jQuery(AJAX) post to php issue when I try to insert data into DB

Question

when I echo the php variable it work properly , but when I try to insert the data into database it doesn't work , what is the solution please I get stuck

I got this error on console

  POST http://localhost/validate.php 500 (Internal Server Error)

        send @ jquery-3.1.1.min.js:4
        ajax @ jquery-3.1.1.min.js:4
        (anonymous) @ jquery.PHP:26
        dispatch @ jquery-3.1.1.min.js:3
        q.handle @ jquery-3.1.1.min.js:3

HTML/JQUERY

<form action="" id="myForm">
    <input type="text" id="name" ><br/> 
    <input type="text" id="age" ><br/> 
    <input type="submit" value="Submit"> 
</form> 
<div id="result"></div>

   <script> 

   $(function() {
    $("#myForm").submit(function(e) {
        e.preventDefault();
        var name = $('#name').val(); 
        var age = $('#age').val();

        $.ajax({
        url: 'validate.php',
        method: 'POST',
        data: {postname:name, postage:age},
        success: function(res) {
            $("#result").append(res);
        } 
    });
    });
}); 
   </script>

PHP

 <?php 

include 'mysqldb.php';


$name = $_POST['postname']; 
$age =  $_POST['postage']; 


$sql = "insert into uss (first, last) values('".$name."','".$age."')";
$result = $conn->query($sql);

echo $result ;

?>

mysqldb.php

<?php

$conn = mysql_connect('localhost', 'root', 'password' , 'datab');

if (!$conn) {

    die("Connection failed: ".mysqli_connect_error());

}

?>

Sahil · Accepted Answer · 2017-03-09 04:29:38Z

1

Please add the details of the error message you get.

Make little changes to your code so that it can show the query error if any

<?php 

include 'mysqldb.php';


$name = $_POST['postname']; 
$age =  $_POST['postage']; 

$sql = "INSERT INTO `uss` (`first`, `last`) VALUES('{$name}','{$age}')";
if($conn->query($sql))
{
  echo "Record inserted";
}
else
{
 echo $conn->error;
}
?>

Sugesstions: Your query have the chances of the SQL Injection. Make it secure.

answered Mar 9, 2017 at 4:29

Sahil

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1 Comment

zexurity Over a year ago

I've edit the post I put the error from the console , have a look thanks

Oops · Accepted Answer · 2017-03-09 04:29:50Z

0

if you are using ajax , try the following,

<form >
<input type="text" id="name" ><br/> 
<input type="text" id="age" ><br/> 
<input type="submit" value="Submit" id="submit"> 
</form> 
<div id="result"></div> 

 $("#submit").click(function(){
 var name = $('#name').val(); // getting name
 var age = $('#age').val();
 $.ajax({
 url : "validate.php",
 type: "POST",
 data: {name:name, age:age},
 success: function(data)
 {
 $("#result").html(data);
 }
});
});

in your controller function,echo the result

<?php 

include 'mysqldb.php';


$name = $_POST['postname']; 
$age =  $_POST['postage']; 

$sql = "insert into uss (first, last) values('$name','$age')";
$result = $conn->query($sql);
echo $result;
?>

answered Mar 9, 2017 at 4:29

Oops

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Comments

Sahil · Accepted Answer · 2017-03-09 04:42:21Z

jQuery Ajax

Form with id myFrom

<form action="" id="myForm">
    <input type="text" id="name" ><br/> 
    <input type="text" id="age" ><br/> 
    <input type="submit" value="Submit"> 
</form> 
<div id="result"></div>

jQuery Ajax section

$(function() {
    $("#myForm").submit(function(e) {
        e.preventDefault();
        var name = $('#name').val(); // getting name
        var age = $('#age').val();   // getting age 

        /* Ajax section */
        $.ajax({
        url: 'validate.php',
        method: 'POST',
        data: {postname:name, postage:age},
        success: function(res) {
            $("#result").append(res);
        } 
    });
    });
});

validate.php

<?php 

    include 'mysqldb.php';


    $name = $_POST['postname']; 
    $age =  $_POST['postage']; 

    //check ajax response by `echo $name` and `$age` 

    $sql = "insert into uss (first, last) values('".$name."','".$age."')";
    $result = $conn->query($sql);

    echo $result ;

?>

g.m.ashaduzzaman · Accepted Answer · 2017-03-09 05:24:40Z

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
 <form >
    <input type="text" id="name" ><br/> 
    <input type="text" id="age" ><br/> 
    <input type="button" value="Submit" onclick="postdata();"> 
    </form> 
    <div id="result"></div> 

    <script type="text/javascript"> 

    function postdata() { 
	alert("ashad");
    var name = $('#name').val(); 
    var age = $('#age').val();
    $.post('validate.php',{postname:name,postage:age},
    function(data){

    $('#result').html(data); 
         }); 
    } 
   </script>

   <?php 
	include 'mysqldb.php';
    $name = $_POST['postname']; 
    $age =  $_POST['postage']; 

    //check ajax response by `echo $name` and `$age` 
	if ($conn->connect_error) {
		die("Connection failed: " . $conn->connect_error);
	} else
	{
		$sql = "insert into uss(first, last) values('$name','$age')";
		$result = $conn->query($sql);
		
	}
    echo $result ;

?>

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