Re: Please help. Is this php script wrong?
From: Mark (drecaise_at_hotmail.com)
Date: 07/19/04
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Date: Mon, 19 Jul 2004 14:36:25 +0200
I haven't read the whole script but you're right about the $ok
variable... it isn't assigned any value before the if statement and that
makes a buggy script...
Mark
John Wrate wrote:
> Hi all,
>
> I have a customer who wants his site to reside in one of my hosts. The
> host I have assigned for him is a Windows NT server with the latest
> binary PHP and MySQL servers.
>
> The people that made the site are a**holes and want to host the site
> themselves and are resorting to all sorts of lies about our server not
> working porperly.
>
> They sent this script. I don't know any PHP but I feel something is
> not right with the script. And I get an error:
>
> Notice: Undefined variable: ok in C:\.....\test.php on line 3
>
> This is the test.php script:
>
> $bdConexion=mysql_pconnect("localhost","xxxx","11122"); // usr & pswd
> mysql_select_db("datos_prueba",$bdConexion);
> if(!$ok){
> $sql="CREATE TABLE Empleados (ID INT PRIMARY KEY,Nombre VARCHAR (50)
> , Apellidos VARCHAR (70), Edad INT(4))";
> if(!($reg=mysql_query($sql,$bdConexion)))
> echo "ERROR CREATING TABLE.";
> else
> {
> ?>
> <form name="forma1" ENCTYPE="multipart/form-data" method="post"
> action="">
> Nombre:<input type="text" name="name"><br>
> Apellido:<input type="text" name="last"><br>
> Edad:<input type="text" name="year"><br>
> Selecciona una Imagen:<input name="imagen" type="file">
> <br>
> <br><input type="submit" name="ok" value="Enviar"><br>
> </form>
> <?}
> }else if($ok){
> $sql2="INSERT INTO Empleados(ID,Nombre,Apellidos,Edad) VALUES
> (".'1'.",'".$name."','".$last."',".$year.")";
> if(!($reg2=mysql_query($sql2,$bdConexion)))
> echo "ERROR AL INSERTAR EN LA TABLA.".$sql2;
> $sql3="UPDATE Empleados SET Nombre='Perlita' WHERE ID=1";
> if(!($reg3=mysql_query($sql3,$bdConexion)))
> echo "ERROR AL ACTUALIZAR LA TABLA";
> $sql4="SELECT * FROM Empleados";
> if(!($reg4=mysql_query($sql4,$bdConexion)))
> echo "ERROR AL SELECCIONAR DE LA TABLA";
> ELSE{
> echo "<BR>TUS DATOS FUERON:<BR><BR>";
> IF($regis=mysql_fetch_array($reg4))
> {
> echo "Nombre:".$regis[Nombre]."<br>";
> echo "Apellido:".$regis[Apellidos]."<br>";
> echo "Edad:".$regis[Edad]."<br>";
> }
> }
> echo "Imagen=".$imagen_name;
> echo "<br>";
> $pathSeccion="../media/";
> $pathfin=$pathSeccion."1";
> echo "<BR>== GRABANDO UNA IMAGEN ==<BR>";
> if(!file_exists($pathfin))
> {
> mkdir($pathfin,0777);
> if($imagen!=""){
>
> copy($imagen,$pathfin."/".$imagen_name);
> }
> }
> ECHO "<BR>LEYENDO LA IMAGEN<BR><br>";
> if(file_exists($pathfin))
> {
> $dir = opendir($pathfin);
> while($arch = readdir($dir)){
> if($arch != '.' && $arch != '..')
> echo "Imagen: <img src=\"".$pathfin."/".$arch."\"><br>";
> }
> closedir($dir);
> }
> echo "<BR><h1>==TEST OK ==</h1><BR>";
> }//ok
>
>
> In the first "If" it is very strange to me to see that if the
> mysql_select_db falied if(!$ok), then a table is being created! How is
> that possible, and should't ok be assigned the value of the
> my_select_db(xxxx,xxx) in order to test it later. Like I said, I get
> the error and although the table is created the first time the
> test.php script is opened/run, the data is not inserted in the second
> run (after filling the form).
>
> Any comments will be very much appreciated.
>
> Wrate
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