Retrieve Value with insert PDO - php

I'm working on a page where the user can add rows with fields of a form, dynamic, depending on the model this page http://www.linhadecomando.com/scripts/input-dinamico/
I'm having problem RECOVER FROM values ​​of the second row inserted as the first I can get quietly. Check my code:
<?php
// exibindo os dados
if ($_POST){
$IDCliente = $_POST['IDCliente'];
$IDBanco = $_POST['Banco'];
$Titular = $_POST['Titular'];
$Agencia = $_POST['Agencia'];
$CC = $_POST['CC'];
$quant_linhas = count($IDBanco);
for ($i=0; $i<$quant_linhas; $i++) {
//$pdo = conectar();
$sql = "INSERT INTO Clientes_Referencias_Bancos (IDCliente, IDBanco, Titular, Agencia, CC) VALUES (:IDCliente, :IDBanco, :Titular, :Agencia, :CC)";
$cadastrar = $pdo->prepare($sql);
$cadastrar ->bindValue(':IDCliente', $IDCliente[$i]);
$cadastrar ->bindValue(':IDBanco', $IDBanco[$i]);
$cadastrar ->bindValue(':Titular', $Titular[$i]);
$cadastrar ->bindValue(':Agencia', $Agencia[$i]);
$cadastrar ->bindValue(':CC', $CC[$i]);
$cadastrar ->execute();
if($cadastrar->rowCount() == 1):
echo "Departamento cadastrado com sucesso!<br/>";
else:
echo "Erro ao cadastrar o Departamento!";
endif;
echo "Cliente: ".$IDCliente[$i]."<br />";
echo "Banco: ".$IDBanco[$i]."<br />";
echo "Titular: ".$Titular[$i]."<br />";
echo "Agência: ".$Agencia[$i]."<br />";
echo "Conta Corrente: ".$CC[$i]."<br />";
}
}
?>
Here is my form...
<select name="Banco[]">
<input type="text" name="Titular[]" id="Titular[]" />
<input type="text" name="Agencia[]" id="Agencia[]" />
<input type="text" name="CC[]" id="CC[]" />
<input name="IDCliente[]" id="IDCliente[]" type="hidden" value="1" />

You can use transactions. You should first prepare the query. In for loop you should bind variables and execute the queries

Related

how to insert multidimensional array using for loop

I have 2 tables. One is for students and one of subjects. I want to fetch data from the students and subjects tables on same page. In front of student 1 subject 1 subject 2 subject 3. Then in front of student 2 subject 1 subject 2 subject 3. It is to submit result.
I have done this.
Than I have to insert this in 3rd table of results. I have successfully inserted students in result table. And subjects. But on the time of marks, I am unable to insert. I used a multidimensional array. I am unable to insert marks data into array. How can I do this?
Let me show some snapshots.
My multidimensional array is not working, and I don't know how to do this. Kindly help me out.
This is a detailed pic: This is the detailed snapshot.
// count students
$sql = "SELECT * FROM tb_students";
$run = mysqli_query($mysqli,$sql);
$total_students = mysqli_num_rows($run);
$numbers=$total_students;
for($i=1;$i<=$numbers;$i++)
{
while ($rows = mysqli_fetch_assoc($run)) {
$id = $rows['student_id'];
$name = $rows['student_name'];
?>
<tr>
<td>Name</td>
<td hidden><input type="text" value="<?php echo $id?>" name="student_id[]" /></td>
<td><?php echo $name ?> </td>
</tr>
<input type="hidden" value="<?php echo $numbers;?>" name="numbers" />
<?php
$sel_sub = "SELECT * FROM subjects WHERE class_name = '1st year'";
$run_sub = mysqli_query($mysqli,$sel_sub);
$total_sub = mysqli_num_rows($run_sub);
for ($k=0; $k < $total_sub ; $k++) {
while ($rows = mysqli_fetch_assoc($run_sub)) {
$sub_id = $rows['sub_id'];
$sub_name = $rows['sub_name'];
?>
<tr>
<td><?php echo $sub_name; ?></td>
<td hidden><input type="" value="<?php echo $sub_id;?>" name="sub_id[]" /></td>
<input type="hidden" value="<?php echo $total_sub;?>" name="subject" />
<td><input type="text" name="marks[][]" placeholder="Marks" /></td>
</tr>
<?php
}
}
?>`
and this is isnert query
<?php
$mysqli = mysqli_connect("localhost","salman","salman1214","shan");
if(mysqli_connect_errno())
die("Connection failed".mysqli_connect_error());
$s = '';
for($i=0;$i<$_POST['numbers'];$i++)
{
for($j=0;$j<$_POST['subject'];$j++)
{
$s = "insert into result(student_id,exam_name, subject_name, sub_marks) values";
$s .="('".$_POST['student_id'][$i]."','".$_POST['exam_name']."','".$_POST['sub_id'][$j]."','".$_POST['marks'][$i][$j]."'),";
$s = rtrim($s,",");
if(!mysqli_query($mysqli,$s))
echo mysqli_error();
else
echo "Records Saved <br />";
$sub_list = $_POST['marks'][$i][$j];
echo $sub_list;
}
}
mysqli_close($mysqli);?>
I don't want to say if this way is the best way.
But, your problem is you are using the lines in the loops which should not. Try this:
<?php
$mysqli = mysqli_connect("localhost","salman","salman1214","shan");
if(mysqli_connect_errno())
die("Connection failed".mysqli_connect_error());
$s = '';
$s = "insert into result(student_id,exam_name, subject_name, sub_marks) values";
for($i=0;$i<$_POST['numbers'];$i++)
{
for($j=0;$j<$_POST['subject'];$j++)
{
$s .="('".$_POST['student_id'][$i]."','".$_POST['exam_name']."','".$_POST['sub_id'][$j]."','".$_POST['marks'][$i][$j]."'),";
}
}
$s = rtrim($s,",");
if(!mysqli_query($mysqli,$s))
echo mysqli_error();
else
echo "Records Saved <br />";
mysqli_close($mysqli);?>

Storing database value into variable

My table category has these columns:
idcategory
categorySubject
users_idusers
I have a form with a simple radio buttons and a textbox.
I have a select all statement for category and need to get the idcategory stored into a variable ($getCatId) so I can use this statement:
$sql="INSERT INTO topic(subject, topicDate, users_idusers, category_idcategory, category_users_idusers) VALUES('($_POST[topic])', '$date', '$_SESSION[userid]', '$getCatId', '$_SESSION[userid]');";
What is the best way to get and store categoryid?
if($_SERVER['REQUEST_METHOD'] != 'POST') //show form if not posted
{
$sql = "SELECT * FROM category;";
$result = mysqli_query($conn,$sql);
?>
<form method="post" action="createTopic.php">
Choose a category:
</br>
</br>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<div class= 'choice'><input type='radio' name='category' value='". $row['idcategory'] . "'>" . $row['categorySubject'] ."</div></br>";
}
echo 'Topic: <input type="text" name="topic" minlength="3" required>
</br></br>
<input type="submit" value="Add Topic" required>
</form>';
}
if ($_POST){
if(!isset($_SESSION['signedIn']) && $_SESSION['signedIn'] == false)
{
echo 'You must be signed in to contribute';
}
else{
$sql="INSERT INTO topic(subject, topicDate, users_idusers, category_idcategory, category_users_idusers) VALUES('($_POST[topic])', '$date', '$_SESSION[userid]', '$getCatId', '$_SESSION[userid]');";
$result = mysqli_query($conn,$sql);
echo "Added!";
If I understand this question correctly, you'll have your $getCatId (id of the category) in $_POST['category'] (after sending form) in your case
The first thing you should do is protect yourself from SQL injection by parameterizing your queries before old Bobby Tables comes to pay you a visit.
You might also look into using PDO as I've demonstrated below because it's a consistent API that works with a lot of different database management systems, so this leads to wonderfully portable code for you. Here's an annotated working example on Github:
<?php
// returns an intance of PDO
// https://github.com/jpuck/qdbp
$pdo = require __DIR__.'/mei_DV59j8_A.pdo.php';
// dummy signin
session_start();
$_SESSION['signedIn'] = true;
$_SESSION['userid'] = 42;
//show form if not posted
if($_SERVER['REQUEST_METHOD'] != 'POST'){
$sql = "SELECT * FROM category;";
// run query
$result = $pdo->query($sql);
?>
<form method="post" action="createTopic.php">
Choose a category:
</br>
</br>
<?php
// get results
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
echo "
<div class= 'choice'>
<input type='radio' name='category' value='$row[idcategory]'/>
$row[categorySubject]
</div>
</br>
";
}
echo '
Topic: <input type="text" name="topic" minlength="3" required>
</br></br>
<input type="submit" value="Add Topic" required>
</form>
';
}
if ($_POST){
if(!isset($_SESSION['signedIn']) && $_SESSION['signedIn'] == false){
echo 'You must be signed in to contribute';
} else {
// simulate your date input
$date = date("Y-m-d");
// bind parameters
$sql = "
INSERT INTO topic (
subject, topicDate, users_idusers, category_idcategory, category_users_idusers
) VALUES(
:subject, :topicDate, :users_idusers, :category_idcategory, :category_users_idusers
);
";
// prepare and execute
$statement = $pdo->prepare($sql);
$statement->execute([
'subject' => "($_POST[topic])",
'topicDate' => $date,
'users_idusers' => $_SESSION['userid'],
// to answer your question, here's your variable
'category_idcategory' => $_POST['category'],
'category_users_idusers' => $_SESSION['userid'],
]);
echo "Added!";
}
}

How do I run multiple SQL Queries using "if(isset($_POST['Submit'])){"

Trying to make a CRUD, everything works except my Update function. I feel like the problem is in the second sql query. When I click on submit it just refreshes and the change is gone. Can anyone show me how to find what I need to change/show me what to change?
<head>
<title>Update</title>
</head>
<body>
</form>
<?php
require_once('dbconnect.php');
$id = $_GET['id'];
$sql = "SELECT * FROM dealers where ID=$id";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<form action="" method="post">';
echo "Company: <input type=\"text\" name=\"CName\" value=\"".$row['CName']."\"></input>";
echo "<br>";
echo "Contact: <input type=\"text\" name=\"Contact\" value=\"".$row['Contact']."\"></input>";
echo "<br>";
echo "City: <input type=\"text\" name=\"City\" value=\"".$row['City']."\"></input>";
echo "<br>";
echo "<input type=\"Submit\" = \"Submit\" type = \"Submit\" id = \"Submit\" value = \"Submit\">";
echo "</form>";
}
echo "</table>";
} else {
echo "0 results";
}
if(isset($_POST['Submit'])){
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where ID=$id";
$result = $conn->query($sql);
}
$conn->close();
?>
Instead of building a form inside PHP, just break with ending PHP tag inside your while loop and write your HTML in a clean way then start PHP again. So you don't make any mistake.
Also you've to submit your $id from your form too.
Try this
<?php
require_once('dbconnect.php');
$id = $_GET['id'];
$sql = "SELECT * FROM dealers where ID=$id";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
?>
<form action="" method="post">
<input type="hidden" name="id" value="<?= $id ?>" />
Company: <input type="text" name="CName" value="<?= $row['CName'] ?>" />
<br>
Contact: <input type="text" name="Contact" value="<?= $row['Contact'] ?>" />
<br>
City: <input type="text" name="City" value="<?= $row['City'] ?>" />
<br>
<input type="Submit" name="Submit" id="Submit" value="Submit" />
</form>
<?php
} // end while loop
echo "</table>";
}
else {
echo "0 results";
}
Note: You are passing undefined variables into your update query. As you are submitting your form you must have to define those variables before you use them.
if (isset($_POST['Submit'])) {
$CName = $_POST['CName'];
$Contact = $_POST['Contact'];
$City = $_POST['City'];
$id = $_POST['id'];
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where ID=$id";
$result = $conn->query($sql);
}
$conn->close();
that loop? ID primary key or not?
maybe u need create more key in table dealer like as_id
<input type="hidden" name="idform" value="$as_id">
in statment
if($_POST){
$idf = $_POST['idform'];
if(!empty($idf)){
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where as_id=$idf";
$result = $conn->query($sql);
}
$conn->close();
}

PHP database not updating, no errors

I have a page of entries with an edit button behind each entry, clicking it brings you to the edit page of that entry, the form has the existing data of that entry as default values. I change the values and click update, it redirects where it should, no errors but also no change in the data entry values.
My form:
<form method="post" action="edit.php" enctype="multipart/form-data">
Item name:</br>
<input type="text" name="item_name" value="<?php echo $row[1]; ?>"></br></br>
<input type="hidden" name="item_id" value="<?php echo $row[0]; ?>">
Item price:</br>
<input type="text" name="item_price" value="<?php echo $row[3]; ?>" ></br></br>
Item image:</br>
<input type="file" name="item_image"value="<?php echo $row[4]; ?>" ></br></br>
Item description:</br>
<textarea type="text" class="txtinput" cols="55" rows="20" name="item_description"><?php echo $row[2]; ?>"</textarea></br></br>
<input type="submit" name="submit" value="Uppdate entry">
</form>
My PHP:
<?php
include("includes/connect.php");
if( isset($_GET['edit']))
{
$id= $_GET['edit'];
$res= mysql_query("SELECT * FROM shop_items WHERE item_id=$id");
$row= mysql_fetch_array($res);
}
if( isset($_POST['submit']))
{
$item_name = $_POST['item_name'];
$id = $_POST['item_id'];
$item_price = $_POST['item_price'];
$item_image = $_FILES['item_image'];
$image_tmp = $_FILES['item_image'] ['tmp_name'];
$item_description = $_POST['item_description'];
if ($item_name=='' or $item_price=='' or $item_image=='' or $item_description==''){
echo "<script>alert('One or more of your fields are blank, please ensure you have entered content in ALL fields.')</script>";
}
else {
move_uploaded_file($image_tmp,"images/$item_image");
$sql = "UPDATE shop_item SET item_name='$item_name', item_price='$item_price,' item_image='$item_description', item_name='$item_description' WHERE item_id='$id'";
echo "<meta http-equiv='refresh' content='0;url=admin_shop.php'>";
}
}
?>
You're not actually running your SQL query to update the database! You've stored the SQL query in a variable, $sql, but you haven't actually called mysql_query($sql);
} else {
move_uploaded_file($image_tmp,"images/$item_image");
$sql = "UPDATE shop_item SET item_name='$item_name', item_price='$item_price,' item_image='$item_description', item_name='$item_description' WHERE item_id='$id'";
// Add this line
mysql_query($sql);
echo "<meta http-equiv='refresh' content='0;url=admin_shop.php'>";
}
However, MySQL functionality is deprecated. You should look into PDO: http://uk3.php.net/pdo or mysqli: http://uk3.php.net/mysqli
Change this -
$sql = "UPDATE shop_item SET
item_name='".mysql_real_escape_string($item_name)."',
item_price='".mysql_real_escape_string($item_price)."',
item_image='".mysql_real_escape_string($item_description)."',
item_name='".mysql_real_escape_string($item_description)."'
WHERE item_id='$id'";
$exe = mysql_query($sql) or die(mysql_error());
NOTE: Avoid using mysql_* function since they are deprecated and use mysql_* or PDO instead.

checked values for the SELECT: WHERE clause

I am trying to create a compare option between selected cars.
<?php
if(isset($_POST['compares'])) {
$id_nums = array($_POST['cBox']);
//$id_nums = array(1,6,12,18,24);
$id_nums1 = implode(", ", $id_nums);
$query = "SELECT * FROM wp_cars WHERE id in ($id_nums1)";
$cCars = mysql_query($query) or mysql_error();
while($car = mysql_fetch_array($cCars)) {
echo $car['cartitle']."<br/>";
echo $car['saleprice']."<br/>";
}
} else {
$query1 = "SELECT * FROM wp_cars";
$allcars = mysql_query($query1) or die(mysql_error()); `
while($car1 = mysql_fetch_array($allcars)) {
echo "<input type='checkbox' value=".$car1['id']." name='cBox[]' />";
echo $car1['cartitle']."<br/>";
echo $car1['saleprice']."<br/>";
}
}
?>
How to pass the checkbox name(cBox[]) array based on checkboxes selection.
<form action="compares.php" method="post">
<button name="compares">Select Cars to Compare</button>
</form>
$id_nums = array($_POST['cBox']);
$_POST['cBox'] is already an array, you are making a 2d array. Doing
$id_nums1 = implode(", ", $_POST['cBox']);
would do what you want. Although it is wide open to SQL injection.
From your HTML part, send inputs this way:
<form action="compares.php" method="post">
<?php foreach (mysql_fetch_array($allcars) as $car: ?>
<input type="checkbox" value="<?php echo $car['id']; ?>" name="cBox[]" />
<?php echo $car['cartitle']; ?><br />
<?php echo $car['saleprice']; ?><br />
<?php endforeach; ?>
</form>
And in your PHP part, receive inputs this way:
$id_nums = implode(",", $_POST['cBox']);

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