POST">
S1: <input class="inputfield" type="text" name="s1" size="5"> <br /><br/>
S2: <input class="inputfield" type="text" name="s2" size="5"> <br /><br/>
<b>Year:<b/>
<select name="year"> 
<option value="Choose">Please select..</option>
<option value="2005">2005</option> 
<option value="2006">2006</option>
<option value="2007">2007</option>
<option value="2008">2008</option>
<option value="2010">2010</option>
<option value="2011">2011</option>
<option value="2012">2012</option></select><br/><br/>

<br/>
<input type="submit" value="Save" name="submit">
<input type="reset" name="reset" value="Clear"><br /><br/>

Can you state in your code at which line you seem that notice will generate. And please use mysqli_* or PDOinstead of mysql_* because it is officially deprecated noe. — Alive to die - Anant, May 02 '15 at 14:17
`if(!mysql_query($sql)){` <- Look at this if statement and think about it, when you get the output: `Please fill all the fields` when and where the variable `$sql` gets defined — Rizier123, May 02 '15 at 14:18
@ anant kumar singh At line 31 in `saveITScores.php` I get this `Notice`. — aurora, May 02 '15 at 14:21

score 0 · Accepted Answer · answered May 02 '15 at 14:20

0

<?php 

if(isset($_POST['submit'])){
    $value1=$_POST['s1'];
    $value2=$_POST['s2'];
    $value3=$_POST['year'];

    if(!empty($value1) && !empty($value2) && !empty($value3))
    {
       $sql=mysql_query("INSERT INTO `itservice`(`s1`,`s2`,`year`)VALUES('".$value1."','".$value2."','".$value3."')")or die(mysql_error());
    }
    else{
       echo "Please fill all the fields.";
    }
}

     ?>

answered May 02 '15 at 14:20

Vivek Singh

2,453
1
14
27

1

1. And what should OP do with this code? 2. That won't fix OP's problem! – Rizier123 May 02 '15 at 14:22
@Rizier123 why not? just corrected the code!! – Vivek Singh May 02 '15 at 14:25
And what did you corrected? Why did you corrected it, what was wrong with OP's code? – Rizier123 May 02 '15 at 14:25
@Vicky thanks a lot it works fine, I don't get the `notice` anymore. – aurora May 02 '15 at 14:28
@Rizier123 i think the op's problem has solved!! – Vivek Singh May 02 '15 at 14:28
@Vicky Still I think an explanation for future SO visitors would be helpful. – Rizier123 May 02 '15 at 14:31
1

ok u were talking about the explaination – Vivek Singh May 02 '15 at 14:32

Form validation in php getting Notice?

1 Answers1