Data is not submitted to the database and a message "ERROR" shows - php

Question

There are two php files one contains the form and the other contains the code to insert the form data to the table in a database.
This is the submit form code from the first file
<form name="form" action="insert_dataE.php" onSubmit="return
validation()" method="post" id="formServiceEntry">
<fieldset id="fieldSetServiceEntry">
<legend align="center">Please fill the form</legend>
<p class="FieldHeading"><i>Vehicle No(*): </i></p>
<input id="VehicleNoFieldArea" type="text" name="VehicleNoField"
size="6" maxlength="8"/>
<p class="FieldHeading"><i>Description(*):</i></p>
<textarea id="descriptionFieldArea" name="descriptionField"
rows="2" cols="20" size="15" maxlength="18"></textarea>
<p class="FieldHeading"><i>Total(*):</i></p>
<input id="totalFieldArea" name="totalField" type="text"
size="4" maxlength="4"/>
<p id="amountFieldHeading"><i>Bill(*):</i></p>
<input id="amountFieldArea" name="amountField" type="text"
size="3" maxlength="3" onKeyUp="balance();" />
<br/>
<div id="divisionRadioButton">
<h3 id="radioButtonHeading">Service(*):</h3>
Service
<input class="textFields" type="radio"
name="serviceSelection" value="service" checked />
<br/>
Wash
<input class="textFields" type="radio"
name="serviceSelection" value="wash" />
</div>
<p id="balanceFieldHeading"><i>Balance(*):</i></p>
<input id="balanceFieldArea" name="balanceField" type="text"
size="4" maxlength="4"/>
</fieldset>
<input class="btnsSE" type="submit" name="Button" value="Submit" />
<input class="btnsSE" type="reset" name="Button" value="Reset Form"/>
<input type="button" class="btnsSE" value="Back to the staff
interface" onClick="window.location='staffE.php';"/>
</form>
This is the insert data code from the second file
<?php
// Connects to your Database
$conn=mysql_connect("localhost", "webgeek1_service", "6defyu4642070") or
die(mysql_error());
mysql_select_db("webgeek1_software_order", $conn) or die(mysql_error());
$result = mysql_query("SELECT * FROM application", $conn);
$num_rows = mysql_num_rows($result);
$num_rows = $num_rows + 1;
$id= $num_rows;
$dateAndTime = date('y-m-d H:i:s',time());
$vehicleNo=mysql_real_escape_string($_POST['VehicleNoField']);
$description=mysql_real_escape_string($_POST['descriptionField']);
$amount=mysql_real_escape_string($_POST['amountField']);
$service=mysql_real_escape_string($_POST['serviceSelection']);
// Build an sql statment to add the query details
$sql="INSERT INTO `webgeek1_software_order`.`application`(`serialNo`,
`dateAndTime` , `vehicleNo` , `description` ,`amount`,`service`)
VALUES
('$id',
'$dateAndTime','$vehicleNo','$description','$amount','$service')";
$result = mysql_query($sql, $conn);
if($result)
{
echo "<p id='headingInsertData'>Service Station Web Application</p>";
echo "<p id='receiptHeading'>Receipt</p>";
echo "<div id='mainFieldsInsertData'>";
echo "Serial No: " . " " . $id;
echo "<br/>";
echo "Date and Time: " . " " . $dateAndTime;
echo "<br/>";
echo "Vehicle No: " . " " . $vehicleNo;
echo "<br/>";
echo "Description: " . " " . $description;
echo "<br/>";
echo "Amount: " . " " . $amount;
echo "<br/>";
echo "Service:" . " " . $service;
echo "<br/>";
echo "<br/>";
echo"Thanks for using our services";
echo "</div>";
echo "<div id='footerInsertData'>";
echo "<i>Developed by: Web Geeks - Information Technology (IT)
Company</i>";
echo "</div>";
echo "<div align='center'>";
echo "<input class='btns' type='button' value='Print'
onClick='javascript: window.print();'/>";
echo "<input type='button' class='btns' value='Back to the
Application' onClick='newDoc()'/>";
echo "</div>";
}
else
{
echo "ERROR";
}
// close connection
mysql_close($conn);
?>

Answer 1

The error you're having (Duplicate entry '51' for key 'PRIMARY') seems pretty logical since you're giving your specified ID to row instead of writing automatically. Moreover, you're using ID based off of the amount of rows there currently are. This leads to MySQL error with ID duplication.
To solve this issue:
Modify serialNo column and tick a checkbox on A_I column (AUTO_INCREMENT). This will make sure you will always have unique ID.
Remove entirely this part in your code:
$result = mysql_query("SELECT * FROM application", $conn);
$num_rows = mysql_num_rows($result);
$num_rows = $num_rows + 1;
$id= $num_rows;
Modify your query:
This is modified already
$sql="INSERT INTO `webgeek1_software_order`.`application` (`dateAndTime`, `vehicleNo`, `description`,`amount`, `service`) VALUES('$dateAndTime', '$vehicleNo', '$description', '$amount', '$service')";
I believe database should automatically set what next ID should come after inserting new data. This will prevent your from getting such errors as ID duplication because you're no longer inserting your own number.
A side note (but important): you should use mysqli or PDO statements because mysql extension is deprecated (and is even removed in PHP 7.0.0).

Answer 2

In the structure for my table in phpmyadmin, I needed to set the column 'serialNo' to AI (Autoincrement) and in the insert data code, I needed to comment the lines:
$num_rows = mysql_num_rows($result);
$num_rows = $num_rows + 1;
$id= $num_rows;
Similarly, I needed to remove the 'serialNo' from the insert query in the same file. At last I needed to comment the lines :
echo "Serial No: " . " " . $id;
echo "<br/>";
in the insert data code