Problem in Php --Parse error: syntax error, unexpected '='

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I am trying to make a form that takes the data from a table in phpmyadmin. It gives me error in the following code. Does anyone know the error? 

<?php
    if(isset($_GET["libro"])){
        $resultado = mysqli_query($enlaceBD, "SELECT * FROM libros where id= row[$id]");  
    } else {
        resultado = mysqli_query ($enlaceBD, "SELECT * FROM libros where id = '1'"];
    }
    $row = $resultado ->fetch_assoc();
{
    
asked by Juan 20.11.2018 в 13:25
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1 answer

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Replace this 

$resultado = mysqli_query($enlaceBD, "SELECT * FROM libros where id= row[$id]");

Because of this 

$resultado = mysqli_query($enlaceBD, "SELECT * FROM libros where id= ".row[$id].")";
    
answered by 20.11.2018 в 13:36
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