Performing php insert function with ajax

Question

I have a html form with <input type='radio'>

<form action="results.php" method="post" enctype="multipart/form-data" onsubmit='return false'><br>
 <p>Have you ever turned a client down?</p>
 <div id="q_1">
 <input type="radio" name="q1" id="q_1_yes" value="yes">
 <label for="q_1_yes">Yes</label>
 <input type="radio" name="q1" id="q_1_no" value="no">
 <label for="q_1_no">No</label>
 </div><br>
 <p>Are you comfortable with failure?</p>
 <div id="q_1">
 <input type="radio" name="q2" id="q_2_yes" value="yes">
 <label for="q_2_yes">Yes</label>
 <input type="radio" name="q2" id="q_2_no" value="no">
 <label for="q_2_no">No</label>
 </div><br>
 <input type="submit" onclick='return handleClick();' name="sub_eit" id="sub_eit" value="Submit">
</form>

I have a javascript function to check if the amount of radio buttons with value "yes" is greater than the amount with value "no" as below

function handleClick()
{
var amountYes = 0;
for(var i = 1; i <= 4; i++) {
 var radios = document.getElementsByName('q'+i);
 for(var j = 0; j < radios.length; j++) {
 var radio = radios[j];
 if(radio.value == "yes" && radio.checked) {
 amountYes++;
 }
 }
}
//code to perform php insert function if yes is less than or equal to 2 
if (amountYes <= 2) {
 $.ajax({
 type: "POST",
 url: "results.php",
 dataType: "json",
 success: function (response) {
 }
 });
} else {
 alert("Correct Responses: " + amountYes);
}
}

results.php

if (isset($_POST['q_1']) && isset($_POST['q_2']))
{
$q_1 = $_POST['q_1'];
$q_2 = $_POST['q_2'];
$yes = "yes";
$no = "no";
$id = $_SESSION['id'];
$u_name = $_SESSION['uname'];
 $recommend = "Executive";
 $osql =<<< EOF
INSERT INTO results (id, recomend, u_name, status) VALUES ('$id', '$recommend', '$u_name', '1');
EOF;
 $ret = $db->exec($osql);
}

But the ajax code doesn't seem to work. Please what is the issue. Thanks in advance for your help. Highly appreciate

• John Conde: How does it not work? What are you expecting to happen? What is actually happening? What have you done to troubleshoot this?
• John Conde: Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
• Oke Tega: @JohnConde i want it to insert to database what is in the php code but it is not inserting and i am this is a quick trial for a quiz app i am trying to create and would handle security later
• Vandolph Reyes: Please take a look here. stackoverflow.com/questions/20769364/…
• diagold: Thanks @DanMiller but it didn't work

Answer 1

You have include post data in your ajax request.

 $.ajax({
 type: 'POST',
 url: 'results.php',
 data: $('form').serialize(),
 dataType: 'json',
 success: function( response){
 console.log( 'the feedback from your result.php: ' + response);
 }
});

and in your result.php you have to rename $_POST['q_1'] to _POST['q1'] and so on. q1 and q2 is the name of your input radio.

• diagold: Thanks for pointing out the q_1 issue for me forgot to change that. and your answer works perfectly.