I'm submitting a form using MySQL command inside a PHP file. I'm able to insert the data without any problem.
However, I also, at the same time, want to display the user a "Thank you message" on the same page so that he/she knows that the data has been successfully registered. On the other hand I could also display a sorry message in case of any error.
Therein lies my problem. I've written some lines in Javascript to display the message in the same page. However, I'm stuck on what (and how) should I check for success and failure.
I'm attaching my code below.
Can you please help me on this with your ideas?
Thanks
AB
HTML Form tag:
<form id="info-form" method="POST" action="form-submit.php">
form-submit.php:
<?php
require("database-connect.php");
$name = $_POST['name'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$sql = "INSERT INTO tbl_details ".
"(name,email_id,mobile_number) ".
"VALUES ".
"('$name','$email','$mobile')";
mysql_select_db('db_info');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
return false;
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
submit-logic.js:
$(function ()
{
$('form').submit(function (e)
{
e.preventDefault();
if(e.target === document.getElementById("info-form"))
{
$.ajax(
{
type:this.method,
url:this.action,
data: $('#info-form').serialize(),
dataType: 'json',
success: function(response)
{
console.log(response);
if(response.result == 'true')
{
document.getElementById("thankyou_info").style.display = "inline";
$('#please_wait_info').hide();
document.getElementById("info-form").reset();
}
else
{
document.getElementById("thankyou_info").style.display = "none";
document.getElementById("sorry_info").style.display = "inline";
$('#please_wait_info').hide();
}
}
}
)};
});
}
Per documentation: http://api.jquery.com/jquery.ajax/
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server.
You are explicitly setting this to json but then returning a string. You should be returning json like you are telling the ajax script to expect.
<?php
require("database-connect.php");
$name = mysql_real_escape_string($_POST['name']);
$email = mysql_real_escape_string($_POST['email']);
$mobile = mysql_real_escape_string($_POST['mobile']);
$sql = "INSERT INTO tbl_details ".
"(name,email_id,mobile_number) ".
"VALUES ".
"('$name','$email','$mobile')";
mysql_select_db('db_info');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die(json_encode(array('result' => false, 'message' => 'Could not enter data: ' . mysql_error()));
}
echo json_encode(array('result' => true, 'message' => 'Entered data successfully'));
mysql_close($conn);
?>
I also added code to sanitize your strings, although mysql_* is deprecated and it would be better to upgrade to mysqli or PDO. Without sanitization, users can hack your database..
Nevertheless, returning json properly will ensure that your response in success: function(response) is an object, and response.result will be returned as expected, and you can use response.message to display the message where you want.
Related
I have a simple AJAX function bound to a button that should execute a PostgreSQL query. However, when I click the button that I bound the ajax query to, all I get is the confirmation that the database connection was successful. Nothing seems to happen withe the ajax result (should be printing to console in the handleAjax() function. What am I doing wrong?
This is the javascript code (with jquery):
$(document).ready(function() {
function sendAjax() {
$.ajax({
url: "db/database.php",
success: function (result) {
handleAjax(result);
}
});
}
function handleAjax(result) {
console.log(result);
}
$("#submit-button").on("click", sendAjax);
});
And this it the contents of database.php:
<?php
function dbconn(){
ini_set('display_errors', 1); // Displays errors
//database login info
$host = 'localhost';
$port = 5432;
$dbname = 'sms';
$user = 'postgres';
$password = 'postgres';
// establish connection
$conn = pg_connect("host=$host port=$port dbname=$dbname user=$user password=$password");
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
}
}
$conn = dbconn();
$sql = "SELECT * FROM numbers;";
$result = pg_query( $sql ) or die('Query Failed: ' .pg_last_error());
$count = 0;
$text = 'error';
while( $row = pg_fetch_array( $result, null, PGSQL_ASSOC ) ) {
$text = $row['message'];
//echo $text;
}
pg_free_result( $result );
?>
The problem is in the database.php file, all you get is "Connected." because you don't print your result at the end. Ajax only receive the output of the php file.
So at the end of your php file you should add :
echo $text;
And you also should remove the echo "Connected.";
AJAX is not a magic wand that in magic way reads PHP code. Let's say AJAX is a user. So what does user do.
Open web page
Wait until PHP execute code and display data
Tells you what he sees
If you don't display anything, ajax can't tell you what he saw.
In thi's place is worth to say that the best way to communicate between PHP and AJAX is using JSON format.
Your code generally is good. All you have to do is to display your data. All your data is in your $text var. So let's convert your array ($text) to JSON.
header('Content-Type: application/json');
echo json_encode($text);
First you set content-type to json, so ajax knows that he reads json. Then you encode (convert) your PHP array to js-friendly format (JSON). Also delete unnecessary echoes like 'Conntected' because as I said, AJAX reads everything what he sees.
You should return $conn from dbconn()
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
return $conn;
}
I have two php files that handle a commenting system I have created for my website. On the index.php I have my form and an echo statement that prints out the user input from my database. I have another file called insert.php that actually takes in the user input and inserts that into my database before it is printed out.
My index.php basically looks like this
<form id="comment_form" action="insertCSAir.php" method="GET">
Comments:
<input type="text" class="text_cmt" name="field1_name" id="field1_name"/>
<input type="submit" name="submit" value="submit"/>
<input type='hidden' name='parent_id' id='parent_id' value='0'/>
</form>
<!--connects to database and queries to print out on site-->
<?php
$link = mysqli_connect('localhost', 'name', '', 'comment_schema');
$query="SELECT COMMENTS FROM csAirComment";
$results = mysqli_query($link,$query);
while ($row = mysqli_fetch_assoc($results)) {
echo '<div class="comment" >';
$output= $row["COMMENTS"];
//protects against cross site scripting
echo htmlspecialchars($output ,ENT_QUOTES,'UTF-8');
echo '</div>';
}
?>
I want users to be able to write comments and have it updated without reloading the page (which is why I will be using AJAX). This is the code I have added to the head tag
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script>
// this is the id of the form
$("#comment_form").submit(function(e) {
var url = "insert.php"; // the script where you handle the form input.
$.ajax({
type: "GET",
url: url,
data: $("#comment_form").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>
However, nothing is happening. The alert() doesn't actually do anything and I'm not exactly sure how to make it so that when the user comments, it gets added to my comments in order (it should be appending down the page). I think that the code I added is the basic of what needs to happen, but not even the alert is working. Any suggestions would be appreciated.
This is basically insert.php
if(!empty($_GET["field1_name"])) {
//protects against SQL injection
$field1_name = mysqli_real_escape_string($link, $_GET["field1_name"]);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element){
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0){
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO parentComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
header("Location:index.php");
die();
mysqli_close($link);
}
else{
die('comment is not set or not containing valid value');
it also filters out bad words which is why there's an if statement check for that.
<?php
if(!empty($_GET["field1_name"])) {
//protects against SQL injection
$field1_name = mysqli_real_escape_string($link, $_GET["field1_name"]);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element)
{
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0)
{
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO parentComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
if ($result)
{
http_response_code(200); //OK
//you may want to send it in json-format. its up to you
$json = [
'commment' => $newComment
];
print_r( json_encode($json) );
exit();
}
//header("Location:chess.php"); don't know why you would do that in an ajax-accessed file
//die();
mysqli_close($link);
}
else{
die('comment is not set or not containing valid value');
}
?>
<script>
// this is the id of the form
$("#comment_form").submit(function(e) {
var url = "insert.php"; // the script where you handle the form input.
$.ajax({
type: "GET", //Id recommend "post"
url: url,
dataType: json,
data: $("#comment_form").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
$('#myElement').append( data.comment );
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>
To get a response from "insert.php" you actually need to print/echo the content you want to handle in the "success()" from the ajax-request.
Also you want to set the response-code to 200 to make sure "success: function(data)" will be called. Otherwise you might end up in "error: function(data)".
So I'm building a mobile (cordova) app using a mySQL server, PHP, javascript (jQuery) but I have encountered a rather annoying problem.
When sending a test string with AJAX the recieving end, a PHP script does not receive all the data:
function testJson() {
$.ajax({
type: 'POST',
data: {
text: "hello, this is a test to see if the string is send."
},
url: 'http://a.working.url/save.php',
success: function(data) {
alert('succes');
},
error: function(data) {
alert('There was an error adding your comment');
}
});
return false;
}
then at the serverside I have a small PHP script:
<?php
$server = "localhost:3307";
//changed port to 3307
$username = "user";
$password = "pass";
$database = "test";
$con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
mysql_select_db($database, $con);
$comment = $_POST['text'];
echo $comment;
$sql = "INSERT INTO data (comment) ";
$sql .= "VALUES ('$comment')";
if (!mysql_query($sql, $con)) {
die('Error: ' . mysql_error());
} else {
echo "Comment added";
}
mysql_close($con);
?>
but this PHP script did not recieve the full string, when I check using:
echo $_POST['text']
It only got: "hello, this is a test to see if the" (without quotes).
If anybody can help me it would be great cause I cannot continue working on the app unless the data transfer works.
Thanks in advance.
I am trying to write an insert query with jquery, ajax and php. The record is getting inserted but returns a status error. First I tried to echo the message in php as it didn't work I tried it with print json_encode but both returned the status as error. Why doesn't it return the responseText?
{readyState: 0, responseText: "", status: 0, statusText: "error"}
This is the addmember.php file
<?php
require '../database.php';
function random_password( $length = 8 ) {
$chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!##$%^&*()_-=+;:,.?";
$password = substr( str_shuffle( $chars ), 0, $length );
return $password;
}
$password = random_password(8);
//$regno=$_POST['regNo'];
$adminno=$_POST['adminNo'];
$batch=$_POST['batchText'];
$type=$_POST["memberType"];
$initials=$_POST["initialName"];
$fullname=$_POST["fullName"];
$address=$_POST["address"];
$telephone=$_POST["contact"];
$email=$_POST["email"];
$nic=$_POST["nic"];
$dob=$_POST["birthDate"];
$priv=$_POST["memberType"];
$userid="";
$sql="select username from memberinfo where username='$adminno'";
$result=mysqli_query($con,$sql);
if(mysqli_num_rows($result)==0){
$sql="insert into memberinfo(username,nic_no,class,name_initial,full_name,address,telephone,email,date_of_birth) VALUES ('$adminno','$nic','$batch','$initials', '$fullname', '$address', '$telephone','$email','$dob')";
$result1=mysqli_query($con,$sql);
$sql = "select * from memberinfo where username='$adminno'";
$result = $con->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$userid = $row['user_id'];
}
}
$sql="insert into userlogin(user_id,username,privilege,password) VALUES ('$userid','$adminno','$priv','$password')";
$result2=mysqli_query($con,$sql);
if ($result1 && $result2) {
$message = "<p>New record created successfully</p>";
} else {
$message = "<p>Error: " . $sql . "<br>" . $con->error.".</p>";
}
} else{
$message = "<p>Admission no already exists.</p>";
}
print json_encode($message);
$con->close()
?>
This is the .js file with the ajax function
$(document).ready(function(){
$('#addmember').click(function(){
console.log("addmember");
var adminno=$("#adminNo").val();
var nic=$("#nic").val();
var batch=$("#batchText").val();
var initials=$("#initialName").val();
var fullname=$("#fullName").val();
var address=$("#address").val();
var telephone=$("#contact").val();
var email=$("#email").val();
var dob=$("#birthDate").val();
var priv=$("#memberType").val();
//$("#result").html("<img alt='ajax search' src='ajax-loader.gif'/>");
$.ajax({
type:"POST",
url:"../ajax/addmember.php",
dataType: "json",
data:{'adminNo':adminno, 'nic':nic,'batchText':batch,'initialName':initials, 'fullName':fullname, 'address':address, 'contact':telephone,'email':email,'birthDate':dob,'memberType':priv},
success:function(response){
console.log(response);
$("#result").append(response);
},
error:function(response){
console.log(response);
}
});
});
});
Status zero normally means the page is navigating away. Stop it from happening.
$('#addmember').click(function(evt){ //<--add the evt
evt.preventDefault(); //cancel the click
You are not returning valid JSON from the server. You're json encoding a string, but valid JSON requires an object, or array to encapsulate the day coming back.
So at the very least:
echo json_encode(array($message));
No need for the JSON response. Simply return the message from your PHP script as shown below (note the use of echo and the semicolon following close()):
PHP
$con->close();
echo $message;
Also, remove the JSON filetype from your AJAX call and instead append response.responseText rather than response:
JS
$.ajax({
type:"POST",
url:"../ajax/addmember.php",
data:{'adminNo':adminno,'nic':nic,'batchText':batch,'initialName':initials, 'fullName':fullname, 'address':address, 'contact':telephone,'email':email,'birthDate':dob,'memberType':priv},
success:function(response){
console.log(response);
$("#result").append(response.responseText);
},
error:function(response){
console.log(response);
}
});
<?php
$con = mysqli_connect("localhost", "root", "", "" ) or die("Neuspjelo spajanje");
function InsertUser(){ global $con;
if(isset($_POST['sign_up'])){
$name = mysqli_real_escape_string($con, $_POST['u_name']);
$pass = mysqli_real_escape_string($con,$_POST['u_pass']);
$email = mysqli_real_escape_string($con,$_POST['u_email']);
$country = mysqli_real_escape_string($con,$_POST['u_country']);
$gender = mysqli_real_escape_string($con,$_POST['u_gender']);
$b_day = mysqli_real_escape_string($con,$_POST['u_birthday']);
$date = date("m-d-Y");
$status = "unverified";
$posts = "No";
$get_email = "select * from users where user_email='$email'";
$run_email = mysqli_query($con, $get_email);
$check = mysqli_num_rows($run_email);
$insert = "insert into users (user_name, user_pass, user_email, user_country, user_gender, user_b_day,
user_image, register_date, last_login, status, posts) values
('$name','$pass', '$email', '$country', '$gender', '$b_day', 'default.jpg',
'$date', '$date', '$status', '$posts')";
$run_insert = mysqli_query($con, $insert);
$result = mysql_query($insert);
if($result){
echo "<script>alert ('You're successfully registered!')</script>";
echo "<script>window.open('home.php', '_self')</script>";
}
}
}
?>
You can't echo javascript and run it in a page that's already loaded. This would need to be the result of an ajax call on the client side with your redirects occuring from your ajax callbacks.
If you're ok with ditching the alert, you can just issue a redirect from php:
header('Location: home.php');
To do it ajaxy:
$.ajax({
type: "GET",
url: "your_insert_user.php"
}).success(function(xhr) {
alert ("You're successfully registered!");
window.open('home.php', '_self');
}).fail(function (jqXHR, status, errorThrown) {
//something else here
});
But, why would you want to issue an ajax call just to redirect?
Additionally, you need to issue the appropriate responses from your insert script:
if ($result) { echo ""; } //issues a "200 OK"
else { header("HTTP/1.1 422 Unprocessable Entity"); } //fires the failure callback in ajax
I would pass a conditional GET or POST paramater to home.php with some value flag and display your message there.
Based on what you post above, you are dealing with two separate issues here.
You say "it inserts" so I'm assuming that means that the mysql query to insert the new row into your database completes successfully. Then you send some HTML code, containing a (somewhat mangled) Javascript snippet, to the browser, which is supposed to issue a redirect request to the client's web browser, which doesn't have the desired result, seeing as you write that it "won't redirect".
Keep in mind that redirection is performed by the browser, is dependent on the browser's capabilities and/or settings, and requires proper javascript in the first place.
How do properly request a redirect from the browser has been discussed before on SO.
First of all,remove this line $result = mysql_query($insert); then modify your code and add this, hope it will work:
$run_insert = mysqli_query($con, $insert);
if($run_insert){
echo "<script>alert ('You\'re successfully registered!')</script>";
echo "<script>window.open('home.php', '_self')</script>";
}