Friends,
I'm a newbie to PHP.
I've had a problem to deal with that I couldn't understand, so I posted it in this thread.
I've dynamically created 2 textboxes and a button.
Question ID text field
Question text field
Change Button
for the change button I need to write a 'onclick' javascript to pass Question ID
and Question value to a PHP function (set_id) written inside the Same file. In fact that’s why i
Called Form action $_SERVER[“PHP_SELF”].
Here’s my code.
<html>
<head>
<script>
function getvalue(value)
{
var qid_value = 'qid_'+value.substring(4);
alert('QID = '+ document.getElementById(qid_value).value + ' QUESTION = ' + document.getElementById(value.substring(4)).value);
/*
I created this javascript alert to test the id s of textboxes and their values
*/
}
</script>
</head>
<body>
<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
<!-- These fields are dynamically created -->
<input type="text" id="'.$var_id.'" name="'.$var_id.'" value="'.$row['qid'].'" readonly size="2 px"/>
<input type="text" id="'.$var_question.'" name="'.$var_question.'" value="'.$row['question'].'" style="size:auto"/>
<input type="button" id="'.$var_question.'" name="'.$var_question.'" value="Change" onclick="getvalue(this.name)"/>
<!-- These fields are dynamically created -->
</form>
</body>
</html>
<?php
$msg= "";
function display($qid,$question)
{
require('uni_db_conn.php'); // this is my db connection
$qid = $_POST[$qid];
$question= $_POST[$question];
$query = "UPDATE question SET question='.$question.' WHERE qid='.$qid.'";
$result = mysql_query($query);
if(!$result)
{
$msg= 'Cant Insert Values to the Table !'.mysql_error();
}
else
{
$msg = 'Successfully Added to the Table !';
}
echo '<label>'.$msg.'</label>';
}
function set_id($qid,$question)
{
if(isset($_POST[$question]))
{
display($qid,$question);
}
}
?>
Thank You ! Sorry If there was any mistake.
Try this code
<?php
if(isset($_POST['submit'])){
$QID = $_POST["qid"];
$QUE = $_POST["question"];
echo $QID;
echo $QUE;
}
?>
<html>
<head>
<script language="javascript">
function getvalue()
{
var valid= true;
var id = document.getElementById("ID").value;
var ques = document.getElementById("ques").value;
return valid;
}
</script>
</head>
<body>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" onSubmit=" return getvalue()" >
<input type="text" id="ID" name="qid"/>
<input type="text" id="ques" name="question"/>
<input type="submit" name="submit" value="Change"/>
</form>
</body>
</html>
Related
I have a PHP to verify the HTML form (post method). If the value doesn't match, the error message should be added at the HTML page. I test the embedded JS script in console, it works. But it doesn't work in PHP. How can I fix it? Thank you
Here is my code
<!DOCTYPE html>
<body>
<form action="http://localhost/submitTest.php" method="post">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
<div id="submitField">
<input type="submit">
</div>
</form>
</body>
</html>
<?php
if ($_GET["name"] == 1 && $_GET["email"] == 2) {
header('Location: http://localhost/table.html');
} else {
$url = 'http://localhost/submitTest.html';
echo "<meta http-equiv='Refresh' content='0;URL=$url'>";
echo "
<script>
var messageP = document.createElement('span');
messageP.style.color = 'red';
var parent = document.getElementById('submitField');
parent.appendChild(messageP);
messageP.innerHTML = 'not match';
</script>";
}
?>
I want to do something like thiserror message
This question already has answers here:
How to submit a form with JavaScript by clicking a link?
(9 answers)
Closed 2 years ago.
i have a db records echoed out and formatted in a table with a delete link(icon) on each table row, indexed with a while loop as the data gets fetched from db.
On clicking the delete icon, i expect the particular indexed row to get deleted from db and also from the table .
Here are my codes:
<script>
function submitForm(){
document.getElementById("myform").submit()
}</script>
</head>
<body>
<table>
<tbody>
<?php
$rowIndex = 0;
while($row = mysqli_fetch_row($result){
$user = $row[0];
$address=$row[1];
$phone = $row[2];
echo '<tr>
<td>'.$user.'</td>
<td>'.$address.'</td>
<td>'.$phone.'</td>
<td>Delete(icon)
</tr>'
$rowIndex+=1;
}
?>
</tbody>
</table>
<form method ="post" id ="myform" action ="test.php" >
<input type="hidden" name="del">
</form>
on clicking the 'Delete' Link , i expect the javascript to submit the form and move to 'test.php'(though i intend to remain on same form, but that way i will know my form ACTUALLY got submitted). Please any help will be appreciated.
<!DOCTYPE html>
<html>
<head>
<title>test javascript form test </title>
</head>
<body>
<form action="#" method="post">
<label>Username</label>
<input type="text" id="name"><br>
<label>Password</label>
<input type="password" id="password"><br>
<button onclick="data()" type="button">Submit</button>
</form>
<script src="https://code.jquery.com/jquery-3.5.1.js"></script>
<script>
function data(){
var name = $("#name").val();
var pass = $("#password").val();
alert(name);
alert(pass);
}
</script>
</body>
</html>
I have a HTML TextBox on my web page that show number of current question (A small web page to answer some question), I want to make shortcut to a question that users want by type number of question in TextBox.
I use this code below but this not work correctly.
For example all questions are 8 and when I enter 15 in TextBox and press Enter, if clause don't work and Question variable set with 15.
I use alert function to trace it and I understand that if clause don't work correctly. Can somebody check it and guide me?
This is all of my code:
<?php
$All = 8;
$URL = "http://localhost/Test.php";
if(isset($_GET["edtQuestionNo"])){
$QuestionNo = $_GET["edtQuestionNo"];
}else{
$QuestionNo = 1;
}
?>
<html>
<head>
<title>Test Page</title>
<script type="text/javascript">
function KeyPress(e, URL, All){
if(e.keyCode === 13){
var Question = document.getElementsByName("edtQuestionNo")[0].value;
if(Question > All){
Question = All;
alert(All + " " + Question + " yes");
}
else{
alert(All + " " + Question + " no");
}
window.open(URL + "?edtQuestionNo=" + Question,"_self");
}
}
</script>
</head>
<body>
<form action="Test.php" method="get" name="FRMQuestion">
<label>Enter question number : </label>
<input type="text" name="edtQuestionNo" id="QuestionNo" value="<?php echo $QuestionNo; ?>"
onkeypress="KeyPress(event,'<?php echo $URL; ?>','<?php echo $All; ?>')">
<br>
<label>Question number is : <?php echo $QuestionNo; ?></label>
</form>
</body>
</html>
I solve it
1. I have to use parseInt function for comparing All and Question value. because they are in different type.
2. I put Question value (after computing) in HTML TextBox again and then open URL.
My code is:
<?php
$All = 8;
$URL = "http://localhost/Test.php";
if(isset($_GET["edtQuestionNo"])){
$QuestionNo = $_GET["edtQuestionNo"];
}else{
$QuestionNo = 1;
}
?>
<html>
<head>
<title>Test Page</title>
<script type="text/javascript">
function KeyPress(e, URL, All){
if(e.keyCode === 13){
var Question = document.getElementsByName("edtQuestionNo")[0].value;
if(parseInt(Question) > parseInt(All)){
Question = All;
}
document.getElementsByName("edtQuestionNo")[0].value = Question;
window.open(URL + "?edtQuestionNo=" + Question,"_self");
}
}
</script>
</head>
<body>
<form action="Test.php" method="get" name="FRMQuestion">
<label>Enter question number : </label>
<input type="text" name="edtQuestionNo" id="QuestionNo" value="<?php echo $QuestionNo; ?>"
onkeypress="KeyPress(event,'<?php echo $URL; ?>','<?php echo $All; ?>')">
<br>
<label>Question number is : <?php echo $QuestionNo; ?></label>
</form>
</body>
</html>
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
My code is supposed to store the info that the user inserts and store it in a Database.
This function should be called on a button click, but the problem is that the function runs when the page loads and when I press the button.
here is my code
<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
<head>
<meta charset="UTF-8">
<title></title>
<script src="//code.jquery.com/jquery-2.1.1.min.js"></script>
</head>
<body>
<?php
function write_command($name,$address,$telephone,$cart){
$servername = 'localhost';
$server_username = 'root';
$server_password = '';
$DB_name = 'OnlineGarcon';
$DB_connect = mysqli_connect($servername,$server_username,$server_password,$DB_name);
if ($DB_connect -> connect_error){
echo 'unable to connect to due to server error!';
}else{
$_write_command = "INSERT INTO `orders`(`name`, `address`, `telephone`, `cart`) VALUES ('$name','$address','$telephone','$cart')";
if ($DB_connect->query($_write_command) == TRUE) {
echo "Order was placed,Thanks!\nyour order was:$cart";
} else {
echo "Server error!";
}
}
}
?>
<center>
<form method="POST">
<input name="name" class="field" placeholder="Full name"></input>
<br>
<br>
<input name="address" class="field" placeholder="Address"></input>
</br>
</br>
<input name="telephone" class="field" placeholder="Telephone/Mobile number"></input>
<br>
<br>
<br>
<br>
<button id="submit">Submit order</button>
</br>
</br>
</br>
</br>
</form>
<script>
$("#submit").click(function(){
alert('<?php
$name = $_POST['name'];
$address = $_POST['address'];
$telephone = $_POST['telephone'];
$cart = $_GET['cart'];
write_command($name,$address,$telephone,$cart);
?>');
});
</script>
</body>
</html>
Using an jQuery Ajax call you you can do this quite simply.
form.php
<html>
<head>
<script src="//code.jquery.com/jquery-2.1.1.min.js"></script>
</head>
<body>
<input name="name" id="name" placeholder="Full name">
<input name="address" name="address" placeholder="Address">
<input name="telephone" id="telephone" placeholder="Telephone/Mobile number">
<button id="submit">Submit order</button>
<script type="text/javascript">
$('#submit').click( function() {
$.ajax({
url: "functions.php",
type: "POST",
data: {
name: $('#name').val();
address: $('#address').val();
telephone: $('#telephone').val();
cart: <?php echo $_GET['cart']; ?>
}
});
});
</script>
</body>
</html>
functions.php
<?php
function write_command($name, $address, $telephone, $cart) {
$servername = 'localhost';
$server_username = 'root';
$server_password = '';
$DB_name = 'OnlineGarcon';
$DB_connect = mysqli_connect($servername,$server_username,$server_password,$DB_name);
if( $DB_connect->connect_error ){
echo 'unable to connect to due to server error!';
} else {
$_write_command = "INSERT INTO `orders`(`name`, `address`, `telephone`, `cart`) VALUES ('$name','$address','$telephone','$cart')";
if ($DB_connect->query($_write_command) == TRUE) {
echo "Order was placed,Thanks!\nyour order was:$cart";
} else {
echo "Server error!";
}
}
}
write_command( $_POST['name'], $_POST['address'], $_POST['telephone'], $_POST['cart'] );
This will happen , what you need to do is
Method 1: <form method="POST" onsubmit="callFuntion()">
Method 2:
<form action="demo_form.php" method="post">
put all your php code in file demo_form.php and when the submit is clicked it will go that file , which will help you to call the
function once not twice .
Hi i am trying to save value and alert them using ajax which i am insert using php in my sql table but my alert is not working
Here is my code
demo.php
<html>
<head>
<script>
function my(){
var name = document.getElementById("name").value;
var last_name = document.getElementById("last_name").value;
document.getElementsById('div1').style.backgroundColor = green;
var dataString = 'name='+name+'&last_name='+last_name;
$.ajax({
type:'POST',
data:dataString,
url:'demo.php',
success:function(data) {
alert(data);
}
});
} </script>
</head>
<body>
<form action="" method="post">
<input type="text" name="name" id="name" value="" />
<input type="text" name="last_name" id="last_name" value="" />
<input type="submit" name="Update" id="update" value="Update" onclick="my();" />
</form>
<div id="div1" style="width:300px;height: 50px;background-color: yellow;" >
</div>
</body>
</html>
<?php
include('conn.php');
if (isset($_POST['Update'])) {
$name = $_POST['name'];
$last_name = $_POST['last_name'];
echo $name;
$insert = "insert into ajaxsave values('$name','$last_name')";// Do Your Insert Query
if(mysql_query($insert)) {
echo "Success";
} else {
echo "Cannot Insert";
}
}?>
demo.html
<html>
<head>
</head>
<body>
<div id="div2" style="width:300px;height: 50px;background-color: yellow;" >
</div>
</body>
</html>
here i want when i submit form them div color should change which is in demo.html
where i am wrong in this code
and how can i achieve my goal
Any help will be appreciated
changes you need to make:
add jquery as a dependency as you are using $.ajax utility function which is provided by Jquery.
As you are using Jquery, you could use its selectors for getting values of elements and binding functions to dom elements. I have commented it in the source code.
You are using a form with a submit button and executing the ajax call on click of it. But you need to prevent the page from submitting the form by preventing the default behavior of the submit button. Refer event.preventDefault();
Move the php ajax response part to the top and call exit() once your response is complete. Else your ajax response will include the whole page html source also.
.
<?php
include('conn.php');
if (isset($_POST['Update'])) {
$name = $_POST['name'];
$last_name = $_POST['last_name'];
$insert = "insert into ajaxsave values('$name','$last_name')";// Do Your Insert Query
if(mysql_query($insert)) {
echo "Success";
} else {
echo "Cannot Insert";
}
//Call exit as your ajax response ends here. You dont need the html source along with it.
exit();
}
?>
<html>
<head>
</head>
<body>
<form action="" method="post">
<input type="text" name="name" id="name" value="" />
<input type="text" name="last_name" id="last_name" value="" />
<input type="submit" name="Update" id="update" value="Update" />
</form>
<div id="div1" style="width:300px;height: 50px;background-color: yellow;" >
</div>
<!-- include jquery dependeny before your js code block -->
<script src="https://code.jquery.com/jquery-latest.js"></script>
<script>
$("#update").on("click",function(event) {
//Prevent Default submit button behavour. Prevent the form from submission.
event.preventDefault();
// using Jquery selectors for better readability of code.
var name = $("#name").val();
var last_name = $("#last_name").val();
$("#last_name").css("background-color","green");
$.ajax({
type:'POST',
data:{name:name,last_name:last_name,Update:true},
url:'demo.php',
success:function(data) {
alert(data);
}
});
});
</script>
</body>
</html>
You send two parameters in "dataString" variable, and then in php check undefined variable "Update"
So, just replace string
if (isset($_POST['Update'])) {
to
if (isset($_POST['name']) && isset($_POST['name'])) {
And add this line to tag
<script src="//code.jquery.com/jquery-1.11.0.min.js"></script>