Parse error syntax error, unexpected 'if' in PHP: (Fixed)

in this tutorial I am going to explain how to fix a most common error that every programmer face i.e Parse error: syntax error, unexpected 'if' (T_IF) in php.

[NewMusArt]

Thanks this was EXTREMELY helpful and easy to follow! Much appreciated.

[nataschabjerning]

Omg, so simple and yet i was completely lost until i found this. Thank you!

[indrajitganguly]

THANKS THANKS THANKS.... I CAN'T THANK YOU ENOUGH. I WAS STRUGGLING WITH THIS FOR LAST TWO HOURS ! YOU SAVED MY LIFE ....

[pelesengupta]

Thank you so much for this video.Keep progressing.

[poisedstool]

holy shit. i just missed a semi colon HAHAHAHAHA

[SLC]

I'm not a dev, I just built a wordpress site on my own and I keep getting this error. I can't login to the site? "Parse error: syntax error, unexpected 'if' (T_IF), expecting ')' in on line 406" is the error I'm getting - is this related to the theme itself? I didn't write that code.

[mauriciomenacousin]

Si le aparece este error abajo digo como se soluciona
si el usuario no esta vacio
$usuarioIDActualizado = $_GET[ 'usuarioIDActualizado' ];
$nombreActualizado = $_GET[ 'nombreActualizado' ];
$cliente = mysqli_real_escape_string($conexion, $nombreActualizado );
$resultado = mysqli_query($conexion, "UPDATE users SET name = '$cliente' WHERE id = '$usuarioIDActualizado'");
mysqli_close($conexion);
}
si el id del usuario eliminado no se encuentra vacio

$resultado=mysqli_query($conexion, "DELETE FROM users WHERE id='$usuarioIDEliminado'");
mysqli_close($conexion);
}
el error es
Parse error: syntax error, unexpected 'if' (T_IF) in C:\xampp\htdocs\curso ajax\curso de ajax video23 PARTE 4 - Borrar Usuario DELETE\servidor.php on line 43

Ya encontre la solución es así

$usuarioIDActualizado = $_GET[ 'usuarioIDActualizado' ];
$nombreActualizado = $_GET[ 'nombreActualizado' ];
$cliente = mysqli_real_escape_string($conexion, $nombreActualizado );
$resultado = mysqli_query($conexion, "UPDATE users SET name = '$cliente' WHERE id = '$usuarioIDActualizado'");
mysqli_close($conexion);
}
si el id del usuario eliminado no se encuentra vacio

$resultado=mysqli_query($conexion, "DELETE FROM users WHERE id='$usuarioIDEliminado'");
mysqli_close($conexion);
}

[derpsicles]

Although i placed a semi colon at the end of the variable. There still is an error with the if stament

[ithielragnaros]

o thanks, gracias amiga me sirvio mucho no sabia que pedo jajaja gracias