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

How does it not work? What are you expecting to happen? What is actually happening? What have you done to troubleshoot this? — John Conde
– John Conde, Commented Apr 28, 2017 at 12:41
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. — John Conde
– John Conde, Commented Apr 28, 2017 at 12:41
@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 — Oke Tega
– Oke Tega, Commented Apr 28, 2017 at 12:44
Please take a look here. stackoverflow.com/questions/20769364/… — Vandolph Reyes
– Vandolph Reyes, Commented Apr 28, 2017 at 12:48

Andrew Reyes · Accepted Answer · 2017-04-28 12:49:13Z

2

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.

answered Apr 28, 2017 at 12:49

Andrew Reyes

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1 Comment

diagold Over a year ago

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

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