PHP Form not adding data to mySQL database?

Question

I am having trouble getting this form to successfully submit the data to the database. I have verified that the database/table exists (schema shown below) and for some reason it's not inserting it into the database. When I try to submit sample data, it doesn't change the page and just sits there. What is going on?

<body>
<?php if($_SERVER["REQUEST_METHOD"] != "POST"){ ?>
<h1>My Favorite Foods</h1>

<form action="index.php" method="post" id="foodForm">
Name: <input type="text" name="foodname" id="nameField"></input><br />
Type: <select name="foodtype" id="typeField">
<option value="fruit">Fruit</option>
<option value="vegetable">Vegetable</option>
<option value="dairy">Dairy</option>
<option value="meat">Meat</option>
<option value="grain">Grain</option>
<option value="other">Other</option>
</select><br />
Number of Calories: <input type="text" name="foodcals" id="calsField"></input><br />
Healthy? <input type="checkbox" name="foodhealth" value="healthy" id="healthyField"></input><br />
Additional Notes:<br />
<textarea name="foodnotes" id="notesField"></textarea><br />
<input type="submit" value="Add" onclick="validateForm();return false;"></input>
</form>

<?php }else{ ?>
<!-- form handling and output printing stuff goes here -->
<?php $insert = "INSERT INTO Foods(Name, Type, NumCals, Healthy, Notes) VALUES ('$_POST[foodname]', '$_POST[foodtype]', '$_POST[foodcals]', '$_POST[foodhealth]', '$_POST[foodnotes]'"; 
$con = mysqli_connect("localhost","root");
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_select_db("mydb"); 
$result = mysqli_query($con, $insert); 
if ($result) { 
    echo "Food added successfully."; 
    /*while ($row = mysqli_fetch_array($result)) { 
        echo $row['Name'] . ", " . $row['Type'] . ", " . $row['NumCals'] . ", " . $row['Healthy'] . ", " . $row['Notes']; 
        echo "<br>"; 
     } */
} else { 
     echo "Error adding person";  
     mysqli_error($con); 
} 
} ?>
</body>
</html>

Schema:

Foods(PID INT NOT NULL AUTO_INCREMENT,PRIMARY KEY(PID),Name VARCHAR(20),Type VARCHAR(9),NumCals INT,Healthy BOOL,Notes TEXT)

Barmar · Accepted Answer · 2013-10-06 01:30:03Z

3

This attribute in the submit button:

onclick="validateForm();return false;"

prevents the form from submitting. return false means that the browser should not perform the default action from clicking the button.

Assuming the validateForm() function returns a boolean indicating whether validation was successful, change that to:

onclick="return validateForm();"

Change:

} else { 
     echo "Error adding person";  
     mysqli_error($con);

to:

} else { 
     echo "Error adding person: " . mysqli_error($con);

So that the error message will be displayed.

And you're missing the ) at the end of your INSERT statement.

edited Oct 6, 2013 at 1:30

answered Oct 6, 2013 at 1:19

Barmar

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6 Comments

vsingal5 Over a year ago

I'm now getting an error that says it cannot add food. Is there something wrong with my insert statement? If so, what is it?

vsingal5 Over a year ago

How do I figure out what mysqli_error() shows? It just echos "Food cannot be added."

Barmar Over a year ago

You're missing the close parenthesis at the end of the INSERT statement.

Barmar Over a year ago

@GaryHayes When you're interpolating array values, you leave out the quotes unless you wrap it in {...}.

Barmar Over a year ago

Change your error code to echo "Food cannot be added: " . mysqli_error();. You're calling the error function, but not displaing the message it returns.

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