PHP form is not returning data

Question

Is there any reason why this is not working?

<form method="post" action="CorseProject.php">
    <table width="400" border="0" cellspacing="1" cellpadding="2">
        <tr>
            <td width="100">EmployeeID</td>
            <td><input name="employeeid" type="text" id="employeeid" required oninvalid="this.setCustomValidity('You must enter an employee id')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">ManagementID</td>
            <td><input name="managementid" type="text" id="managementid" required oninvalid="this.setCustomValidity('You must enter a management id')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">First Name</td>
            <td><input name="first_name" type="text" id="first_name" required oninvalid="this.setCustomValidity('You must enter a first name')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Last Name</td>
            <td><input name="last_name" type="text" id="last_name" required oninvalid="this.setCustomValidity('You must enter a last name')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Age</td>
            <td><input name="age" type="text" id="age" required oninvalid="this.setCustomValidity('You must enter an age)" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Employment Period in Months</td>
            <td><input name="employment_period" type="text" id="employment_period" required oninvalid="this.setCustomValidity('You must enter an employment period')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100"> </td>
            <td><input name="find" type="submit" id="add" value="Insert New Record"></td>
        </tr>
    </table>
</form>

My config and opendb file below are working fine, so they are not causing this error.

<?php 
    include 'mod2_config.php';
    include 'mod2_opendb.php';
    $employeeid = (isset($_POST['employeeid'])? $_POST['employeeid']:'');
    $managementid = (isset($_POST['managementid']) ? $_POST['managementid']: '');
    $fname= (isset($_POST['first_name'])    ? $_POST['first_name']   : '');
    $lname= (isset($_POST['last_name'])    ? $_POST['last_name']   : '');
    $age= (isset($_POST['age'])    ? $_POST['age']   : '');
    $employment_period= (isset($_POST['employment_period'])    ? $_POST['employment_period']   : ''); 
    $sql= " INSERT INTO employees (employeeid, managementid, first_name,last_name, age, employement_period)
    VALUES ('$employeeid','$managementid',$fname','$lname','$age', '$employment_period')";
    $result = mysqli_query($con, $sql);
    $sql= " SELECT employeeid, managementid, first_name, last_name, age,employment_period from employees WHERE employeeid = $employeeid LIMIT 1";
    $result = mysqli_query($con, $sql);
    if (mysqli_num_rows($result) > 0) {
        // output data of each row
        while($row = mysqli_fetch_assoc($result)) {
            echo "<b>Record successfully inserted:</b><br>";
            echo "<b>EmployeeID: " . $row["employeeid"]. "</b><br>";
            echo "<b>ManagementID: " . $row["managementid"]. "</b><br>";
            echo "<b>Name: " . $row["first_name"]. " " . $row["last_name"]. "</b><br>";
            echo "<b>Age: " . $row["age"]. "</b><br>";
            echo "<b>Employment Period In Months: " . $row["employment_period"]. "</b><br>";
        }
    }
    else {
        echo "Sorry there are no matches! Please check your entry and try again.";
    }

    //The else statement above keeps running instead of the form inserting information into the database and showing the results.
    mysqli_close($con);
?>

I have tried this insert script for several hours and I still can't get it to insert into the database.

You need me to print the value of my variables? This is a very basic php script pulling information from a database with an html form. If you do not know the error it is okay. — CodingTime
– CodingTime, Commented Apr 14, 2018 at 3:56

Death-is-the-real-truth · Accepted Answer · 2018-04-14 06:01:20Z

1

Try this.

<?php 
include 'mod2_config.php';
include 'mod2_opendb.php';


$employeeid= (isset($_POST['employeeid'])    ? $_POST['employeeid']   : '');
$managementid= (isset($_POST['managementid'])    ? $_POST['managementid']   : '');
$fname= (isset($_POST['first_name'])    ? $_POST['first_name']   : '');
$lname= (isset($_POST['last_name'])    ? $_POST['last_name']   : '');
$age= (isset($_POST['age'])    ? $_POST['age']   : '');
$employment_period= (isset($_POST['employment_period'])    ? $_POST['employment_period']   : ''); 


$sql_insert = " INSERT INTO employees (employeeid, managementid, first_name, last_name, age, employement_period)
                    VALUES ('$employeeid','$managementid','$fname','$lname','$age', '$employment_period')";
$result_insert = mysqli_query($con, $sql_insert);

$sql_fetch = " SELECT * from employees WHERE employeeid = $employeeid LIMIT 1";

$result_fetch = mysqli_query($con, $sql_fetch);

if (mysqli_num_rows($result_fetch) > 0) {
  // output data of each row
  while($row = mysqli_fetch_assoc($result_fetch)) {
    echo "<b>Record successfully inserted:</b><br>";
    echo "<b>EmployeeID: " . $row["employeeid"]. "</b><br>";
    echo "<b>ManagementID: " . $row["managementid"]. "</b><br>";
    echo "<b>Name: " . $row["first_name"]. " " . $row["last_name"]. "</b><br>";
    echo "<b>Age: " . $row["age"]. "</b><br>";
    echo "<b>Employment Period In Months: " . $row["employment_period"]. "</b><br>";
  }
}   
    else {
    echo "Sorry there are no matches! Please check your entry and try again.";
}

edited Apr 14, 2018 at 6:01

Death-is-the-real-truth

72.3k10 gold badges59 silver badges105 bronze badges

answered Apr 14, 2018 at 4:01

Ricky Singh

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3 Comments

Death-is-the-real-truth Over a year ago

Why try? Any explanation? Always add an explanation about what changes you made so that code will work properly.

Death-is-the-real-truth Over a year ago

You need to add explanation now in your current answer itself.

degr Over a year ago

I'm sure that explanation no need, because @CodingTime have no idea about code. He got that sample somewhere, he have no idea how it works, and he need decision here and now and for free /thread.

Gyandeep Sharma · Accepted Answer · 2018-04-14 06:51:25Z

0

The error is in your insert query at $fname

Replace $fname' to '$fname'

$sql= " INSERT INTO employees (managementid, first_name,last_name, age, employement_period)
VALUES ('$managementid','$fname','$lname','$age', '$employment_period')";

Do this

Make an auto-incrementing primary key. it may be employee_id.
While inserting don't add that primary key in the query, because it is auto-increment.

Then write the select query like this.

$sql= " SELECT employeeid, managementid, first_name, last_name, age,employment_period from employees ORDER BY employeeid DESC LIMIT 1";

It will gives you the result you want....

edited Apr 14, 2018 at 6:51

answered Apr 14, 2018 at 4:21

Gyandeep Sharma

2,3274 gold badges24 silver badges40 bronze badges

9 Comments

CodingTime Over a year ago

Thanks that helped, but now it tells me that I inserted successfully but the information pulled is that of the person I am trying to insert over, and it does not allow me to make an insert into a number that is not in the list of employee id's. Any thoughts?

CodingTime Over a year ago

Basically, it does not actually insert the information I am trying to insert, it just says it did. and pulls up information of another person.

Gyandeep Sharma Over a year ago

This is not possible... Check your database... Wait I'm giving you some suggestion..

CodingTime Over a year ago

For example, I have 20 people in my database. When I enter a number 21, I get the else message that tells me that person is not in the database, but I want to insert a new person with the number 21. It's a very weird error.

CodingTime Over a year ago

Now it just shows me the one person that is below the number I put, it still does not insert. By the way, the primary key is employeeid and the foreign key is managementid, if that helps.

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