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Realizando la función de inserción de php con ajax

Tengo un formulario html con <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>

Tengo una javascript function para verificar si la cantidad de botones de radio con valor "sí" es mayor que la cantidad con valor "no" como se muestra a continuación

 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); } }

resultados.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); }

Pero el código ajax no parece funcionar. Por favor cual es el problema. Gracias de antemano por tu ayuda. Muy apreciado

over 4 years ago · Santiago Trujillo
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0

Ha incluido datos de publicación en su solicitud de ajax.

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

y en su resultado.php debe cambiar el nombre de $_POST['q_1'] a _POST['q1'] y así sucesivamente. q1 y q2 es el nombre de su radio de entrada.

over 4 years ago · Santiago Trujillo Report
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