I am beginner in PHP and i want to insert value into my database table for my drop-down value using PHP insert query, I am having an error to insert data and also want to update the existence value of database table when insert new value into database table, here is my code help me to solve it, My Code is below :
<?php
include 'dbconfig.php';
?>
<html>
<head>
<title>Sales Panel App</title>
</head>
<body>
<form method="POST" name="myForm" action="<?php $_PHP_SELF ?>">
<table border="1">
<tr>
<td>Sales Person List</td>
<td>
<?php
$sql="select name,id from sales order by name";
echo "<select name = 'salelist' value=''>Sales Person Name</option>";
echo "<option value = 'Select Sales Person' selected>Select Sales Person</option>";
{
foreach ($conn->query($sql) as $row){
echo "<option value=$row[id]>$row[name]</option>";
}
echo "</select>";
}
echo "<br>";
?>
</td>
</tr>
<tr>
<td>No of Panels</td>
<td><input type="text" name="panelnumber" size="1"></td>
</tr>
</table>
<p><input type="submit" value="Submit" name="submit" id="submitdata"></p>
</form>
<?php
$salelist = $_POST['salelist'];
$panelnumber = $_POST['panelnumber'];
$sql2 = "select * from sales where name = $salelist";
$result = mysqli_query($sql2);
if ($row = mysqli_fetch_array($result)) {
$oldvalue = $panelnumber;
$newvalue = $oldvalue + $panelnumber;
$query = 'update sales where name = $panelnumber';
}
else {
$qry = "insert into sales (id, name, panelnumber) values ('', $salelist, $panelnumber)";
}
?>
</body>
</html>
Your problem is here
$query = 'update sales where name = $panelnumber';
You are not saying what to update. You have to use one of the methods available to update an SQL row.
For example, a correct one should be:
$query = "update sales set city='rio' where name = '$panelnumber'";
Here is the official MySQL Documentation.
And you have one more error, if you write this:
$number = 5;
$variable = '$number';
echo $variable;
The output of $variable will not be "5", will be "$number", literally. If you want to set a variable in a string, you must us double quote (") not single quote (') because single quote will literally save what is between them, and double will try to "translate" variables and other stuff.
The correct way would be:
$number = 5;
$variable = "$number";
echo $variable;
Here you have a detailed PHP official explanation about what I am describing.
Related
i am trying to bulid an online attendace in php, i am fetching student list from student table, and want to put all chkboxs if checkd as present and if unchecked as absent,
i am unable to do that.
<div class="attendance">
<form accept="att.php" method="POST">
<?php
$sel_sql = "SELECT * FROM student";
$run_sql = mysqli_query($conn,$sel_sql);
while($rows = mysqli_fetch_assoc($run_sql)){
$id = $rows['id'];
echo '<input type="checkbox" name="status" value="'.$id.'" checked>'.ucfirst($rows['f_name']).'';
}
?>
<input type="submit" name="submit_attendance" value="Post Attendance">
<?php echo $msg; ?>
</form>
</div>
it shows prefect students list, but i dont know how to set insert query for all of these chkboxes
try this query to insert from another table
SELECT * INTO TABLE2 FROM student
use where condition on student table as where student.column value to select checked values
Applied same above things with checkbox input field
echo '<input type="checkbox" name="status" value="'.$id.'" 'if($row['present_field_of_database']) ? 'checked' : ''
'>'.ucfirst($rows['f_name']).'';
it's fine then updated code with your problems i hope it's work for you
<div class="attendance">
<form accept="att.php" method="POST">
<?php
$sel_sql = "SELECT * FROM student";
$run_sql = mysqli_query($conn,$sel_sql);
while($rows = mysqli_fetch_assoc($run_sql)){
$id = $rows['id'];
// if your $id value is right from $rows['id'] then
// change your student table name to the another table where available status of the student
$wh_sql = "SELECT * FROM student WHERE id=".$id;
$run_wh_sql = mysqli_query($conn, $wh_sql);
$Wh_rows = mysqli_fetch_assoc($run_wh_sql);
// add student present or absent value to the rows data
$rows['status'] = $Wh_rows['status'];
}
// set value as A or P respectively absent or present add jquery plugins for onclick event while client click on checkbox change value respectively
echo '<input type="checkbox" name="status" value="'.$rows['status'].'" 'if($rows['status'] == "A") ?'checked': '' '>'.ucfirst($rows['f_name']).' onclick= "$(this).val(this.checked ? P : A)"';
?>
<input type="submit" name="submit_attendance" value="Post Attendance">
<?php echo $msg; ?>
</form>
</div>
if (isset($_POST['send'])) {
$s_id = $_POST['status'];
$id = $_POST['student'];
if ($s_id) {
foreach ($s_id as $s) {
foreach ($id as $i) {
if (mysqli_query($conn, "INSERT INTO attendence(s_id,status) VALUES('".
mysqli_real_escape_string($conn, $s)."','".
mysqli_real_escape_string($conn, $i)."')")) {
$msg = "success";
}else{
$msg = "failed";
}
}
}
}
}
i have 3 students. when i press send it sends 9 entries. i am unable to understand how to insert all students data
This is attendance table
This is attandance table
i want to put entries like this if check box chekd it wil post p and if not it wil post a. i just need how to insert all sutdent at once quert
I'm trying to update this database, and I've verified within this script that the update is completed, and that the $nw and $p variables are correct.
<?php
session_start();
$num = (int) $_SESSION["cart"];
$cart = $num + 1;
$_SESSION["cart"] = (string) $cart;
$nme = $_POST['nameofitem'];
$pst = $_SESSION["user"];
$db = new mysqli('localhost', 'spj916', "cs4501", 'spj916');
$query = "select * from Items where Items.Id = '$nme'";
$result = $db->query($query) or die ($db->error);
$item = $result->fetch_array();
$nw = $item[5] - 1;
$p = (int) $pst;
echo $p;
$query3 = "update Items set Quantity = '$nw' where Id = '$p'";
$db->query($query3) or die ("Invalid insert " . $db->error);
$query2 = "insert into Bought (Name, Cost, BuyerID) values ('$item[1]', '$item[4]', '$pst')";
$db->query($query2) or die ("Invalid insert " . $db->error);
header("Location: store.php");
?>
However, when it redirects to this script, it echoes the information as if it weren't updated. What is the problem?
<?php
session_start();
$db = new mysqli('localhost', 'spj916', "cs4501", 'spj916');
$user = $_SESSION["user"];
$pw = $_SESSION["pw"];
# determines number of items in cart to display
if (!isset($_SESSION["category"]))
$_SESSION["category"] = "Book";
if (isset($_POST["Ccategory"])) {
$cat = $_POST["Ccategory"];
$_SESSION["category"] = $cat;
}
if (!isset($_SESSION["cart"]))
$_SESSION["cart"] = "0";
$cart = $_SESSION["cart"];
?>
<!DOCTYPE html>
<html>
<?php # setting up table with items to buy ?>
<table border = "1" border-spacing = "5px" >
<caption><h2> UVA Bookstore 2.0</h2>
<p align=right> Items in cart: <?php echo $cart?> </p> <br />
<b><i>Welcome to the new and improved bookstore with a better selection than ever</i></b>
<br/><br/>
</caption>
<tr align = "center">
<th>Item</th>
<th>Description</th>
<th>Price</th>
<th>Number left</th>
<th>Buy</th>
</tr>
<?php
$category = $_SESSION["category"];
$query = "select * from Items where Items.Category = '$category'";
$result = $db->query($query) or die ($db->error);
$rows = $result->num_rows;
for ($i = 0; $i < $rows; $i++)
{
$row = $result->fetch_array();
?>
<form action="addtocart.php"
method="POST">
<tr align = "center">
<td>
<?php
echo $row[1];
?>
</td>
<td> <?php echo $row[3];?> </td>
<td> <?php echo $row[4];?> </td>
<td> <?php echo $row[5];?> </td>
<?php # sets up add to cart button that adds item to cart ?>
<td> <input type = "hidden" name ='nameofitem'
value= "<?php echo $row[0]?>">
<input type='submit' value='Add to Cart'> </input> </td>
</tr>
</form>
<?php
}
# form to check out and go to summary page ?>
<form action = "store.php"
method = "POST">
<tr align = "center"> <td>
<select name = "Ccategory">
<option value = "Book">Books</option>
<option value = "Music">Music</option>
<option value = "Car">Cars</option>
</select>
<input type = "hidden" name = "cat"> </td>
<td> <input type = "submit" value = "Switch Category"> </td>
</form>
<form action="summary.php"
method="POST">
<td> <input type = "submit" value = "Check out"> </td> </tr>
</table><br/>
</form>
</html>
Have you tried changing
$query3 = "update Items set Quantity = '$nw' where Id = '$p'";
to
$query3 = "update Items set Quantity = '$nw' where Id = $p";
The best way to determine if an UPDATE should work is to replace it with a SELECT containing the same WHERE clause. This way you can see what rows would be changed if you were to run the original query.
Otherwise, it seems to be the case that your changes in the current transaction are never committed. Is this the only script that has an issue with updates to the database? Please see the PHP manual for more information:
//mysqli::commit -- mysqli_commit — Commits the current transaction
bool mysqli::commit ([ int $flags [, string $name ]] )
A commit should be issued when you are done doing all updates that have dependencies (or for those that are atomic), however, you don't always have to commit depending on the configuration of your server. Also, it looks like your script has SQL injection vulnerabilities as other have mentioned. It would probably be best to use prepared statements or sanitize your inputs.
I have built a checkbox group using forums and everything works on the standalone script that a created but when i add it to my user system afte selecting more than one I get array echoed where the list is supposed to be below is the production script
<?php
session_start();
include 'dbconfig.php';
$chkbox = array('option1', 'option2', 'option3');
if (isset($_POST['submit']))
{
$chkbox = $_POST['chkbox'];
$chkNew = "";
foreach($chkbox as $chkNew1)
{
$chkNew .= $chkNew1 . ",";
}
$query = "INSERT INTO demo_table_5 (MultipleValue) VALUES ('$chkNew')";
mysqli_query($conn,$query) or die(mysqli_error());
}
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>PHP Insert/Store multiple selected checkbox values in MySQL database</title>
</head>
<body>
<h2>PHP Insert multiple checkbox values in MySQL database</h2>
<form method="post" action="">
<table>
<tr>
<td><input type="checkbox" name="chkbox[]" value="option1"><label>option1</label></td>
<td><input type="checkbox" name="chkbox[]" value="option2"><label>option2</label></td>
<td><input type="checkbox" name="chkbox[]" value="option3"><label>option3</label></td>
</tr>
<tr>
<td colspan="4"><input type="submit" name="submit" Value="submit"></td>
</tr>
</table>
</form>
<?php
$sql = "SELECT MultipleValue FROM demo_table_5";
$result = mysqli_query($conn, $sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>Interests</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>".$row["MultipleValue"]."</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
so when I added it to script I put
<table>
<tr>
<td><input type="checkbox" name="chkbox[]" value="option1"><label>option1</label></td>
<td><input type="checkbox" name="chkbox[]" value="option2"><label>option2</label></td>
<td><input type="checkbox" name="chkbox[]" value="option3"><label>option3</label></td>
</tr>
in the register form its self
then I put the following in the php block at the top of the registration page
$chkbox = array('option1', 'option2', 'option3');
and I added a column to the existing userinfo table in database and named it multipleValue and added the following line to update database
MultipleValue = '".$_POST['chkbox']."',
then I added the following bit to the profile page
<?php
$sql = "SELECT MultipleValue FROM user_info";
$result = mysqli_query($conn, $sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>Interests</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>".$row["MultipleValue"]."</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
Since your inputs have name="chkbox[]", this means that $_POST['chkbox'] is an array containing the values of all the checked boxes. In PHP, when you try to use an array as a string, as in
MultipleValue = '".$_POST['chkbox']."',
the array is converted to the string "Array". You need to use implode to convert it to a comma-separated list:
MultipleValue = '".implode(',', $_POST['chkbox'])."',
However, I recommend you change your table design. Comma-separated values in a table cell are a bad way to represent lists, because it's difficult to search and update them -- indexes can't be used to search for individual values in the list, for instance. You should use a many-to-many table where there's a row for each user+option combination.
You are inserting in the db a concatenated string such as "option1,option2,option3".
Retrieving data by the SQL select statement will return one row containing the concatenation
I am writing a basic CMS system and have come across something which should be seemingly simple -but is beginning to frustrate me.!
I am trying to pass an array through a select option field to populate a list of categories in which I can save a post.
I have a 'posts' form which comprises of 3 fields. Title, content and Category ID (CatID).
When the user creates a post, they can select the category they wish to assign the post assigned to by using a drop down list - (this is populated by using a different form).
So the technical bit; -
MySQL DB:-
categories = catname (char60 PRIMARY), catid (INT10, AI)
posts = id (bigint20 PRIMARY), catid (int10 PRIMARY), title (text), content (varchar255)
Example of categories populates: catname = Home / catid = 1 ...etc
Output.php ;
<?php
function display_post_form($post = '') {
$edit = is_array($post);
?>
<form action="<?php echo $edit ? 'edit.php' : 'add.php' ; ?>" method="post">
<table border="0">
<tr>
<td> Title:</td>
<td> <input type="text" name="title" value="<?php echo $edit ? $post['title'] : '' ; ?>" size="60" /> </td>
</tr><tr>
<td> Content:</td>
<td> <textarea id="editor1" name="content" value="<?php echo $edit ? $post['content'] : '' ; ?>"> </textarea> </td>
</tr><tr>
<td> Category:</td>
<td><select name="catid">
<?php
$cat_array = get_categories($catid, $catname);
foreach($cat_array as $thiscat) {
echo "<option value=\"".$thiscat['catid']."\" ";
if (($edit) && ($thiscat['catid'] == $post['catid'])) {
echo " selected";
}
echo ">".$thiscat['catname']."</option>";
}
?>
</select>
</td>
</tr><tr>
<td> Button:</td>
<td <?php if (!$edit) { echo "colspan=2"; } ?> align="center">
<?php
if ($edit)
echo "<input type=\"hidden\" name=\"_id\" value=\"". $post['id'] ."\" />";
?>
<input type="submit" value="<?php echo $edit ? 'Update' : 'Add' ; ?> Post" />
</form></td>
</td>
</tr>
</table>
</form>
<?php
}
?>
Functions.php ;
function get_categories($catid, $catname) {
$conn = db_connect();
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL " .mysqli_connect_error();
}
$sql = "SELECT catname, catid FROM categories";
$result = mysqli_query($conn, $sql) or die(" Could not query database");
while($row = mysqli_fetch_assoc($result)) {
printf("\n %s %s |\n",$row["catname"],$row["catid"]);
}
mysqli_close($conn);
}
I am able to call in the 'get_cattegories()' function which generates a flat data of categories and their respective id's. I then combined this with the Select Option Field in the Output.php file and it doesn't generate anything.
Can anyone give some useful tips or advice? Many thanks :)
You are not returning the array but printing a string to the output. Change printf to return:
function get_categories($catid, $catname) {
$conn = db_connect();
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL " .mysqli_connect_error();
}
$sql = "SELECT catname, catid FROM categories";
$result = mysqli_query($conn, $sql) or die(" Could not query database");
$categories = array();
while($row = mysqli_fetch_assoc($result)) {
$categories[] = $row;
}
mysqli_close($conn);
return $categories;
}
Also I agree for the comments to your question. The arguments are useless.
You also may refactor the code, actually... alot. Move the mysql_connect() to the other place, probably at the beginning of your script.
I suggest to use some frameworks. I think KohanaPHP will be a good start. You will learn about architecture and some design patterns. Keep the good work and improve your skills ;-)
I am trying to delete , edit and add new recodes on the same page but it seems am failing to make it work .And I do not want to do it using ajax jquery or java script but only php .I need some help please below are my code :
<?php
include_once('con.php');
$strSQL = "SELECT film_id, name
from
filmsbox";
$rs = mysql_query($strSQL);
echo "<table border='1' ><tr bgcolor='#eeeeee'><td>Name</td> <td colspan='2'>Action</td></tr>";
while($row = mysql_fetch_assoc($rs))
{
$film_id = $row['film_id'];
$name = $row['name'];
$hometeam= mysql_real_escape_string($name);
echo "<tr bgcolor='#eeeee'><td>$name</td> <td><a href='index.php?film_id=$film_id' name ='edit'>Edit</a></td><td><a href='index.php?film_id=$film_id' name ='delete'>Delete</a></td></tr>";
}
?>
<?php
$strSQL = "SELECT film_id, name
from
filmsbox";
$rs = mysql_query($strSQL);
$row = mysql_fetch_assoc($rs);
$film_id= $row['film_id'];
$name = $row['name'];
$name = mysql_real_escape_string($name);
$film_id= $_GET['film_id'];
?>
<?php
if(isset($_POST['edit'])){
?>
<table>
<form action="index.php" method="post">
<tr>
<td>
Name
</td>
<td>
<input type = "text" name = "name" value="<?php echo $name;?>">
</td>
</tr>
<input name="film_id" type="hidden" id="film_id" value="<?php echo $film_id; ?>">
<tr>
<td>
<input type = "submit" name = "submit" value="update">
</td>
</tr>
<?php
$name = (isset($_POST['name']))? trim($_POST['name']): '';
$film_id = $_POST['film_id'];
$sql = "UPDATE filmsbox SET name='$name'
WHERE film_id ='$film_id'";
$result = mysql_query($sql);
if($result)
{
echo "Success";
}
else
{
echo "Error";
}
}
?>
<?php
/*Delete section*/
if(isset($_POST['delete']))
{
$film_id = $_GET['film_id'];
$delete = "DELETE FROM filmsbox WHERE film_id = '$film_id'";
$result = mysql_query($delete);
if($result)
{
echo "Record deleted successfuly ";
}
else
{
echo "No data deleted";
}
}
?>
Couple of pointers:
You only need to escape values before they go into the database, not when they come out and are used in HTML i.e $hometeam = mysql_real_escape_string($name);
You are pulling the same query from the database twice in quick succession which is not needed. You can remove one of the 2 $strSQL = "SELECT film_id, name
from
filmsbox";
$rs = mysql_query($strSQL); sections from the top of your code
You need to run any update/delete queries on the data before you then do your select query to pull out the records for the page, otherwise your changes will not be shown
You should be escaping the values for your update and delete queries to prevent SQL injection
Edit:
To reload the page in an edit mode, you need to change the link URL in the table to something like
<a href='index.php?film_id=$film_id&edit=1' name ='edit'>Edit</a>
Then your edit block needs to be
if ($_GET['edit']) {
I want to be clear this is not in any way a secure method of editing values, as anyone can put ?edit=1 on the url and get to the form