CODEWITHSUNDEEP
phpmysqlsqldatabase
Hans
Hans

Reputation: 11

sql command INSERT INTO not working

I'm creating a website and I'm running into an error with inserting data from a form into my database (using phpMyAdmin). It is simply coming up with the self defined error 'Error', 'Sorry, your registration failed. Please go back and try again.'

I've looked at many other questions with the same issue but I cannot find an answer which corresponds to my problem. 

<?php include "base.php"; ?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">  

<div id="main">
 <?php

if(!empty($_SESSION['LoggedIn']) && !empty($_SESSION['Email']))
{
    $PlanSize = mysql_real_escape_string($_POST['PlanSize']);
    $DatePaid = mysql_real_escape_string($_POST['DatePaid']);
    $AmountPaid = mysql_real_escape_string($_POST['AmountPaid']);
    $MonthUsage = mysql_real_escape_string($_POST['MonthUsage']);
     
     	// $currentUser = mysql_query('SELECT * FROM User WHERE Email = "'. mysql_real_escape_string($_SESSION['Email']) . '"') or trigger_error(mysql_error());
     
     	$currentUser = ($_SESSION['Email']);
        $registerquery = mysql_query("INSERT INTO Plan (PlanSize, DatePaid, AmountPaid, MonthUsage, PlanUserID) VALUES('".$PlanSize."','".$DatePaid."','".$AmountPaid."', '".$MonthUsage."','".$currentUser."')");
        if($registerquery)
        {
            echo "<h1>Success</h1>";
            echo "<p>Your account was successfully created. Please <a href=\"index.php\">click here to login</a>.</p>";
        }
        else
        {
            echo "<h1>Error</h1>";
            echo "<p>Sorry, your registration failed. Please go back and try again.</p>";    
        }       
     }

    ?>
    <form method="post" action="Plans.php">
	<table class="bill">
	<input type="hidden" name="data" value="true" />
    <fieldset>
    	<p>
    	<select name="PlanSize" id="PlanSize" placeholder="Plan Size"/>
			<option value="Small">Small</option>
			<option value="Medium">Medium</option>
			<option value="Large">Large</option>
			<option value="Extra Large">Extra Large</option>
		</select><p>

    	<input type="text" name="DatePaid" id="DatePaid" placeholder="Date Paid (YYYY-MM-DD)"/><br />
        <input type="text" name="AmountPaid" id="AmountPaid" placeholder="Amount Paid"/><br />
      	<input type="text" name="MonthUsage" id="MonthUsage" placeholder="Data Used"/><br />
        <input type="submit" name="Submit" id="Submit" value="Submit" />
    </fieldset>
	</table>
    </form>
  
</center>

</div>
</body></html>

Note: my base.php document is used to connect to the database and is fully functional; I have implemented other database data insertions using the base.php so I know it is not the root of the problem.

Many thanks in advance for any help

Upvotes: 0

Views: 120

Answers (3)

volkinc
volkinc

Reputation: 2128

try it like this 

$registerquery = mysql_query("INSERT INTO `Plan` (`PlanSize`, `DatePaid`, `AmountPaid`, `MonthUsage`, `PlanUserID`) VALUES('$PlanSize','$DatePaid','$AmountPaid', '$MonthUsage','$currentUser')");

Upvotes: 0

DirtyBit
DirtyBit

Reputation: 16772

Give this a go:

$registerquery = mysql_query("INSERT INTO `Plan` (`PlanSize`, `DatePaid`, `AmountPaid`, `MonthUsage`, `PlanUserID`) 
 VALUES('$PlanSize', '$DatePaid', '$AmountPaid', '$MonthUsage','$currentUser')");

Moreover, Are you sure you're inserting the correct values to their respective columns? Is $currentUser really suppose to go inside PlanUserID column?

After the discussion in the comments:

Since you have the FK Constraint enable and it should work but let's check a few things:

  • The Parent and Child table columns should have same data types.
  • Make sure you don't have any values in your child table's column that do not exist in your parent table's column.
  • Both tables should have identical collation.
  • Both of your tables should be InnoDB.

You can do so by:

SHOW TABLE STATUS WHERE Name =  'table1';
ALTER TABLE table1 ENGINE=InnoDB;

And if all fails you can always take backup of your parent and child table first, then truncate child table and try to make a relation again.

You can always study more at the similar questions available here: MySQL - Cannot add or update a child row: a foreign key constraint fails

Upvotes: 1

RomanPerekhrest
RomanPerekhrest

Reputation: 92854

Looks like you haven't made check for existance of 'PlanSize' and others in $_POST array. Consider that you can just reload page for authenticated user without sending data via post method. So you get into your first condition, but indexes such as 'PlanSize' and other won't be created in POST array. Make check for isset beforehand:

$planSize = isset($_POST['PlanSize'])? $_POST['PlanSize'] : null;

Upvotes: 1

Related Questions