Solved

If Else Delima, if query doen's return antying, can't get the else part to dosplay

Posted on 2012-03-16
4
283 Views
Last Modified: 2012-06-26
I'm getting frusted over an if else statememt that won't work.  It is calling one variable from the table.  If the variable is in the table it will display one thing, if the variable is absent, it should display something else.  I've tried !empty, isset, even $num_rows =1, but the content in the else doesn't display.  Here's the code:

Code: [Select]


<?
$result = mysql_query( "SELECT * FROM codeWords WHERE Mycode = '$MyCode' " )
or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
  while ($row = mysql_fetch_array($result))
  {
     extract($row);
     
if (!empty( $MyCode  )) {    
echo "here is the code word";
} else {
echo "you need to go back and get the code";
}
$row_count++;
}
mysql_close;
?>


When the code variable is correct the "here is the code word" displays.  when the code variable is incorrect, what should appear as "you need to go back and get the code" is just blank.

Any thoughts on how to structure the query in such a way that eh incorrect text appears?

Thanks,
0
Comment
Question by:TecTaoMC
  • 2
4 Comments
 
LVL 14

Expert Comment

by:Scott Madeira
Comment Utility
It looks like $MyCode will always have a value.  That is what you are passing in to look at.  What you need to look at in your if else statement is one of the fields from your query result or check to see how many rows you get back.

If the codeword is in your database then $num_rows will be greater than 0.  If the codeword is not in the database then $num_rows would be 0.
0
 
LVL 14

Expert Comment

by:Scott Madeira
Comment Utility
A couple other things:  your query will only return rows that have the $MyCode codeword in it.  Not sure what you are trying to do with the row_count++ variable.  That should have the same value as $num_roes;


And, if you want to look at the value of the MyCode field from your database rows then you will want to reference $row['MyCode'] and not $MyCode.
0
 

Author Comment

by:TecTaoMC
Comment Utility
Thanks for your quick reply.  I read and made some changes but still can't seem to get the variable to whow 0 if no result is returned which fowls up the if else statememt.  The if else have been written a couple of different ways if $row_count == 1 {....}else {....} and if @num_row == 1 {..} If $row_count == 0 {..}

here's the code using
<?php
include("include/session.php");
$MyCodePassed = $_POST['MyCode'];

$result = mysql_query( "SELECT * FROM codeWords WHERE Mycode = '$MyCodePassed' " )
or die("SELECT Error: ".mysql_error());
$row_count = mysql_num_rows($result);
  while ($row [Mycode]= mysql_fetch_array($result))
  {
     extract($row [Mycode]);
     
if ($row_count == 1) {    
 echo show the page for this}
 if ($row_count == 0) {    
 echo show the page for no result}  
?>

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0
 
LVL 108

Accepted Solution

by:
Ray Paseur earned 500 total points
Comment Utility
Try it a little more like this.  It will enable you to see the data and that means you will have a better idea of what the programming is doing.  Once you have this working, then you can use the if() statement to test the possible different values of $num
<?php
error_reporting(E_ALL);

// DOES THIS CONNECT AND SELECT THE DATA BASE?
include("include/session.php");

// ESCAPE THE EXTERNAL DATA FOR SAFE USE IN A QUERY
$MyCodePassed = mysql_real_escape_string($_POST['MyCode']);

// PREPARE THE QUERY STRING
$sql = "SELECT * FROM codeWords WHERE Mycode = '$MyCodePassed' ";

// ACTIVATE THIS TO SHOW THE QUERY STRING
// var_dump($sql);

// RUN THE QUERY AND TEST FOR SUCCESS
$res = mysql_query( $sql );
if (!$res)
{
    echo "FAIL: $sql<br/>";
    die ( mysql_error() );
}

// GET THE NUMBER OF ROWS AND DISPLAY THE NUMBER
$num = mysql_num_rows($res);
echo "<br/>QUERY FOUND $num ROWS";

// USE AN ITERATOR TO SHOW WHAT THE QUERY FOUND
while ($row = mysql_fetch_assoc($res))
{
    echo "<br/>";
    print_r($row);
}

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