I fetch total_amount value and insert into another table but i want to insert total_amount in one field and seperated by ",". How to do that?
I know Mysqli is latest version . but here mysql is working properly.
<?php
include('database/db.php');
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
$sql1="Insert into test(`total_amount`) values ('{$_POST['i']}')";
$result1=mysql_query($sql1);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit Page</title>
</head>
<body>
<form name="" action="<?php $_SERVER['PHP_SELF'] ?>" method="post">
<table>
<tr>
<td>Total Amount</td>
</tr>
<?php
$sql="Select * from tes1";
$result=mysql_query($sql);
while($dtset=mysql_fetch_array($result))
{
?>
<tr>
<td><input type="hidden" name="i" value="<?php echo $dtset['total_amount'];
?>"><?php echo $dtset['total_amount']; ?></td>
</tr>
<?php } ?>
</table>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
<?php
$sql="Select * from tes1";
$result=mysql_query($sql);
// Empty array
$amounts = [];
while($dtset=mysql_fetch_array($result))
{
// Push the amounts to the array.
$amounts[] = $dtset['total_amount'];
}
// Implode will create a string from your array and seperates it with
// the chosen character
$combined = implode(",",$amounts);
?>
<tr>
<td>
<input type="hidden" name="i" value="<?php echo
$combined; ?>">
<?php echo $combined; ?>
</td>
</tr>
The while loop is now not part of the table row, but instead creates the array with all values first and then you can add the string created by implode to your hidden form field.
Related
I need to display an HTML table with the data that i get from a PhP form. I did a thing like this but It doesn't work properly. It shows the data row on top of the title of the columns. Here's the code:
<html>
<body>
<form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
Nome: <input type="text" name="nome">
Ragione sociale: <input type="text" name="rag">
Indirizzo: <input type="text" name="ind">
Partita IVA: <input type="text" name="iva">
<input type="submit">
</form>
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = htmlspecialchars($_REQUEST['nome']);
$rag = htmlspecialchars($_REQUEST['rag']);
$ind = htmlspecialchars($_REQUEST['ind']);
$iva = htmlspecialchars($_REQUEST['iva']);
}
?>
<table>
<tbody>
<tr>
<?php
$a = array($name,$rag,$ind,$iva);
for ($i=0; $i<count($a); $i++){
print_r($a[$i]);
echo " ";
}
?>
</tr>
<tr>
<td>|Nome|</td>
<td>Ragione Sociale|</td>
<td>Indirizzo|</td>
<td>Partita IVA|</td>
</tr>
</tbody>
</table>
</body>
</html>
You print your data in the first row, without outputting any <td> for your data. Then in the second row you print your titles/headers. Change your table like so, see if that's what you want:
<table>
<thead>
<tr>
<th>|Nome|</th>
<th>Ragione Sociale|</th>
<th>Indirizzo|</th>
<th>Partita IVA|</th>
</tr>
</thead>
<tbody>
<tr>
<?php
$a = array($name,$rag,$ind,$iva);
for ($i=0; $i<count($a); $i++){
echo "<td>";
echo $a[$i];
echo "</td>";
}
?>
</tr>
</tbody>
</table>
More on tables: https://developer.mozilla.org/en-US/docs/Web/HTML/Element/table
Edit: I'd also suggest to place your table inside the if ($_SERVER["REQUEST_METHOD"] == "POST") { so it only gets printed after you submitted the form
<?php
include("dbFunctions.php");
$query ="SELECT * FROM `physical_examination` WHERE `PE_Opt_ans`= 0";//select form options name
$result = mysqli_query($link,$query);
?>
<div id="tabs-3">
<table>
<form action="<?php echo $_SERVER['PHP_SELF']?>" method="post" name="tab0">
<?php while ($arrayResult = mysqli_fetch_array($result)){ ?>
<tr>
<td><label for="input"><b><?php echo $arrayResult['PE_Opt_name']?></label></td>
<?php if ($arrayResult['PE_Opt_type'] == "textarea") { ?>
<td><textarea rows="8" cols="45"name = "other1"></textarea></td>
<?php } else { ?>
<td><input type="<?php echo $arrayResult['PE_Opt_type']?>" name="input<?php echo $arrayResult['id']?> " ></td>
</tr>
<?php } ?>
<?php } ?>
<br> <input value="Submit" type="submit" name="submit1">
</form>
</table>
<?php
include "dbFunctions.php";
if(isset($_POST['submit1'])) {
$number = $_POST['other1'];
I am stuck here.
How do I $_POST the form based on the while loop after I click on submit button? Do I need another while loop again for the name value of input <?php echo $arrayResult['id']?>
I didn't quiet get what you want but i assume you want the contents of the textarea generated by the while loop.
Change in your loop the names from input to input[] and other1 to other1[]
If you var_dump($_POST) afterwards you should see that they are now an array which you can loop over with foreach
PS: You should NEVER EVER eat raw $_POST/$_GET inputs and PHP_SELF is another security breach. But that's just my two cents
i have this code in PHP and a database sql.. the situation is .. if i type the 1, 2 or 3 (productID) .. the textbox will be populated and field with database values.. but when i run the program.. fortunately it has no errors.. but when i type the id or 1 and click the submit button.. it doesnt get the neccessary values.. sorry for this im a complete newbie and im practicing PHP for a while now.. any help will do.. thank you..
<?php
session_start();
include_once 'dbconnect.php';
if(!isset($_SESSION['user'])){
header("Location: index.php");
}
$res = mysql_query("SELECT * FROM users WHERE user_id=".$_SESSION['user']);
$userRow = mysql_fetch_array($res);
?>
<?php
require('dbconnect.php');
$id = (isset($_REQUEST['productID']));
$result = mysql_query("SELECT * FROM tblstore WHERE productID = '$id'");
$sql = mysql_fetch_array($result);
if(!$result){
die("Error: Data not found");
} else {
$brandname = $sql['brandname'];
$price = $sql['price'];
$stocks = $sql['stocks'];
}
?>
<html>
<body>
<p>
hi' <?php echo $userRow['username']; ?> Sign Out
</p>
<form method="post">
<table align="center">
<tr>
<td>Search Apparel:</td>
<td><input type="text" name="search" name="productID" /></td>
</tr>
<tr>
<td>Brandname:</td>
<td><input type="text" name="brandname" value="<?php echo $brandname; ?>"/ </td>
</tr>
<tr>
<td>Price:</td>
<td><input type="text" name="price" value="<?php echo $price; ?>"/></td>
</tr>
<tr>
<td>Stocks:</td>
<td><input type="text" name="stocks" value="<?php echo $stocks; ?>"/></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="submit" value="Search" /></td>
</tr>
</table>
</form>
</body>
</html>
your getting the id incorrectly, you have:
<?php
$_REQUEST['productID']=8; //for testing
$id = (isset($_REQUEST['productID']));
if you check it you will find the output is true\false as returned by isset
var_dump($id); //true
what you should use is:
<?php
if(isset($_REQUEST['productID'])){ //maybe also check its a number and or valid range
$id=$_REQUEST['productID'];
}
I'm struggling now for a few days to get the value of a checkbox in my code.
Basically I have an admin-page where the customer can select and deselect images that will put online.
You can select and deselect images that will be shown on the homepage, and separate on the gallery-page. Both checked is also possible.
I have another checkbox that can be selected to remove the image from the list(image_deleted).
There is still a database entry and the images are still on file-system but later on I'll create a cleanup-job.
Here is my code:
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
ob_start();
require('../../lib/dbconnection.php');
require("../../lib/checklogin.php");
require("includes/upload.inc.php");
$query = 'SELECT * FROM gallery where image_deleted != 1 order by id desc';
$result=$conn->query($query);
$count=$result->num_rows;
?>
<!DOCTYPE html>
<html>
<head>
<title>Classic Nails - CMS</title>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="ClassicNails">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="../css/screen.css">
<link rel="stylesheet" href="../css/libs/magnific-popup.css">
<script src="../js/libs/min/jquery-min.js" type="text/javascript"></script>
<script src="../js/min/custom-min.js" type="text/javascript"></script>
<script src="js/jquery.magnific-popup.js"></script>
<script>
$(document).ready(function() {
$('.image-link').magnificPopup({
type:'image',
gallery:{
enabled:true
}
});
});
</script>
</head>
<body>
<?php include('includes/header.inc.php'); ?>
<?php include('includes/nav.inc.php'); ?>
<div class="wrapper">
<article class="content">
<h1>Foto gallery</h1>
<?php
if (isset($uploadResult)) {
echo "<p><strong>$uploadResult</strong></p>";
}
?>
<form action="" method="post" enctype="multipart/form-data" name="uploadImage" id="uploadImage">
<p>
<label for="image">Upload image:</label>
<input type="hidden" name="MAX_FILE_SIZE" value="<?php echo MAX_FILE_SIZE; ?>" />
<input type="file" name="images" id="imagesd" />
</p>
<p>
<input type="submit" name="upload" id="upload" value="Upload" />
</p>
</form>
<div id="maincontent">
<h2>Foto informatie</h2>
<form name="FotoInformatie" id="fotoInformatie" method="post" action="">
<table>
<tr>
<td align="center"><strong>Foto<strong></td>
<td align="center"><strong>Titel</strong></td>
<td align="center"><strong>Beschrijving</strong></td>
<td align="center"><strong>Homepage</strong></td>
</tr>
<?php
while ($rows=$result->fetch_assoc()) {
?>
<tr>
<td class="hide" align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td>
<td><img src="../img/thumbs/<?php echo $rows['filename']; ?>"></td>
<td align="center"><input name="title[]" type="text" id="title" value="<?php echo $rows['title']; ?>"></td>
<td align="center"><input name="caption[]" type="text" id="caption" value="<?php echo $rows['caption']; ?>"></td>
<td><input type="checkbox" name="checkboxHome[]" id="checkBoxHome" value="<?php echo ($rows['home'] == 1) ? 'checked="checked"' : ''; ?>"/></td>
</tr>
<?php
}
?>
<tr>
<td colspan="4" align="center">
<input type="submit" name="submit" value="Submit">
</tr>
</table>
</form>
</div>
</article> <!-- end of content -->
</div> <!-- end of container -->
<?php include('includes/footer.inc.php'); ?>
</body>
</html>
<?php
if(isset($_POST['submit'])) {
$title = $_POST['title'];
$caption = $_POST['caption'];
if ($_POST['checkboxHome'] == "") {
$checkboxHome[] = '0';
} else {
$checkboxHome[] = '1';
}
for($i=0;$i<$count;$i++){
$result1=mysqli_query($conn, "UPDATE gallery SET title='$title[$i]', caption='$caption[$i]', home='$checkboxHome[$i]' WHERE id='$id[$i]'");
header("location:/admin/foto-admin.php");
}
}
?>
The checkbox only works on the first row in my DB. When I select another record, only the first record in my db will be updated.
Another issue is that my checkbox won't be checked so I don't know based on my screen when a image is online or not. in the database I see a 1 of a 0.
I know that sql-injection is possible and I have to prepare the statements, but that is the next step when I get this checkbox-issue working.
Hope someone can help me with my code. It's giving me a headache.
Check these
Attribute name="id[]" for id field is not given. And it should get inside
if(isset($_POST['submit'])) {
$id = $_POST['id'];
}
Incorrect spelling in getting Post value
change
$checkboxHome = $_POST['checkboxHome'];
$checkboxFotoboek= $_POST['checkboxFotoboek'];
$checkboxDelete = $_POST['image_deleted'];
to
$checkboxHome = $_POST['checkBoxHome'];
$checkboxFotoboek= $_POST['checkBoxFotoboek'];
$checkboxDelete = $_POST['checkboxDelete'];
You are trying to get wrong value.
Your check-box name is checkBoxHome and you are trying to get $_POST['checkboxHome'] instead of $_POST['checkBoxHome'] .
Try $_POST['checkBoxHome'] and print it as print_r('checkBoxHome')
Same mistake in checkBoxFotoboek check-box.
try this
if(isset($_POST['submit'])) {
$title = $_POST['title'];
$caption = $_POST['caption'];
$checkboxHome = $_POST['checkBoxHome'];
$checkboxFotoboek= $_POST['checkBoxFotoboek'];
$checkboxDelete = $_POST['checkboxDelete'];
for($i=0;$i<$count;$i++){
$result1=mysqli_query($conn, "UPDATE gallery SET title='$title[$i]', caption='$caption[$i]', home='$checkboxHome[$i]', fotoboek='$checkboxFotoboek[$i]', image_deleted='$checkboxDelete[$i]' WHERE id='$id[$i]'");
header("location:/admin/foto-admin.php");
}
}
?>
I want to call php function in form action and i want to pass id as a argument. What I am doing is, in html form database column values will be displayed in text boxes, If I edit those values and click 'update' button values in database should be updated and 'Record updated successfully'message should be displayed in same page. I tried below code but not working. Let me know the solution. Thanks in advance.
<html>
<head>
<link rel="stylesheet" type="text/css" href="cms_style.css">
</head>
<?php
$ResumeID = $_GET['id'];
$con = mysql_connect("localhost", "root", "");
mysql_select_db("engg",$con);
$sql="SELECT * from data WHERE ResumeID=$ResumeID";
$result = mysql_query($sql);
$Row=mysql_fetch_row($result);
function updateRecord()
{
//If(!isset($_GET['id']))
//{
$NameoftheCandidate=$_POST[NameoftheCandidate];
$TelephoneNo=$_POST[TelephoneNo];
$Email=$_POST[Email];
$sql="UPDATE data SET NameoftheCandidate='$_POST[NameoftheCandidate]', TelephoneNo='$_POST[TelephoneNo]', Email='$_POST[Email]' WHERE ResumeID=$ResumeID ";
if(mysql_query($sql))
echo "<p>Record updated Successfully</p>";
else
echo "<p>Record update failed</p>";
while ($Row=mysql_fetch_array($result)) {
echo ("<td>$Row[ResumeID]</td>");
echo ("<td>$Row[NameoftheCandidate]</td>");
echo ("<td>$Row[TelephoneNo]</td>");
echo ("<td>$Row[Email]</td>");
} // end of while
} // end of update function
?>
<body>
<h2 align="center">Update the Record</h2>
<form align="center" action="updateRecord()" method="post">
<table align="center">
<input type="hidden" name="resumeid" value="<? echo "$Row[1]"?>">
<? echo "<tr> <td> Resume ID </td> <td>$Row[1]</td> </tr>" ?>
<div align="center">
<tr>
<td> Name of the Candidate</td>
<td><input type="text" name="NameoftheCandidate"
size="25" value="<? echo "$Row[0]"? >"></td>
</tr>
<tr>
<td>TelephoneNo</td>
<td><input type="text" name="TelephoneNo" size="25" value="<? echo "$Row[1]"?>"></td>
</tr>
<tr>
<td>Email</td>
<td><input type="text" name="Email" size="25" value="<? echo "$Row[3]"?>">
</td>
</tr>
<tr>
<td></td>
<td align="center"><input type="submit" name="submitvalue" value="UPDATE" ></td>
</tr>
</div>
</table>
</form>
</body>
</html>
try this way
HTML
<form align="center" action="yourpage.php?func_name=updateRecord" method="post">
PHP
$form_action_func = $_POST['func_name'];
if (function_exists($form_action_func)) {
updateRecord();
}
write form action="" and then write your php code as below
note : use form method as get
<?php
if(isset($_GET['id']))
{
call your function here
}
?>
in function access all values using $_GET['fieldname']
simple way make your "Submit " and "Update" action performed on same page then
if(isset($_POST['update']))
{
//perform update task
update($var1,var2,$etc); // pass variables to function
header('Location: http://www.example.com/');// link to your form
}
else if(isset($_POST['submit']))
{
//perform update task
submit($var1,$var2,$etc);// pass variables to function
header('Location: http://www.example.com/'); // link to next page after submit successfully
}
else
{
// display form
}