Call Variable for filename php [closed]

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I have a problem naming the variable for the name of the Excel file. I need to put the variable name in the $ file. The variable name comes from Select * ... 

 //--- Fazendo o preenchimento inicial --------------------------
 $row=mysql_fetch_array($sql);
 //- Definimos o nome do arquivo que será exportado -------------
 $arquivo = 'Empresa Não Autorizada '.$row[1]'.xls';

Give me this error: Parse error: syntax error, unexpected '' .xls ''

    
asked by anonymous 30.05.2014 / 11:03

1 answer

3

One dot is missing before xls : 

$arquivo = 'Empresa Não Autorizada ' . $row[1] . '.xls';
    
30.05.2014 / 12:06
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