PHP URL parameter error is not taking

Question

I'm getting an error " Undefined index: tableName in C:\xampp\htdocs\online test\study_question1.php on line 8"

My URL is :"http://localhost/online%20test/study_question1.php?tableName=computer_science"

This is PHP code for below html code

    <?php
        $connection = mysqli_connect('localhost','root','','userquestion') or die(mysqli_error($connection));

        if(isset($_POST['next'])){
            $table=$_POST['table_name'];
            $query= "CREATE TABLE `userquestion`.`$table` ( `id` INT(15) NOT NULL AUTO_INCREMENT , `question` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option1` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option2` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option3` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option4` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `true_ans` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , PRIMARY KEY (`id`)) ENGINE = InnoDB;";
           $data = mysqli_query($connection,$query);
           header('Location: study_question1.php?table_name=' . $table);

        }


?>

HTML code from where I,m taking table_name

<html> 

<link rel="stylesheet" type="text/css" href="style.css">
<head> 
<title>Online</title> 
</head> 
<body id="body-color"> 


<div id="Sign-Up"> 
<fieldset style="width:30%">
<legend>Online</legend> 
<table border="0"> 
<tr> 
<form method="POST" action="online.php">
 <tr> 
 <td>table</td>
 <td> <input type="text" name="table_name"></td> 
 </td>

 </tr> 
 <tr> 
 <td><input id="button" type="submit" name="next" value="next"></td> 
 </tr> 

 </form> 
 </table> 
 </fieldset> 
 </div>

 </body>

My PHP code is

 <?php
$connection = mysqli_connect('localhost','root','','userquestion') or die(mysqli_error($connection));
 if(isset($_POST['next'])) 
{
$table = $_GET['table_name'];
 $question = $_POST["question"];
 $option1 = $_POST['option1'];
 $option2 = $_POST['option2'];
 $option3 = $_POST['option3']; 
  $option4 = $_POST['option4'];
   $true_ans = $_POST['true_ans'];
 $query = "INSERT INTO `$table` (question,option1,option2,option3,option4,true_ans) VALUES ('$question','$option1','$option2','$option3','$option4','$true_ans')"; 
 $data = mysqli_query($connection,$query);

 if($data) 
 {
 echo "ab dusra dal..."; 
 }
 } 

 ?>

can you post your html tag which has `tableName` id. Looks like you are not having the id "tableName" in your html tag. — Punith R Kashi, Feb 28 '17 at 17:11
I'm getting the error on line number 5 and the whole PHP code is written in the file study_question1.php — Shubham Gosewade, Feb 28 '17 at 17:11
@ShubhamGosewade: The HTML code is vital to the answer. Most likely there is a typo in "tableName" but only the HTML code will tell for sure. — geeves, Feb 28 '17 at 17:13
@ShubhamGosewade, please put the code in your question. Edit the question. — Jonathan M, Feb 28 '17 at 17:13
@ShubhamGosewade: There is an error with the submit element on the form. It should have a name of "submit" not "next". It needs to match the php $_POST['submit']. — geeves, Feb 28 '17 at 17:21

score 0 · Answer 1 · answered Feb 28 '17 at 19:15

This should help. GET and POST come from different places. GET is from the URL and POST is from the form. The code below has all unnecessary code removed so the answer is more transparent.

<?php
if (isset($_POST['next'])) {
    $table = $_POST['table_name'];
    echo "You picked: " . $table . " in the form";
} else if (isset($_GET['table_name'])) {
    $table = $_GET['table_name'];
    echo "You picked: $table in the url";
} else {
    echo "first call to page";  // you can delete this line after testing    
}
?>
<html> 
    <head> 
        <title>Online</title> 
    </head> 
    <body id="body-color"> 
        <div id="Sign-Up"> 
            <fieldset style="width:30%">
                <legend>Online</legend> 
                <form method="POST" action="online.php">
                    <table border="0"> 
                        <tr> 
                        <tr> 
                            <td>table</td>
                            <td> <input type="text" name="table_name"></td> 

                        </tr> 
                        <tr> 
                            <td>&nbsp;</td>
                            <td><input id="button" type="submit" name="next" value="next"></td> 
                        </tr> 

                    </table> 
                </form> 
            </fieldset> 
        </div>
    </body>
</html>

This is the URL for your question above.

http://localhost/online%20test/study_question1.php??table_name=fred

Getting same error. please someone help me if you want the whole file of this code the send me your E-mail id on shubham1835@gmail.com and I will send you the whole file.. — Shubham Gosewade, Mar 01 '17 at 13:30

PHP URL parameter error is not taking

1 Answers1