Json return null - php

I have the next problem doing a json to read in Andorid
(the credentials are hidden but connection going good in others files)
class reportes
{
var $parametro;
var $conexion;
function __construct(){
$host = "IP"; $DBName = "DbName";
$usuario="user"; $contrasena="pass";
$driver = "DRIVER={iSeries Access ODBC Driver};
SYSTEM=$host;Uid=$usuario;
Pwd=$contrasena;Client_CSet=UTF-8;";
$this->conexion = odbc_connect($driver, $usuario, $contrasena);
}
function consulta($parametro){
$query=
"SELECT OHSNME,OHTOT$,OHREPÑ
FROM MYDB.SANFPRD.FOMHDR
WHERE OHORDÑ= $parametro ";
echo $query."<br><br>";
if ($this->conexion == 0) {echo "Ha fallado la conexion a la BBDD </br>";}
else{
$datos = array();
$result=odbc_exec($this->conexion,$query);
while($row = odbc_fetch_object($result)){
$datos[]= $row;
}
echo json_encode($datos);
}
}//Fin funcion consulta()
}//Fin de la clase
$consultar = new reportes();
$nota_venta = $_REQUEST['parametro'];
$consultar->consulta($nota_venta);
the response JSON that i get is:
SELECT OHSNME,OHTOT$,OHREPÑ FROM DELLORTO.SANFPRD.FOMHDR WHERE OHORDÑ= 366
[{"OHSNME":"E.C. GM. ","OHTOT$":"1861.00",null:" A07"}]
you can see that OHORDÑ is probably the problem with the 'Ñ'
but this table are part a productive database and i can't update

Solution #1, alias the column name to a name without non-ascii characters:
$query=
"SELECT OHSNME,OHTOT$,OHREPÑ AS OHREPN
FROM MYDB.SANFPRD.FOMHDR
WHERE OHORDÑ= $parametro ";
Solution #2, manually serialize using utf8_encode():
$result=odbc_exec($this->conexion,$query);
while($row = odbc_fetch_object($result)){
$_row_fix = array();
foreach ($row as $field => $val) {
$_row_fix[utf8_encode($field)] = utf8_encode($val);
}
$datos[]= $_row_fix;
}

Related

How to save a file name in the MySQL database?

I have developed this code below to the user upload a file and save the name of this file in the database, to be able to access it later, the upload is done normally, it goes to the designated folder, but the name is not saved in the database, does anyone know what's wrong with the code? Especially below the move_uploaded_file, because so far it works, then it goes wrong.
<?php
if (isset($_POST['enviar'])) {
$arq = $_FILES['arquivo']['name'];
$arq = str_replace(" ", "_", $arq);
$arq = str_replace("ç", "c", $arq);
if (file_exists("uploads/$arq")) {
$a = 1;
while (file_exists("uploads/[$a]$arq")) {
$a++;
}
$arq = "[".$a."]".$arq;
}
if (move_uploaded_file($_FILES['arquivo']['tmp_name'], 'uploads/'.$arq)) {
$objDb = new db();
$link = $objDb->conecta_mysql();
$sql = "insert into arquivos (email_vol, nomearq) values ('$email', '$arq')";
if (mysqli_query($link, $sql)){
echo 'Plano de aula 1 enviado com sucesso!';
} else {
echo (mysqli_error($link));
echo 'Erro ao enviar o plano de aula!';
}
} else {
echo "Nenhum arquivo selecionado!";
}
}
?>
That is the code used to connect with the database:
class db {
//host
private $host = 'localhost';
//usuario
private $usuario = '111111';
//senha
private $senha = '11111111';
//banco de dados
private $database = 'dsfadsfasd';
public function conecta_mysql(){
//criar a conexão
$con = mysqli_connect($this->host, $this->usuario, $this->senha, $this->database);
//ajustar a charser de cominicação entre a aplicação e o bd
mysqli_set_charset($con, 'utf8');
//verificar se houve erro de conexão
if (mysqli_connect_errno()) {
echo 'Erro ao tentar se conectar com o banco de dados'.mysqli_connect_error();
}
return $con;
}
}
?>
Don't have the privilege to comment right now but shouldn't this be like this plus you don't have a semicolon at the end of your sql script
$sql = "insert into arquivos (email_vol, nomearq) values ('" . $email . "', '"
.$arq . "');";
and also this
if (file_exists("uploads/" . $arq)) {
$a = 1;
while (file_exists("uploads/". $a . ".". $arq)) {
$a++;
}
$arq = $a.".".$arq;
}
with a full stop between your file number and name

Loop though database with foreach

I want to browse data from my postgre database with a "foreach". So I made my request like that :
$conn_string = "host=localhost port=5432 dbname=test_postgre user=postgres password='1234'";
$dbconn = pg_connect($conn_string);
$sql = "SELECT id_traitement FROM public.traitement WHERE id_essai='.$id_essai.';";
$res = pg_query($sql) or die("Pb avec la requete: $sql");
$data = pg_fetch_all($res);
And I get my values with "pg_fetch_all".
After that, I'm looking for compare the data in my database (get with the request) and the data in my web page. So I created this loop :
foreach($array as $ligne_web)
{
foreach($data['id_traitement'] as $ligne_base)
{
if(($ligne_web[0] == $ligne_base) and ($flag))
{
//update de la ligne
update_traitement($id_traitement,$traitement,$code_traitement,$id_essai);
$flag2 = false;
break 1;
}
}
if(($flag) and ($flag2))
{
insert_traitement($id_traitement,$traitement,$code_traitement,$id_essai);
}
}
When I try to run it, firebug tells me : Invalid argument supplied for foreach(). So I don't know how to browse the rows in the database. Certainly my problem is in my foreach, but I don't find what's wrong.
Help please !
It seems your second foreach needs to be '$data' instead of $data['id_traitement']
So your code need to changed to ,
foreach($arr as $ligne_web)
{
foreach($data as $ligne_base) // <-- Here is the correction
{
if(($ligne_web[0] == $ligne_base) and ($flag))
{
------ REST of your Codes ------
Ok, I found an answer. Instead of an array $data from my database, and directly after the request, I created a new array.
Here is my code :
$conn_string = "host=localhost port=5432 dbname=test_postgre user=postgres password='1234'";
$dbconn = pg_connect($conn_string);
$sql = "SELECT id_traitement FROM public.traitement WHERE id_essai='.$id_essai.';";
$res = pg_query($sql) or die("Pb avec la requete: $sql");
$tableau_database_final = array();
while ($data = pg_fetch_all($res)) //Here is my array
{
$tableau_database = array('id_traitement'=>$data['id_traitement']);
array_push($tableau_database_final,$tableau_database);
}
$flag2 = true;
foreach($array as $ligne_web)
{
foreach($tableau_database_final as $ligne_base)
{
echo ($ligne_web[0]);
echo ($ligne_base);
if(($ligne_web[0] == $ligne_base)) //Si il existe une ligne ayant déjà le même id traitement
{
//update de la ligne
update_traitement($id_traitement,$traitement,$code_traitement,$id_essai);
$flag2 = false;
break 1;
}
}
if(($flag) && ($flag2))
{
//insert_traitement($id_traitement,$traitement,$code_traitement,$id_essai);
}
}

How to create a SQL query in PHP with differents variables

I have a problem when I create a sql query which depends on many variables that the user select in different checkboxes.
I make a httprequest GET and them, when I proposed to create the query, I check the variable and I create the query, gradually. I show you the PHP code:
$link = mysql_connect($hostname, $username, $password) or die('No se pudo conectar: ' . mysql_error());
//echo 'Conectado satisfactoriamente';
mysql_select_db('Agenda Juvenil') or die('No se pudo seleccionar la base de datos');
mysql_query('SET CHARACTER SET utf8');
$query="SELECT id, title, barrio_smultiple, coordenadas_p_0_coordinate, coordenadas_p_1_coordinate, gratuita_b FROM eventosDiarios WHERE";
// check for post data
if (isset($_GET['franjas0'])){
$franja0 = $_GET['franjas0'];
$query.="franja_smultiple IN ('$franja0'";
}
if (isset($_GET['franjas1'])){
$franja1 = $_GET['franjas1'];
$query.=",'$franja1'";
}
if (isset($_GET['franjas2'])){
$franja2 = $_GET['franjas2'];
$query.=",'$franja2'";
}
$query.=")";
// get a product from products table
$result = mysql_query($query) or die('Consulta fallida: ' . mysql_error());
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["eventos"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$evento = array();
$evento["id"] = $row["id"];
$evento["title"] = $row["title"];
$evento["barrio_smultiple"] = $row["barrio_smultiple"];
$evento["coordenadas_p_0_coordinate"] = $row["coordenadas_p_0_coordinate"];
$evento["coordenadas_p_1_coordinate"] = $row["coordenadas_p_1_coordinate"];
$evento["gratuita_b"] = $row["gratuita_b"];
// push single product into final response array
array_push($response["eventos"], $evento);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No se han encontrado eventos";
// echo no users JSON
echo json_encode($response);
}
The query must return something, but I have nothing. The variables are sent throught
List<NameValuePair> params = new ArrayList<NameValuePair>();
I am using Android.
Can someone help me please??
Thanks in advance.
Regards.
Your code needs to have a space after the WHERE. It needs to say
...b FROM eventosDiarios WHERE ";
rather than
....b FROM eventosDiarios WHERE";
or you'll get an illformed query.

Why is mysql_num_rows() returning 0?

I have read in another threads that pasting the query in phpMyAdmin returns the amount of rows that you have, but in my case it doesn't even returns a value in phpMyAdmin, it's always 0! Here is the query:
$query = "SELECT nom_usu FROM usuarios WHERE nom_usu = '$usu' AND pass = '$pass';";
I open the .php file and run the query and it returns empty values, like this:
SELECT nom_usu FROM usuarios WHERE nom_usu = '' AND pass = '';
I also tried to echo the value that returns and is, guess what? Zero!
Here is the complete .php file (is for a login):
<?php
include('dbConfig.php');
$usu = $_POST["usu"];
$pass = $_POST["pass"];
$query = "SELECT nom_usu FROM usuarios WHERE nom_usu = '$usu' AND pass = '$pass';";
echo $query."\n";
if($resultado = mysql_query($query)){
if(mysql_num_rows($resultado) > 0){
echo mysql_num_rows($resultado);
echo "Todo está bien, no te preocupes :)";
echo true;
} else {
echo mysql_num_rows($resultado);
echo "Hay algo mal aquí :(";
}
} else {
echo false;
}
?>
And the dbConfig.php file:
<?php
$host="localhost";
$user="neosoftw_lambda";
$contra="myPass"; <- This is not actually my password...
$db="neosoftw_lambdaMovil";
$ms = mysql_connect($host, $user, $contra) or die("No se pudo conectar :(");
mysql_select_db($db, $ms) or die ("No hay base de datos :(");
?>
Hope someone can help me figure out where I have gone wrong?
Here is what I want to make, is a login in jQuery Mobile, but it doesn't work!
Username: bryan
Password: azul
http://android.neosoftware.org.mx/
Edit
Maybe it could be because of my javascript?
$.post("http://android.neosoftware.org.mx/PHP/login.php", {usu : usu, pass : pass}, function(respuesta_login){
if(!($("#txt_usuario").val() || $("#txt_password") == '')){
$.mobile.changePage("#campos_vacios");
} else {
if(respuesta_login == 1){
$("#txt_usuario").val('');
$("#txt_password").val('');
$.mobile.changePage("#pag_principal");
} else {
$.mobile.changePage("#error_login");
}
}
});
Use mysqli or PDO statements, as mysql is deprecated. Try this:
$usu = mysqli_real_escape_string($_POST["usu"]);
$pass = mysqli_real_escape_string($_POST["pass"]);
$query = "SELECT nom_usu FROM usuarios WHERE nom_usu = '".$usu."' AND pass = '".$pass."'";
And also your condition is incorrect.It should be like this:
if($resultado == mysqli_query($query)){
Try this query.
$query = "SELECT `nom_usu` FROM usuarios WHERE `nom_usu` = '".$usu."' AND `pass` = '".$pass."'";
Also this type of problem occurre when $_POST variable is empty.. So, double check your code.
Note : mysql_* function is deprecated, move on mysqli_* function asap.

pg_close() error? is a sub close closing the 2 connections?

I am currently getting this error: pg_close(): 12 is not a valid PostgreSQL link resource. This is the code that is being executed:
function getProyectosDeUsuarioDAO($idUsuario)
{
$conexion = conectar();
$consulta = "sql things here;";
$resultado = pg_query($consulta) or die('Consulta fallida: ' . pg_last_error());
$lista = array();
$lista = pg_fetch_all($resultado);
var_dump($lista);
$listaProyectos = array();
if(!empty ($lista))
{
foreach ($lista as $p)
{
$proyecto = new Proyecto();
$proyecto->setNombre($p['nombre']);
$proyecto->setFechaInicio($p['fechainicio']);
$proyecto->setFechaFin($p['fechafin']);
$proyecto->setId($p['id']);
//Cargar roles del usuario para cada proyecto
$proyecto->setRoles(getRolesByProyecto($proyecto->getId(),$idUsuario));
array_push($listaProyectos, $proyecto) ;
}
if($conexion)
{
pg_close($conexion); //##############error here#################
}
pg_free_result($resultado);
var_dump($listaProyectos);
return $listaProyectos;
}
var_dumps are made to check if the query is working and it does work; I am not sure how to debug this.
I am using php_pgsql and WAMP server
($conexion) works for checking, however I'm not sure if connection is live or not?

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