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Parse error: syntax error, unexpected end of file in C:\xampp\htdocs\clinica\val...

 2 years ago
source link: https://www.codeproject.com/Questions/5318145/Parse-error-syntax-error-unexpected-end-of-file-in
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Hi everyone! I have a problem in My system when I send My data to My database, and I do not know why it happens, by the way my queries I have to change them to parameterized (THE ERROR IS IN MY QUESTION 😉)

This is my code(THE ERROR IT´S IN THE LAST LINE):

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<?php 
  include ("conex.php");
  //logeo
 
session_start();
include("conex.php"); 
if(isset($_POST['ingre'])){
  
  $name = $_POST["name"];
  $password = $_POST["pass"];

  $sql = "SELECT COUNT(*) as contar FROM usuario  WHERE name ='$name' AND pass ='$password'";
   $consult=mysqli_query($conn,$sql);
   $arreglo=mysqli_fetch_array($consult);
   
  if($arreglo['contar']> 0) {
	  $_SESSION['usuario']=$name;
	 echo "<script>alert('Bienvenido: $name'); window.location='home.php'</script>";
  }else{
	  echo "<script>alert('Los datos estan mal'); window.location='login.php'</script>";
  }
 
  //Registrar
$name = $_POST["name"];
$apellido = $_POST["apec"];
$dad = $_POST["namep"];
$mom = $_POST["namem"];
$tipeCI = $_POST["tipeid"];
$ci = $_POST["ci"];
$phone = $_POST["te"];
$correo = $_POST["email"];
$add = $_POST["dirre"];
$day = $_POST["diana"];
$gradoI = $_POST["gradoin"];
$esC = $_POST["estado"];
$sex = $_POST["sex"];
$pass = $_POST["pass"];
  
if(isset($_POST["butt"]))
{
	$sqlgrabar = "INSERT INTO usuario(name,last,nameDad,nameMom,tipoIdentificacion,identificacion,telefono,email,address
	,born,rango,estadoC,sex,pass)
	values ('$name','$apellido','$dad','$mom','$tipeCI','$ci','$phone','$correo','$add','$day','$gradoI','$esC',
	'$sex','$pass')"; 
	
	//EVITAR REPETIDO
	 $veri_correo=mysqli_query($conn,"SELECT * FROM usuario WHERE email='$correo'");
	 $veri_pass=mysqli_query($conn,"SELECT * FROM usuario WHERE pass='$pass'");
 
	  if(mysqli_num_rows($veri_correo) > 0){
		  echo '<script>window.location="dise/errorEgistro.php";</script>';
		  
		  exit();
	  }
	  
	  if(mysqli_num_rows($veri_pass) > 0){
		   echo '<script>window.location="dise/errorEgistro.php";</script>';
		  exit();
	  }
	
	if(mysqli_query($conn,$sqlgrabar))
	{
		echo "<script> alert('Felicidades el usuario fue registrado con exito: $name'); window.location='login.php' </script>";
	}else 
	{
		echo "Error: ".$sqlgrabar."<br>".mysqli_error($conn);
	}
}

?>  


What I have tried:

I tried some thing but nothing, I Am a little NOOB, and I don't why the error happened

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