Hi
I have one form, named, exemplo.html with one form and action=teste.php
In my file teste.php i need to insert one record in my table with one sql statment. I have one error, please help me. Here is my code:
<?php
//----liga a abase de dados
/* conexão à base de dados vendas*/
$mysql_id = mysql_connect('localhost', 'curso', '123');
mysql_select_db('vendas',$mysql_id);
if(!mysql_select_db('vendas',$mysql_id)){
die ('Erro'.mysql_error());
}
//--
$query='insert into negocios set cliente=$_GET["nome"]';
$result=mysql_query($query);
if(!$result){
echo(mysql_error());
}
else{
echo("Adicionado o Registo");
}
?>
Newbie - Insert Record from one form
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mariolopes
- Forum Contributor
- Posts: 102
- Joined: Sun May 22, 2005 7:08 am
-
mariolopes
- Forum Contributor
- Posts: 102
- Joined: Sun May 22, 2005 7:08 am
Re: Newbie - Insert Record from one form
Never mind. Found it
<?php
/* conexão à base de dados vendas*/
$mysql_id = mysql_connect('localhost', 'curso', '123');
mysql_select_db('mariolopes',$mysql_id);
//--
$query="INSERT INTO negocios (Codigo,Cliente)
VALUES ('$_GET[codigo]','$_GET[nome]')";
$result=mysql_query($query);
if(!$result){
echo(mysql_error());
}
else{
echo("O registo foi adicionado à Tabela.");
}
?>
<?php
/* conexão à base de dados vendas*/
$mysql_id = mysql_connect('localhost', 'curso', '123');
mysql_select_db('mariolopes',$mysql_id);
//--
$query="INSERT INTO negocios (Codigo,Cliente)
VALUES ('$_GET[codigo]','$_GET[nome]')";
$result=mysql_query($query);
if(!$result){
echo(mysql_error());
}
else{
echo("O registo foi adicionado à Tabela.");
}
?>