Parse error: syntax error, unexpected T_IF in index.php on line 46

jtomasrl

Hi, im having a problem here
Line 47 is if ('%s' == 0)

      <TR><TD>Tipo</TD><TD>Cantidad</TD><TD>Fecha</TD></TR> 
<?php       

   while($row = mysql_fetch_array($result)) { 
      printf('<tr><td>'
			 if ('%s' == 0)
			 {
			 printf("Deposito");
			 }
			 else {
			 printf("Gasto");
			 }
			 '</td><td>%s</td><td>%s</td></tr>', $row["tipo"],$row["cantidad"],$row["fecha"]); 
   } 
   mysql_free_result($result); 
   mysql_close($link); 
?>

Pikachu2000

The problem is in the line above it. The printf isn't closed . . .

jtomasrl

$row["fecha"])

there is closed

bradgrafelman

Why do you have a bunch of PHP code in the middle of a call to printf()? Also, your if() statement doesn't make sense... a percent sign followed by the letter 's' is never equal to 0.

Sounds like you should either explain what you're trying to do, or look at the manual to see how [man]printf/man works.

jtomasrl

the printf shows html code (its a table) but inside a table i want to show a text from a Boolean, so if it is 0 i shows a text, if it is 1 is hows another text, the %s place the $row["tipo"], but i guess %s doesnt work for boolean tipe :S

bradgrafelman

Well for one, you can't throw an if() statement and other PHP code into the middle of a string. For another, '%s' is just a placeholder that is evaluated internally in the printf() function; outside of that function (e.g. in an if() statement), it's just a literal percent sign followed by the letter 's'.

Since you don't gain any benefit by using printf() here, the easiest solution would just be to get rid of it:

while($row = mysql_fetch_array($result)) {
  echo "<tr><td>";

  if ($row["tipo"] == 0) 
  { 
    echo "Deposito";
  } else { 
    echo "Gasto";
  } 

  echo "</td><td>$row[cantidad]</td><td>$row[fecha]</td></tr>\n";
}

Or, more compactly (using the ternary operator):

while($row = mysql_fetch_array($result))
  echo "<tr><td>" . ($row["tipo"] == 0 ? "Deposito" : "Gasto") . "</td><td>$row[cantidad]</td><td>$row[fecha]</td></tr>\n";

jtomasrl

yeah ty =)
i created a code at the sime time like this

<?php

while($row = mysql_fetch_array($result)) {

if ($row["tipo"] == 0)
{
$tipo = 'Deposito';
}
else
{
$tipo = 'Gasto';
}
  printf('<tr><td>&#37;s</td><td>%s</td><td>%s</td></tr>', $tipo,$row["cantidad"],$row["fecha"]); 
}
mysql_free_result($result);
mysql_close($link);
?>