php and mysql insert into multiple tables from one form
0
I am creating a simple site to keep records of users, customers, suppliers, etc.
I have created the forms to register clients and users using single forms capturing data through $ _POST form method
So far I have not had problems since the INSERTS are done on singles table wich PK is an AutoIncremented field
In the code shown below, my goal is to let users create a vendor (name, phone number, email, address) as well as asign 1 or many vendor-category. This information is contained in one form
vendors, vendors-categories, users etc are stored in a mysql database with the structure shown in attached file
structure
There is something wrong about my code after succesfully create vendor and i am not sure how to solve the part that manages insertions in the join table (tblprovxrubro)
Code
<?php
session_start();
//available for admin, power-users and role-4
if ($_SESSION['rol'] ==2) {
header("location: ./");
}
include "../conexion.php";
if (!empty($_POST)) {
$alert='';
if (empty($_POST['razonSocial'])|| empty($_POST['email'])) {
$alert='<p class="msg_error">Vendor name and email must not be blank.</p>';
}else{
$razonSocial = $_POST['razonSocial'];
$email = $_POST['email'];
$domicilio = $_POST['domicilio'];
$telefono = $_POST['telefono'];
$usuario_id = $_SESSION['iduser'];
$query = mysqli_query($conection,"SELECT * FROM tblprov WHERE razonSocial = '$razonSocial'");
$result = mysqli_fetch_array($query);
if ($result >0)
{
$alert='<p class="msg_error">vendor already exists.</p>';
}else{
$query_insert = mysqli_query($conection, "INSERT INTO tblprov(razonSocial, numero, domicilio, email, idUsuario)
VALUES('$razonSocial','$telefono','$domicilio','$email', '$usuario_id')");
if ($query_insert)
{
$alert='<p class="msg_save">Vendor created succesfully.</p>';
//Once vendor is created i need his ID in order to insert in table
//tblprovxrubro as many records as vendor-types selected in form
//Not sure about how to achieve this
$queryBuscaprov = mysqli_query($conection, "SELECT id FROM tblprov WHERE razonSocial = '$razonSocial'");
$result_prov = mysqli_fetch_array($queryBuscaProv);
$idProv = $result_prov['id'];
foreach ($_POST['idRubro'] as $opcionSeleccionada)
{
//This INSERT should execute as many times as vendor-types selected in form
$query2 = mysqli_query($conection, "INSERT INTO tblprovxrubro (idRubro, idProv) VALUES ('$opcionSeleccionada', '$idProv')");
}
}else{
$alert='<p class="msg_error">Error creating vendor.</p>';
}
}
}
mysqli_close($conection);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<?php include "includes/scripts.php"; ?>
<title>Registro de Proveedores</title>
</head>
<body>
<?php include "includes/header.php"; ?>
<section id="container">
<div class="form_register" name="form_register">
<h1><i class="fas fa-building"></i> Vendor List</h1>
<hr>
<form action="" method="post" class="form_register">
<div class="alert"><?php echo isset($alert) ? $alert : ''; ?></div>
<label for="razonSocial">Razón Social</label>
<input type="text" name="razonSocial" id="razonSocial" placeholder="Razón Social">
<label for="telefono">Número Telefónico</label>
<input type="text" name="telefono" id="telefono" placeholder="Número Telefónico">
<label for="domicilio">Domicilio</label>
<input type="text" name="domicilio" id="domicilio" placeholder="Domicilio">
<label for="email">Email</label>
<input type="text" name="email" id="email" placeholder="Email">
<?php
$query_rubro = mysqli_query($conection,"SELECT * FROM tblrubros");
mysqli_close($conection);
$result_rubro = mysqli_num_rows($query_rubro);
?>
<select name="rubro" id="rubro" multiple size="12">
<?php
if($result_rubro > 0)
{
while ($rubro = mysqli_fetch_array($query_rubro))
{
?>
<option value="<?php echo $rubro["idRubro"]; ?>"><?php echo $rubro["rubroDescripcion"] ?></option>
<?php
}
}
?>
</select>
<p>press Ctrl in order to select multiple options.</p>
<button name="submit" type="submit" class="btn_save"><i class="fas fa-save"></i> Crear Proveedor</button>
</form>
</div>
</section>
<?php include "includes/footer.php"; ?>
</body>
</html>
conection.php
<?php
$host = 'localhost';
$user = 'root';
$password = 'mypass';
$db = 'compras';
$conection = @mysqli_connect($host,$user,$password,$db);
mysqli_set_charset($conection,"utf8");
if(!$conection){
echo "connection error";
}
?>
php mysql insert
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
add a comment |
0
I am creating a simple site to keep records of users, customers, suppliers, etc.
I have created the forms to register clients and users using single forms capturing data through $ _POST form method
So far I have not had problems since the INSERTS are done on singles table wich PK is an AutoIncremented field
In the code shown below, my goal is to let users create a vendor (name, phone number, email, address) as well as asign 1 or many vendor-category. This information is contained in one form
vendors, vendors-categories, users etc are stored in a mysql database with the structure shown in attached file
structure
There is something wrong about my code after succesfully create vendor and i am not sure how to solve the part that manages insertions in the join table (tblprovxrubro)
Code
<?php
session_start();
//available for admin, power-users and role-4
if ($_SESSION['rol'] ==2) {
header("location: ./");
}
include "../conexion.php";
if (!empty($_POST)) {
$alert='';
if (empty($_POST['razonSocial'])|| empty($_POST['email'])) {
$alert='<p class="msg_error">Vendor name and email must not be blank.</p>';
}else{
$razonSocial = $_POST['razonSocial'];
$email = $_POST['email'];
$domicilio = $_POST['domicilio'];
$telefono = $_POST['telefono'];
$usuario_id = $_SESSION['iduser'];
$query = mysqli_query($conection,"SELECT * FROM tblprov WHERE razonSocial = '$razonSocial'");
$result = mysqli_fetch_array($query);
if ($result >0)
{
$alert='<p class="msg_error">vendor already exists.</p>';
}else{
$query_insert = mysqli_query($conection, "INSERT INTO tblprov(razonSocial, numero, domicilio, email, idUsuario)
VALUES('$razonSocial','$telefono','$domicilio','$email', '$usuario_id')");
if ($query_insert)
{
$alert='<p class="msg_save">Vendor created succesfully.</p>';
//Once vendor is created i need his ID in order to insert in table
//tblprovxrubro as many records as vendor-types selected in form
//Not sure about how to achieve this
$queryBuscaprov = mysqli_query($conection, "SELECT id FROM tblprov WHERE razonSocial = '$razonSocial'");
$result_prov = mysqli_fetch_array($queryBuscaProv);
$idProv = $result_prov['id'];
foreach ($_POST['idRubro'] as $opcionSeleccionada)
{
//This INSERT should execute as many times as vendor-types selected in form
$query2 = mysqli_query($conection, "INSERT INTO tblprovxrubro (idRubro, idProv) VALUES ('$opcionSeleccionada', '$idProv')");
}
}else{
$alert='<p class="msg_error">Error creating vendor.</p>';
}
}
}
mysqli_close($conection);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<?php include "includes/scripts.php"; ?>
<title>Registro de Proveedores</title>
</head>
<body>
<?php include "includes/header.php"; ?>
<section id="container">
<div class="form_register" name="form_register">
<h1><i class="fas fa-building"></i> Vendor List</h1>
<hr>
<form action="" method="post" class="form_register">
<div class="alert"><?php echo isset($alert) ? $alert : ''; ?></div>
<label for="razonSocial">Razón Social</label>
<input type="text" name="razonSocial" id="razonSocial" placeholder="Razón Social">
<label for="telefono">Número Telefónico</label>
<input type="text" name="telefono" id="telefono" placeholder="Número Telefónico">
<label for="domicilio">Domicilio</label>
<input type="text" name="domicilio" id="domicilio" placeholder="Domicilio">
<label for="email">Email</label>
<input type="text" name="email" id="email" placeholder="Email">
<?php
$query_rubro = mysqli_query($conection,"SELECT * FROM tblrubros");
mysqli_close($conection);
$result_rubro = mysqli_num_rows($query_rubro);
?>
<select name="rubro" id="rubro" multiple size="12">
<?php
if($result_rubro > 0)
{
while ($rubro = mysqli_fetch_array($query_rubro))
{
?>
<option value="<?php echo $rubro["idRubro"]; ?>"><?php echo $rubro["rubroDescripcion"] ?></option>
<?php
}
}
?>
</select>
<p>press Ctrl in order to select multiple options.</p>
<button name="submit" type="submit" class="btn_save"><i class="fas fa-save"></i> Crear Proveedor</button>
</form>
</div>
</section>
<?php include "includes/footer.php"; ?>
</body>
</html>
conection.php
<?php
$host = 'localhost';
$user = 'root';
$password = 'mypass';
$db = 'compras';
$conection = @mysqli_connect($host,$user,$password,$db);
mysqli_set_charset($conection,"utf8");
if(!$conection){
echo "connection error";
}
?>
php mysql insert
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
1
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
1
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57
add a comment |
0
0
0
I am creating a simple site to keep records of users, customers, suppliers, etc.
I have created the forms to register clients and users using single forms capturing data through $ _POST form method
So far I have not had problems since the INSERTS are done on singles table wich PK is an AutoIncremented field
In the code shown below, my goal is to let users create a vendor (name, phone number, email, address) as well as asign 1 or many vendor-category. This information is contained in one form
vendors, vendors-categories, users etc are stored in a mysql database with the structure shown in attached file
structure
There is something wrong about my code after succesfully create vendor and i am not sure how to solve the part that manages insertions in the join table (tblprovxrubro)
Code
<?php
session_start();
//available for admin, power-users and role-4
if ($_SESSION['rol'] ==2) {
header("location: ./");
}
include "../conexion.php";
if (!empty($_POST)) {
$alert='';
if (empty($_POST['razonSocial'])|| empty($_POST['email'])) {
$alert='<p class="msg_error">Vendor name and email must not be blank.</p>';
}else{
$razonSocial = $_POST['razonSocial'];
$email = $_POST['email'];
$domicilio = $_POST['domicilio'];
$telefono = $_POST['telefono'];
$usuario_id = $_SESSION['iduser'];
$query = mysqli_query($conection,"SELECT * FROM tblprov WHERE razonSocial = '$razonSocial'");
$result = mysqli_fetch_array($query);
if ($result >0)
{
$alert='<p class="msg_error">vendor already exists.</p>';
}else{
$query_insert = mysqli_query($conection, "INSERT INTO tblprov(razonSocial, numero, domicilio, email, idUsuario)
VALUES('$razonSocial','$telefono','$domicilio','$email', '$usuario_id')");
if ($query_insert)
{
$alert='<p class="msg_save">Vendor created succesfully.</p>';
//Once vendor is created i need his ID in order to insert in table
//tblprovxrubro as many records as vendor-types selected in form
//Not sure about how to achieve this
$queryBuscaprov = mysqli_query($conection, "SELECT id FROM tblprov WHERE razonSocial = '$razonSocial'");
$result_prov = mysqli_fetch_array($queryBuscaProv);
$idProv = $result_prov['id'];
foreach ($_POST['idRubro'] as $opcionSeleccionada)
{
//This INSERT should execute as many times as vendor-types selected in form
$query2 = mysqli_query($conection, "INSERT INTO tblprovxrubro (idRubro, idProv) VALUES ('$opcionSeleccionada', '$idProv')");
}
}else{
$alert='<p class="msg_error">Error creating vendor.</p>';
}
}
}
mysqli_close($conection);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<?php include "includes/scripts.php"; ?>
<title>Registro de Proveedores</title>
</head>
<body>
<?php include "includes/header.php"; ?>
<section id="container">
<div class="form_register" name="form_register">
<h1><i class="fas fa-building"></i> Vendor List</h1>
<hr>
<form action="" method="post" class="form_register">
<div class="alert"><?php echo isset($alert) ? $alert : ''; ?></div>
<label for="razonSocial">Razón Social</label>
<input type="text" name="razonSocial" id="razonSocial" placeholder="Razón Social">
<label for="telefono">Número Telefónico</label>
<input type="text" name="telefono" id="telefono" placeholder="Número Telefónico">
<label for="domicilio">Domicilio</label>
<input type="text" name="domicilio" id="domicilio" placeholder="Domicilio">
<label for="email">Email</label>
<input type="text" name="email" id="email" placeholder="Email">
<?php
$query_rubro = mysqli_query($conection,"SELECT * FROM tblrubros");
mysqli_close($conection);
$result_rubro = mysqli_num_rows($query_rubro);
?>
<select name="rubro" id="rubro" multiple size="12">
<?php
if($result_rubro > 0)
{
while ($rubro = mysqli_fetch_array($query_rubro))
{
?>
<option value="<?php echo $rubro["idRubro"]; ?>"><?php echo $rubro["rubroDescripcion"] ?></option>
<?php
}
}
?>
</select>
<p>press Ctrl in order to select multiple options.</p>
<button name="submit" type="submit" class="btn_save"><i class="fas fa-save"></i> Crear Proveedor</button>
</form>
</div>
</section>
<?php include "includes/footer.php"; ?>
</body>
</html>
conection.php
<?php
$host = 'localhost';
$user = 'root';
$password = 'mypass';
$db = 'compras';
$conection = @mysqli_connect($host,$user,$password,$db);
mysqli_set_charset($conection,"utf8");
if(!$conection){
echo "connection error";
}
?>
php mysql insert
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
I am creating a simple site to keep records of users, customers, suppliers, etc.
I have created the forms to register clients and users using single forms capturing data through $ _POST form method
So far I have not had problems since the INSERTS are done on singles table wich PK is an AutoIncremented field
In the code shown below, my goal is to let users create a vendor (name, phone number, email, address) as well as asign 1 or many vendor-category. This information is contained in one form
vendors, vendors-categories, users etc are stored in a mysql database with the structure shown in attached file
structure
There is something wrong about my code after succesfully create vendor and i am not sure how to solve the part that manages insertions in the join table (tblprovxrubro)
Code
<?php
session_start();
//available for admin, power-users and role-4
if ($_SESSION['rol'] ==2) {
header("location: ./");
}
include "../conexion.php";
if (!empty($_POST)) {
$alert='';
if (empty($_POST['razonSocial'])|| empty($_POST['email'])) {
$alert='<p class="msg_error">Vendor name and email must not be blank.</p>';
}else{
$razonSocial = $_POST['razonSocial'];
$email = $_POST['email'];
$domicilio = $_POST['domicilio'];
$telefono = $_POST['telefono'];
$usuario_id = $_SESSION['iduser'];
$query = mysqli_query($conection,"SELECT * FROM tblprov WHERE razonSocial = '$razonSocial'");
$result = mysqli_fetch_array($query);
if ($result >0)
{
$alert='<p class="msg_error">vendor already exists.</p>';
}else{
$query_insert = mysqli_query($conection, "INSERT INTO tblprov(razonSocial, numero, domicilio, email, idUsuario)
VALUES('$razonSocial','$telefono','$domicilio','$email', '$usuario_id')");
if ($query_insert)
{
$alert='<p class="msg_save">Vendor created succesfully.</p>';
//Once vendor is created i need his ID in order to insert in table
//tblprovxrubro as many records as vendor-types selected in form
//Not sure about how to achieve this
$queryBuscaprov = mysqli_query($conection, "SELECT id FROM tblprov WHERE razonSocial = '$razonSocial'");
$result_prov = mysqli_fetch_array($queryBuscaProv);
$idProv = $result_prov['id'];
foreach ($_POST['idRubro'] as $opcionSeleccionada)
{
//This INSERT should execute as many times as vendor-types selected in form
$query2 = mysqli_query($conection, "INSERT INTO tblprovxrubro (idRubro, idProv) VALUES ('$opcionSeleccionada', '$idProv')");
}
}else{
$alert='<p class="msg_error">Error creating vendor.</p>';
}
}
}
mysqli_close($conection);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<?php include "includes/scripts.php"; ?>
<title>Registro de Proveedores</title>
</head>
<body>
<?php include "includes/header.php"; ?>
<section id="container">
<div class="form_register" name="form_register">
<h1><i class="fas fa-building"></i> Vendor List</h1>
<hr>
<form action="" method="post" class="form_register">
<div class="alert"><?php echo isset($alert) ? $alert : ''; ?></div>
<label for="razonSocial">Razón Social</label>
<input type="text" name="razonSocial" id="razonSocial" placeholder="Razón Social">
<label for="telefono">Número Telefónico</label>
<input type="text" name="telefono" id="telefono" placeholder="Número Telefónico">
<label for="domicilio">Domicilio</label>
<input type="text" name="domicilio" id="domicilio" placeholder="Domicilio">
<label for="email">Email</label>
<input type="text" name="email" id="email" placeholder="Email">
<?php
$query_rubro = mysqli_query($conection,"SELECT * FROM tblrubros");
mysqli_close($conection);
$result_rubro = mysqli_num_rows($query_rubro);
?>
<select name="rubro" id="rubro" multiple size="12">
<?php
if($result_rubro > 0)
{
while ($rubro = mysqli_fetch_array($query_rubro))
{
?>
<option value="<?php echo $rubro["idRubro"]; ?>"><?php echo $rubro["rubroDescripcion"] ?></option>
<?php
}
}
?>
</select>
<p>press Ctrl in order to select multiple options.</p>
<button name="submit" type="submit" class="btn_save"><i class="fas fa-save"></i> Crear Proveedor</button>
</form>
</div>
</section>
<?php include "includes/footer.php"; ?>
</body>
</html>
conection.php
<?php
$host = 'localhost';
$user = 'root';
$password = 'mypass';
$db = 'compras';
$conection = @mysqli_connect($host,$user,$password,$db);
mysqli_set_charset($conection,"utf8");
if(!$conection){
echo "connection error";
}
?>
php mysql insert
php mysql insert
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
asked Nov 20 '18 at 21:23
Hugo JastrzebskiHugo Jastrzebski
81
81
1
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
1
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57
add a comment |
1
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
1
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57
1
1
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
1
1
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57
add a comment |
0
active
oldest
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1
Hugo please use prepared statements, you code right now is prone to sql injection.
– Hackerman
Nov 20 '18 at 21:25
can you expand on "something wrong"? What are the results/errors?
– danblack
Nov 20 '18 at 21:33
will come back in a while and answer this
– Hugo Jastrzebski
Nov 20 '18 at 21:47
1
first you need to prone your results against sql injection. Secondly instead of using else if statement to check if your first query is ok before applying the second one. I suggest you use the transaction mechanism with multi queries that only commit your queries if all are correct, If not you rollback. multi_queries, transaction
– Loic L.
Nov 20 '18 at 21:57