Im doing a Crud with PHP and Oracle, adding the info and deleting the info works fine. But Updating is not saving on the oracle database. Im sure that is something related to the DATE format, because I had the same project with other database and doesnt have any problem. Any guess, whats it happening?
<?php
require_once 'conexion.php';
$idautor = $_POST['ID_AUTOR'];
$nameautor = $_POST['NOMBRE_AUTOR'];
$bdate = date("m-d-Y", strtotime($_POST['FECHA_NACIMIENTO']));
$query = "UPDATE AUTOR SET NOMBRE_AUTOR ='".$nameautor."', FECHA_NACIMIENTO ='".$bdate."' WHERE ID_AUTOR = '".$idautor."' ";
// $query = "UPDATE AUTOR SET NOMBRE_AUTOR ='".$nameautor."' WHERE ID_AUTOR = '".$idautor."' ";
$statement = oci_parse($conexion,$query);
$r = oci_execute($statement,OCI_DEFAULT);
$res = oci_commit($conexion);
if ($res) {
// Mensaje si los datos cambian
echo "<script>alert('Los libros se actualizaron con exito'); window.location.href='sistema.php'</script>";
header('Location: sistema.php');
} else {
// Mensaje si los datos no cambian
echo "<script>alert('Los datos no se pudieron actualizar'); window.location.href='sistema.php'</script>";
// echo oci_error();
}
} else {
// si intenta acceder directamente a esta página, será redirigido a la página de índice
header('Location: sistema.php');
}
Your code is an open door for SQL-Injection. It should be like this:
$query = "UPDATE AUTOR SET
NOMBRE_AUTOR = :nameautor, FECHA_NACIMIENTO = TO_DATE(:bdate, 'YYYY-MM-DD')
WHERE ID_AUTOR = :idautor";
$statement = oci_parse($conexion, $query);
oci_bind_by_name($statement, ':nameautor', $nameautor, 1000, SQLT_CHR);
oci_bind_by_name($statement, ':bdate', date_format($bdate, 'Y-m-d'), 30, SQLT_CHR);
oci_bind_by_name($statement, ':idautor', $idautor, 100, SQLT_INT);
$r = oci_execute($statement, OCI_DEFAULT);
You have to use TO_DATE(...), because type like SQLT_DATE does not exist. Otherwise you rely on current session NLS_DATE_FORMAT which may change at any time.
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Hello I'm currently trying to create a page based on a database under mysql that would update itself for a client. However what I'm trying to do loops and returns the first value of the database each time and indefinetely when I want it to go on to another object in the database. Here is the code, I'm a beginner so the error might be flagrant, thanks for the help.
<?php
try
{
$db = new PDO('mysql:host=localhost;dbname=labase', 'root' ,'');
$db->exec(" SET CHARACTER SET utf8 ");
$db->setAttribute(PDO::ATTR_CASE, PDO::CASE_LOWER);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(Exception $e){
echo'une erreur est survenue';
die();
}
for ($i = 1; $i < 10; $i++) {
if ($i=1){
$select = $db->prepare("Select profession from contact where affiliation='nord' group by profession"); // je récupère les professions de la bdd
$select->execute();
}
$data = $select->fetch(PDO::FETCH_OBJ);
$profess=$data->profession; // je prends la prochaine profession
$selectionner = $db->prepare("Select nomcontact, count(*) as nbrcontact from contact where affiliation='nord' and profession='$profess'"); // je prends les contacts qui ont cette profession ainsi que leur nombre
$selectionner->execute();
$prendre = $selectionner->fetch(PDO::FETCH_OBJ);
$nbrcontact=$prendre->nbrcontact;// je récupère leur nombre
echo $profess;
echo $nbrcontact;
}
?>
I am not a PHP expert and never use PDO, but in Msqli, there is a fetch_array() to get multiple result (instead of fetch for single result), maybe in PDO you have a fetch_array too. Then, you can loop on the result array
Something like that (using msqli)
$sql = "SELECT... FROM ..";
$result = $link->query($sql);
while($row =mysqli_fetch_array($result))
{
}
if ($i=1) { // here is should be == or ===
You're causing an infinite loop by declaring $i=1
<?php
try
{
$db = new PDO('mysql:host=localhost;dbname=labase', 'root' ,'');
$db->exec(" SET CHARACTER SET utf8 ");
$db->setAttribute(PDO::ATTR_CASE, PDO::CASE_LOWER);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(Exception $e){
echo'une erreur est survenue';
die();
}
for ($i = 1; $i < 10; $i++) {
if ($i == 1){ // added code
$select = $db->prepare("Select profession from contact where affiliation='nord' group by profession"); // je récupère les professions de la bdd
$select->execute();
}
$data = $select->fetch(PDO::FETCH_OBJ);
$profess=$data->profession; // je prends la prochaine profession
$selectionner = $db->prepare("Select nomcontact, count(*) as nbrcontact from contact where affiliation='nord' and profession='$profess'"); // je prends les contacts qui ont cette profession ainsi que leur nombre
$selectionner->execute();
$prendre = $selectionner->fetch(PDO::FETCH_OBJ);
$nbrcontact=$prendre->nbrcontact;// je récupère leur nombre
echo $profess;
echo $nbrcontact;
}
?>
Use == for comparison
I'm new with PHP and SQL...
I'm trying to create new tables based on the url, but it's only working the first time I use it. After that, it's not possible.
Here is my PHP code:
if(isset($_GET['id'])){
$tabela = $_GET['tabela'];
$_GET['id'];
$criar = $tabela . $nivel . $page_id;
// Se clicar no botão 'confirmar', então ele faz o seguinte:
if(isset($_POST['submit'])){
$titulo = $_POST['titulo'];
$_FILES['imagem']['tmp_name'];
$texto = $_POST['texto'];
// Se um destes campos estiver vazio:
if($titulo=='' or $imagem=='' or $texto==''){
echo "Preencha todos os campos para o menu!";
exit(); }
// Se não houver campos vazios, ele faz: else {
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "site";
// Ligação à base de dados:
$conn = new mysqli($servername, $username, $password, $dbname);
// Verifica a ligação:
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Cria a nova tabela:
$sql = "CREATE TABLE IF NOT EXISTS $criar (
id INT(9) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
titulo VARCHAR(255),
imagem LONGBLOB,
texto TEXT,
grupo INT(9),
FOREIGN KEY (grupo) REFERENCES $tabela(id)
)";
// Se conseguir ligar-se à base de dados e criar uma nova tabela, ele insere os dados na nova tabela:
if ($conn->query($sql) === TRUE) {
include("includes/connect.php");
mysql_query("SET NAMES 'utf8'");
move_uploaded_file($image_tmp,"../imagens/$imagem");
$insert_query = "INSERT INTO $criar (titulo, imagem, texto, grupo) VALUES ('$titulo','$imagem','$texto','$page_id')";
// Se inserir os dados na nova tabela, ele dá uma mensagem de sucesso:
if(mysql_query($insert_query)){
echo "<script>alert('Menu inserido com sucesso!')</script>";
echo "<script>window.open('index.php','_self')</script>";
}
else{
echo "Erro: " . $insert_query . "<br>" . $conn->error;
}
}
// Caso ele não consiga criar uma nova tabela (porque já existe), ele insere os dados na tabela já existente:
else {
include("includes/connect.php");
mysql_query("SET NAMES 'utf8'");
// Cria a nova tabela:
$sql = "CREATE TABLE IF NOT EXISTS $criar (
id INT(9) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
titulo VARCHAR(255),
imagem LONGBLOB,
texto TEXT,
grupo INT(9),
FOREIGN KEY (grupo) REFERENCES $tabela(id)
)";
if(mysql_query($sql)){
echo "sim!";
}
else {
echo "não!";
}
move_uploaded_file($image_tmp,"../imagens/$imagem");
$insert_query = "INSERT INTO $criar (titulo, imagem, texto, grupo) VALUES ('$titulo','$imagem','$texto','$page_id')";
// Caso consiga inserir os dados na tabela já existente, dá uma mensagem de sucesso:
if(mysql_query($insert_query&&$sql)){
echo "<script>alert('Menu inserido com sucesso!')</script>";
echo "<script>window.open('index.php','_self')</script>";
}
else{
echo "isto não está a correr bem!";
}
// Fecha a ligação à base de dados:
$conn->close();
} } }$nivel = $_GET['grupo']; $page_id =$imagem = $_FILES['imagem']['name']; $image_tmp =
What do you mean 1st time, 2nd time? Do you try to create another table with the same name you just have created? You'll get a table already exists error.
CREATE TABLE IF NOT EXISTS $criar
This explicitly tells MySql not to create the table if it exists, so if you pass the same query parameters to your php then it'll not be able to create the same table again.
Probably you could change it to:
$page_id = $_GET['id'];
$criar = $tabela . $nivel . $page_id;
And then pass a different id and/or different tabela every time.
this code looks behind the issue to me :
$tabela = $_GET['tabela'];
$_GET['id']; //this line looks odd
$criar = $tabela . $nivel . $page_id;
and maybe it's not creating new tables because you give the same $criar every time.
After hours of trying to solve this issue, i finally got good news! :) i was working in my local server (via xampp) and decided to put it online to check if something changes... and dont know why, but it works perfectly online and dont work with xampp! What a mess!! but its working!! thanks all for your help!! ;)
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
this is a searcher program php with MySqL this give me a error, and i need a bit help on this...
This is the php Code:
<?php
if ($_POST['buscar'])
{
// Tomamos el valor ingresado
$buscar = $_POST['palabra'];
// Si está vacío, lo informamos, sino realizamos la búsqueda
if(empty($buscar))
{
echo "No se ha ingresado una cadena a buscar";
}else{
//Conexión a la base de datos
$servidor = "localhost"; //Nombre del servidor
$usuario = "root"; //Nombre de usuario en tu servidor
$password = "1234"; //Contraseña del usuario
$base = "db_maquinas"; //Nombre de la BD
$con = mysql_connect($servidor, $usuario, $password) or die("Error al conectarse al servidor");
mysql_select_db($base, $con) or die("Error al conectarse a la base de datos");
$sql= mysql_query("SELECT * FROM repuestos WHERE id LIKE '%$buscar%' AND descripcion LIKE '%$buscar%' ORDER BY id", $con) or die(mysql_error($con));
$result = mysql_query($sql, $con); //<----LINE 32!!!
// Tomamos el total de los resultados
if($result) { $total = mysql_num_rows($result); } else { die('Invalid query' . mysql_error($con)); }
echo "<table border = '1'> \n";
//Mostramos los nombres de las tablas
echo "<tr> \n";
while ($field = mysql_fetch_field($result)){
echo "<td>$field->name</td> \n";
}
echo "</tr> \n";
do {
echo "<tr> \n";
echo "<td>".$row["id"]."</td> \n";
echo "<td>".$row["descripcion"]."</td> \n";
echo "<td>".$row["cantidad"]."</td> \n";
echo "</tr> \n";
} while ($row = mysql_fetch_array($result));
echo "</table> \n";
echo "¡ No se ha encontrado ningún registro !";
}
}
?>
The error is --> Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\maquinas2000\paginas\buscarepuestos.php on line 32
Invalid query {Line 32 is -> $result = mysql_query($sql, $con); }
i work with a Localhost xampp ofc, this give me a lot of troubles this code, i need only this and i'll finish 100% the work, so if anyone can give me the answer of this error i'll be very grateful for that, thx!
You have already executed the query. mysql_query return true or false and you are passing this return value again in mysql_query , make changes this :
$sql= "SELECT * FROM repuestos WHERE id LIKE '%$buscar%' AND descripcion LIKE '%$buscar%' ORDER BY id";// remove mysql_query from this line
$result = mysql_query($sql, $con);
Important : mysql_ is depricated use mysqli instead of that
i have a trouble with this T_T this is a searching/consult script for cars parts
this is the php
<?php
if ($_POST['buscar'])
{
// Tomamos el valor ingresado
$buscar = $_POST['palabra'];
// Si está vacío, lo informamos, sino realizamos la búsqueda
if(empty($buscar))
{
echo "No se ha ingresado una cadena a buscar";
}else{
//Conexión a la base de datos
$servidor = "localhost"; //Nombre del servidor
$usuario = "root"; //Nombre de usuario en tu servidor
$password = "1234"; //Contraseña del usuario
$base = "db_maquinas"; //Nombre de la BD
$con = mysql_connect($servidor, $usuario, $password) or die("Error al conectarse al servidor");
$sql= mysql_query("SELECT * FROM repuestos WHERE 'id','descripcion' LIKE '%$buscar%' ORDER BY id", $con) or die(mysql_error($con));
mysql_select_db($base, $con) or die("Error al conectarse a la base de datos");
$result = mysql_query($sql, $con);
// Tomamos el total de los resultados
if($result) { $total = mysql_num_rows($result); } else { die('Invalid query' . mysql_error($con)); }
echo "<table border = '1'> \n";
//Mostramos los nombres de las tablas
echo "<tr> \n";
while ($field = mysql_fetch_field($result)){
echo "<td>$field->name</td> \n";
}
echo "</tr> \n";
do {
echo "<tr> \n";
echo "<td>".$row["id"]."</td> \n";
echo "<td>".$row["descripcion"]."</td> \n";
echo "<td>".$row["cantidad"]."</td> \n";
echo "</tr> \n";
} while ($row = mysql_fetch_array($result));
echo "</table> \n";
echo "¡ No se ha encontrado ningún registro !";
}
}
?>
that when on the form i press "Send or Search" this send me to another page that says "NO DATABASE SELECTED"
I hope somebody can help me with that..
PD: i'm using a localhost DB with PhpMyAdmin i have items on tables, i verified tables not empty...
You select your database after you attempt to run queries. Place the code before you attempt to run queries:
mysql_select_db($base, $con) or die("Error al conectarse a la base de datos");
$sql= mysql_query("SELECT * FROM repuestos WHERE 'id','descripcion' LIKE '%$buscar%' ORDER BY id", $con) or die(mysql_error($con));
FYI, you shouldn't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
place the mysql_selectdb() before the mysql_query().
I have a problem with my function db_modif();. Everytime, if the value are correct and exist, the message is the first one (the if condition).
echo "Il n'y a aucun compte au nom de: ".$username." au site: ".$site." dans la base de données";
So, there is no modification in my Database
Here is my code form the manipulation :
<?php
$username = $_POST["user_search"];
$site = $_POST["adr_search"];
$fonction = $_POST["fonction"];
$modif = $_POST["modif_value"];
$prep ="";
if(!$username)
echo 'Nom d\'utilisateur manquant..';
elseif(!$site)
echo 'Site manquante..';
else{
require("db_action.php"); //Require in the database connection.
$bd = db_open(); // Open DATABASE
if($fonction == "usernameOp")
$prep = "username";
if($fonction == "adresseOp")
$prep = "adresse";
if($fonction == "passwdOp")
$prep = "password";
if($fonction == "siteOp")
$prep = "siteWeb";
if($fonction == "fonctionOp")
$prep = "fonction";
db_modif($prep, $username, $site, $modif);
db_close($bd);
}//ELSE
And from the function db_modif();:
function db_modif($prep, $username, $site, $modif){
error_reporting(-1); //Activer le rapport de toutes les genres d'erreurs
$querycon = "UPDATE info_compte SET $prep = '$modif' WHERE username = '$username' AND siteWeb = '$site'";
if($response = mysql_query($querycon) or trigger_error(mysql_error())){
echo "<pre>";
echo "Il n'y a aucun compte au nom de: <b>".$username."</b> au site: <b>".$site."</b> dans la base de données";
echo "</pre>";
}
else{
mysql_query($querycon);
echo "<pre>\n";
echo "Le compte <b>".$username."</b> du site : <b>".$site."</b> a été supprimé avec succès\n";
echo "</pre>";
}//ELSE
}//db_modif
Change AND WHERE to just AND on this line:
$querycon = "UPDATE info_compte SET $prep = '$modif'
WHERE username = '$username'
AND WHERE siteWeb = '$site'";
I'd suggest that you use mysql_error() to help debugging this sort of problem in future.
$response = mysql_query($querycon) or trigger_error(mysql_error());
You may also have an SQL injection vulnerability if any of those variables can contain quotes. Consider using mysql_real_escape_string or PDO with prepared statements.