I have a form on one page linking to a PHP file (action), now the PHP result is being displayed in this PHP file/page. But I want the result to be displayed on the page with the form. I have searched thoroughly and couldn't find it anywhere. Perhaps any of you can help?
Code: /citizens.php (main page)
<form method="post" action="/infoct.php">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="submit">
</form>
Code: /infoct.php
<!DOCTYPE html>
<html>
<head>
<!-- <meta http-equiv="refresh" content="0; url=/citizens.php" /> -->
</head>
<body>
<?php {
$ID2 = isset($_POST['ID']) ? $_POST['ID'] : false;
}
$connect = mysql_connect('localhost', 'root', 'passwd');
mysql_select_db ('inhabitants');
$sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID2";
$res = mysql_query($sql);
echo "<P1><b>Citizen Identification number is</b> $ID2 </p1>";
while($row = mysql_fetch_array($res))
{
echo "<br><p1><b>First Name: </b></b>", $row['Name'], "</p1>";
echo "<br><p1><b>Surname: </b></b></b>", $row['Surname'], "</p1>";
echo "<br><p1><b>Date of birth: </b></b></b></b>", $row['DOB'], "</p1>";
echo "<br><p1><b>Address: </b></b></b></b></b>", $row['Address'], "</p1>";
echo "<br><p1><b>Background information: </b><br>", $row['RPS'], "</p1>";
}
mysql_close ($connect);
?>
</body>
</html>
My fixed code thanks to Marc B
<form method="post">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="submit">
</form>
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$ID = isset($_POST['ID']) ? $_POST['ID'] : false;
$connect = mysql_connect('fdb13.biz.nf:3306', '1858208_inhabit', '12345demien12345');
mysql_select_db ('1858208_inhabit');
$sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID";
$res = mysql_query($sql);
if ($ID > 0) {
echo "<p><b>Citizen Identification number is</b> </p>";
while($row = mysql_fetch_array($res))
echo "<br><p><b>Surname: </b></b></b>", $row['Surname'], "</p>";
echo "<br><p><b>First Name: </b></b>", $row['Name'], "</p>";
echo "<br><p><b>Date of birth: </b></b></b></b>", $row['DOB'], "</p>";
echo "<br><p><b>Address: </b></b></b></b></b>", $row['Address'], "</p>";
echo "<br><p><b>Background information: </b><br>", $row['RPS'], "</p>";
mysql_close ($connect);
}
else {
echo "<p>Enter a citizen ID above</p>";
}
}
?>
DB Snap
A single-page form+submit handler is pretty basic:
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
... form was submitted, process it ...
... display results ...
... whatever else ...
}
?>
<html>
<body>
<form method="post"> ... </form>
</body>
</html>
That's really all there is.
Use code on the same page (citizens.php)
<?php
if (isset($_POST)) {
Do manipulation
}
?>
Else use ajax and remove action method from form.
<form method="post" id="contactForm">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="buttom" id="submitId">
</form>
<script>
$("#submitId").click(function(){
var Serialized = $("#contactForm").serialize();
$.ajax({
type: "POST",
url: "infoct.php",
data: Serialized,
success: function(data) {
//var obj = jQuery.parseJSON(data); if the dataType is not specified as json uncomment this
// do what ever you want with the server response
},
error: function(){
alert('error handing here');
}
});
});
</script>
And in your infact.php in the end Echo the data so that ajax will have the data in return.
You could just put everything in infoct.php, like this:
<!DOCTYPE html>
<html>
<head>
<!-- <meta http-equiv="refresh" content="0; url=/infoct.php" /> -->
</head>
<body>
<form method="post" action="/infoct.php">
<input type="text" name="ID" placeholder="ID" value="<?php isset($_POST['ID']) ? $_POST['ID'] : '' ?>">
<input name="set" type="submit">
</form>
<?php
if (isset($_POST['ID'])) {
$ID2 = $_POST['ID']; // DO NOT FORGET ABOUT STRING SANITIZATION
$connect = mysql_connect('localhost', 'root', 'usbw');
mysql_select_db ('inhabitants');
$sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID2";
$res = mysql_query($sql);
echo "<P1><b>Citizen Identification number is</b> $ID2 </p1>";
while($row = mysql_fetch_array($res))
{
echo "<br><p1><b>First Name: </b></b>", $row['Name'], "</p1>";
echo "<br><p1><b>Surname: </b></b></b>", $row['Surname'], "</p1>";
echo "<br><p1><b>Date of birth: </b></b></b></b>", $row['DOB'], "</p1>";
echo "<br><p1><b>Address: </b></b></b></b></b>", $row['Address'], "</p1>";
echo "<br><p1><b>Background information: </b><br>", $row['RPS'], "</p1>";
}
mysql_close ($connect);
}
?>
</body>
</html>
Do not forget about string sanitization !
I have found the solutions to the folowing problems:
Display results on same page
Thanks to Marc B
A single-page form+submit handler is pretty basic:
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
... form was submitted, process it ...
... display results ...
... whatever else ...
}
?>
<html>
<body>
<form method="post"> ... </form>
</body>
</html>
That's really all there is.
Only first value is showing
I resolved this problem by adding this to my code:
while($row = mysql_fetch_array($res)) {
$surname=$row['Surname'];
$name=$row['Name'];
$dob=$row['DOB'];
$address=$row['Address'];
$RPS=$row['RPS'];
Now all the values are being displayed instead of only the first one.
Display results on same page
Well I've stumbled upon this with the same problem
and I found out you can simply require the other file.
include_once("PATH_TO_FILE")'.
in /citizens.php
<?php include_once="infoct.php" ?>
<form> ... </form>
<div>
<?php $yourdata ?>
</div>
$yourdata should be html.
Do not forget about string sanitization !
Make sure to remove action from the form
Better than having all logic and Html in one file.
Related
So I have a script using HTML, PHP, and mysql, and I want to display a button under certain circumstances.
Here is my script:
<?php
include_once('dbconnect.php');
$q = $_POST['q'];
$q = $_GET['query'];
$query = mysqli_query($conn,"SELECT * FROM `Persons` WHERE `id` LIKE '%$q%'");
$count = mysqli_num_rows($query);
if($count != "1"){
$output = '<h2>No result found!</h2>';
}else{
while($row = mysqli_fetch_array($query)){
$s = $row['name'];
$output .= '<h2>Found: '.$s.'</h2><br>';
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Search</title>
</head>
<body>
<form method="POST" action="index.html">
<input type="submit" name="return" value="Return">
</form>
<?php echo $output; ?>
</body>
</html>
Specifically, I want to display the return button only when the output is "No results found", when the amount of rows in the SQL database matching the given query is not 1. How could I go about accomplishing this? I'm relatively new to PHP and mySQLi, but from my research I couldn't figure out how to do such a task, any ideas?
<?php
if ($count==0) {
echo '<input type="submit" name="return" value="Return">';
}
?>
If you want a much cleaner html code, do this:
<form method="POST" action="index.html">
<?php if ($count!= "1") : ?>
<input type="submit" name="return" value="Return">
<?php else : ?>
<!-- put your other button here -->
<?php endif; ?>
</form>
You can read more about escaping from HTML here.
<?php
include_once('dbconnect.php');
$q = $_POST['q'];
$q = $_GET['query'];
$query = mysqli_query($conn,"SELECT * FROM `Persons` WHERE `id` LIKE '%$q%'");
$results = mysqli_fetch_array($query);
?>
<!DOCTYPE html>
<html>
<head>
<title>Search</title>
</head>
<body>
<form method="POST" action="index.html">
<input type="submit" name="return" value="Return">
</form>
<?php if(0 < count($results)) ?>
<?php foreach($results AS $row) : ?>
<h2><?= $row['name'] ?></h2>
<?php endforeach; ?>
<?php else : ?>
<H2> No results found!</h2>
<?php endif; ?>
</body>
</html>
I displayed a set of questions.
On selecting one of those questions for which I required the answers to be posted on the same page, but the problem is I am not getting id parameter from url which I am passing on selectiong the question.
Now when I am trying to answer the question which is selected, I will need the id of question to post the answer of that perticular question which I already have in url as a parameter for Example: id=1.
Here is the body section of html page:
<?php
include("menu/menu.php");
$sqli = "SELECT * FROM forum_question where id='$id'";
$result=mysqli_query($conn,$sqli);
?>
<form action="submit_answer.php" method="post" name="answers">
<br> <br> <br>
<?php
while($row = mysqli_fetch_array($result))
echo "Q".$row['detail'];
?>
<br>
answers:<br>
<textarea class="tinymce" name="answers"></textarea>
<input type="hidden" name="id" value="<?php echo $id;?>">
<br> <br>
<input type="submit" value="submit" name="submit">
After submit the page "submit_answer.php", Code is:
<?php
include'config.php';
if($conn){
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$answers = $_REQUEST['answers'];
$id= $_GET ['id'];
}
$sqli= "INSERT INTO answers (answers)
VALUES ('$answers')";
if (mysqli_query( $conn,$sqli))
{
echo "New record created successfully";
header("location:answer.php?id='$id'");
} else {
echo "Error: " . $sqli . "<br>" . $conn->error;
}
}else{
}
mysqli_close($conn);
?>
Basically I am very much fresher in Php I just want to know how should I get the id of question and submit it to the "submit_answer.php" with the answer content.
just take a hidden field below the answer field and get the url parameter to that hidden field on page load, as said by user buivankim2020 and submit submit_answer.php,
after submit get the value of that field in variable like what you do for getting the answer..
You should change it (add hidden input for id in markup html)
<?php
include'config.php';
//session_start();
$id= $_GET ['id'];
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/style.css">
<script type="text/javascript" src="tinymce/js/jquery.min.js"></script>
<script type="text/javascript" src="tinymce/plugin/tinymce/tinymce.min.js"></script>
<script type="text/javascript" src="tinymce/plugin/tinymce/init-tinymce.js"></script>
</head>
<body>
<div id="container">
<div id="main">
<?php
include("menu/menu.php");
$sqli = "SELECT * FROM forum_question where id='$id'";
$result=mysqli_query($conn,$sqli);
?>
<form action="submit_answer.php" method="post" name="answers">
<br> <br> <br>
<?php
while($row = mysqli_fetch_array($result))
echo "Q".$row['detail'];
?>
<br>answers:<br>
<textarea class="tinymce" name="answers"></textarea>
<input type="hidden" name="id" value="<?php echo $id;?>">
<br> <br>
<input type="submit" value="submit" name="submit">
</form>
</body>
</html>
submit_answer.php
<?php
include'config.php';
if($conn){
if (isset($_POST['answers']) && isset($_POST['id'])) {
$answers = $_POST['answers'];
$id= $_POST['id'];
$sqli= "INSERT INTO answers (answers) VALUES ('$answers')";
if (mysqli_query( $conn,$sqli))
{
echo "New record created successfully";
header("location:answer.php?id='$id'");
} else {
echo "Error: " . $sqli . "<br>" . $conn->error;
}
}
mysqli_close($conn);
}
?>
Hi i want to update my sql table field without page refresh in php how can i achieve i tried but my code is not working i do not know where i am wrong
index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(document).ready(function(){
$('#myForm').on('submit',function(e) {
$.ajax({
url:'assignlead.php',
data:$(this).serialize(),
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000);
}
});
e.preventDefault();
});
});
</script>
</head>
<body>
<?php
include('conn.php');
$per_page = 3;
if($_GET)
{
$page=$_GET['page'];
}
$start = ($page-1)*$per_page;
$select_table = "select * from clientreg order by id limit $start,$per_page";
$variable = mysql_query($select_table);
?>
<form class="form2" action="" method="POST" name="myForm" id="myForm">
<div style="width:100%;">
<?php
$i=1;
while($row = mysql_fetch_array($variable))
{
?>
<input type="checkbox" name="users[]" value="<?php echo $row["id"]; ?>" >
<?php
}
?>
<div class="buttons"> <span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<input class="greybutton" type="submit" value="Send" />
</div>
<?php
$sql = mysql_query("SELECT *FROM login where role=1");
while ($row = mysql_fetch_array($sql)){
?>
<input type="checkbox" name="eid[]" value="<?php echo $row["eid"]; ?>" ><?php echo $row["username"]; ?>
<?php
}
?>
</div>
</form>
</div>
</body>
</html>
assignlead.php
<?php
$conn = mysql_connect("localhost","root","root");
mysql_select_db("helixcrm",$conn);
if(isset($_POST["submit"]) && $_POST["submit"]!="") {
$usersCount = count($_POST["id"]);
for($i=0;$i<$usersCount;$i++) {
mysql_query("UPDATE clientreg set eid='" . $_POST["eid"][$i] . "' WHERE id='" . $_POST["id"][$i] . "'");
}
}
?>
<?php
$rowCount = count($_POST["users"]);
for($i=0;$i<$rowCount;$i++) {
$result = mysql_query("SELECT * FROM clientreg WHERE Id='" . $_POST["users"][$i] . "'");
$row[$i]= mysql_fetch_array($result);
$id=$row[$i]['id'];
?>
<input type="hidden" name="id[]" class="txtField" value="<?php echo $row[$i]['id']; ?>"></td>
<?php
$rowCoun = count($_POST["eid"]);
for($j=0;$j<$rowCoun;$j++) {
$result = mysql_query("SELECT * FROM login WHERE eid='" . $_POST["eid"][$j] . "'");
$row[$j]= mysql_fetch_array($result);
$eid=$row[$j]['eid'];
?>
<input type="hidden" name="eid[]" class="txtField" value="<?php echo $row[$j]['eid']; ?>">
<?php
}
}
?>
i tried a lot but i am not able to get my output
How can i achieve my output
Thanks in advance
Update your javascript like this :
$(document).ready(function(){
$('#myForm').submit(function(){
$.ajax({
url : 'assignlead.php',
data : $(this).serialize(),
type : 'POST',
success : function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000);
}
});
// !important for ajax form submit
return false;
});
});
My aim to is to Update Value in Database By using Update Query . On my first page i have just displayed database table in webpage. Then by using hyperlink i have to click on Edit to second page "edit.php".While on first page i have to get the value of id and send it to second page. Where a input form is displayed which gets Value casually but Id through hidden tag. On third page getting the values query is implented but the value of id is missing.
First Page
<html>
<head>
<title>Assignment</title>
</head>
<body>
<?php
$con=mysql_connect("localhost","root","");
// Check connection
if (!mysql_connect()) {
echo "Failed to connect to MySQL: " . mysql_connect_error();
}
$db=mysql_select_db("assignment",$con);
$result = mysql_query("SELECT * FROM teacher ",$con);
?><table cellpadding="2px" border="2px"><?php
while($row = mysql_fetch_array($result)) {
?> <tr>
<td><a href="edit.php?id=<?php
echo $row['id']; ?>">Edit</a > Delete
</td><td>
<?php
echo $row['id']; ?></td><td> <?php echo $row['name'];?></td><td><?php echo $row['program']; ?></td>
<?php }
?></table><?php
mysql_close($con);
?>
</body>
</html>
Secnod Page edit.php
<html>
<head>
<title>Assignment Edit</title>
</head>
<body>
<?php
$id = $_GET['id'];
?>
<form action="update.php" method="get">
Address <input type="text" name="program"><br>
<input type="hidden" name="id" value='<?php $id?>'>
<input type="submit" name="submit">
</form>
</body>
</html>
Third Page update.php
<html>
<head>
<title>Update Page</title>
</head>
<body>
<?php
$add=$_GET['program'];
$id=$_GET['id'];
$con=mysql_connect("localhost","root","");
// Check connection
if (!mysql_connect()) {
echo "Failed to connect to MySQL: " . mysql_connect_error();
}
$db=mysql_select_db("assignment",$con);
$query = "UPDATE teacher SET program='$add' WHERE id =".$id;
echo $query;
$result = mysql_query($query,$con);
/* while($row = mysql_fetch_array($result)) {
echo $row['id'] ." " . $row['name']." ". $row['address']."<br>";
}
mysql_close($con);
*/
?>
</body>
</html>
output
UPDATE teacher SET program='openSource' WHERE id =
you need to change this
<input type="hidden" name="id" value='<?php $id?>'>
to
<input type="hidden" name="id" value='<?php echo $id?>'>
(or)
<input type="hidden" name="id" value='<?=$id?>'>
I have a problem with this statement:
after this i just get blank page.
so the whole code looks like this :
im tyrying to use html in php and than php again in html:
the whole rest works and if i replace '' with add2.php it works but it writes something before i even pick something
<?php
function login()
{
echo "?";
}
$get = $_GET['Login'];
$get = $_POST['Login'];
echo $get;
var_dump($get);
?>
<!DOCTYPE html>
<html>
<head>
<title>Login</title>
</head>
<body>
<div class="content">
<?php
require 'connection.php';
include 'user_verify.php';
include 'access_verify.php';
mysql_select_db("idoctor_db") or die("Bląd podczas wybierania bazy danych");
$select = 'SELECT * FROM users;';
$query = mysql_query($select);
// Ustaw domyślny element; tutaj są ustawione kreseczki, żeby nic nie sugerować ;P
echo '<form action="'<?php login(); ?>'" method="post">
Jezyk <select name="Login"><option value="0">------------------</option>';
while ($language = mysql_fetch_object($query))
{
echo '<option value="'.$language->Login.'" selected>'.$language->Login.'</option>';
}
echo '</form>';
?>
<input type="submit" value="Login" name="submit"/>
</div>
</body>
</html>
The HTML form element doesn't render anything on screen... add some content and you will see it is working.
<?php
function login()
{
echo "?";
}
?>
<?php
echo '<form action="<?php login(); ?>" method="post">';
echo 'Hello World!';
echo '</form>';
?>
Make sure your quotes are properly closed too. (I'm not sure if the missing closing single quote in your sample was a copy/paste error or not)
<?php
echo '<form action="<?php login(); ?>" method="post">
?>
Is a syntax error, you forgot the closing single quote and semicolon.
That said, I do not think this code will do what you expect: By using echo you will send "<?php login(); ?>" to the browser, not run it in the PHP interpreter.
This:
<?php
echo '<form action="<?php login(); ?>" method="post">
?>
Should be:
<?php
echo "<form action='" . login() . "' method='post'>";
?>
But then your login function needs to be fixed because printing out ? is not going to work.
function login()
{
return "login.php";
}