How to return all rows if using if (empty($variable))?

Question

I am using this query below and it only returns the first query in the entry if i use only the if (empty($field1)) to display. If i fill in the print(""); it works but i want to use the if (empty($field1)) snippet to display. How can i do it?

$sql="SELECT field1, field2 FROM table WHERE p_id='$pid'
      and k_id='$kid' ORDER BY id DESC";  
$result=mysql_query($sql);

$query = mysql_query($sql) or die ("Error: ".mysql_error());

if ($result == "")
{
    echo "";
}

echo "";

$rows = mysql_num_rows($result);

if($rows == 0)
{
    print("");
}
elseif($rows > 0)
{
    while($row = mysql_fetch_array($query))
    {
        $field1 = $row['field1'];
        $field2 = $row['field2'];
        print("");
    }
}

if (empty($field1)) {
    echo ""; //Thats right, i don't want anything to show for this portion
} else {
   echo "<div id=comments>Comments</div><br>
   <div id=entries>$field1 and $field2</div>";
}

are you executing your query twice or is it just because of an inconsistancy in the example? — Snicksie
– Snicksie, Commented Sep 22, 2011 at 14:04
How many rows do you expect it to return? What are p_id and k_id? — Michael Berkowski
– Michael Berkowski, Commented Sep 22, 2011 at 14:04
Check the values of $pid and $kid - possibly they aren't what you expect? — StuartLC
– StuartLC, Commented Sep 22, 2011 at 14:05
You are selecting field1 and field2, but later you are using $row['field1_name'] and $row['field2_id']. — Karolis
– Karolis, Commented Sep 22, 2011 at 14:07
Please consider reposting your actual code. The code provided doesn't show where you use $field1 and $field2 — Alfwed
– Alfwed, Commented Sep 22, 2011 at 14:15

Alon Eitan · Accepted Answer · 2011-09-22 14:22:06Z

2

What are you trying to do? somthing like this:

$sql="SELECT field1, field2 FROM table WHERE p_id='$pid' and k_id='$kid' ORDER BY id DESC";  
$result=mysql_query($sql)  or die ("Error: ".mysql_error());
$rows = mysql_num_rows($result);

if ($rows > 0)
  echo "here are your entries\n";

while($row = mysql_fetch_array($result))
{
    echo $row['field1']." ";
    echo $row['field2']."\n";
}

another way

$sql="SELECT field1, field2 FROM table WHERE p_id='$pid' and k_id='$kid' ORDER BY id DESC";  
$result=mysql_query($sql)  or die ("Error: ".mysql_error());
$rows = mysql_num_rows($result);

if ($rows > 0)
  echo "here are your entries\n";

while($row = mysql_fetch_array($result))
{
    if (empty($row['field1'])) {
        echo " ";
    } else {
    echo $row['field1']." ";
    echo $row['field2']."\n";
    }
}

edited Sep 22, 2011 at 14:22

answered Sep 22, 2011 at 14:15

Alon Eitan

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1 Comment

AAA Over a year ago

that would work but i need to have to dislay using f (empty($field1)) { echo ""; } else { echo "here are your entries"; }

owen gerig · Accepted Answer · 2011-09-22 14:27:22Z

1

i believe mysql_fetch_array only returns one row http://www.w3schools.com/PHP/func_mysql_fetch_array.asp

also ur sure that neither p_id and k_id are not unique?

i would also try $sql="SELECT * FROM table WHERE p_id='$pid' and k_id='$kid' ORDER BY id DESC";

just to see if that yields any different results, you can always parse out just the two fields from the return data

TRY THIS TO START WITH (the $results variable is just confusing things):

  $sql="SELECT field1, field2 FROM table WHERE p_id='$pid'
      and k_id='$kid' ORDER BY id DESC";  

      $query = mysql_query($sql) or die ("Error: ".mysql_error());

   $rows = mysql_num_rows($query);

  if($rows == 0)
  {
  print("");

  }else{

  while($row = mysql_fetch_array($query))
   {
          if ($row['field1'] == "")
            {
               print("");
            }else{
           $field1 = $row['field1'];
           print($field1)
            }

if ($row['field2'] == "")
            {
               print("");
            }else{
           $field1 = $row['field2'];
           print($field2)
            }
  }

}

edited Sep 22, 2011 at 14:27

answered Sep 22, 2011 at 14:08

owen gerig

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8 Comments

AAA Over a year ago

i have checked that works. the query works i use the print("") to display but if you notice i am using the if empty snipet so maybe thats why?

owen gerig Over a year ago

well per the other post about running your query twice and not really using the results in $results i would say clean up the code, run it, see what happens, then update us here. maybe update the code block in your post too since it looks a little messy

owen gerig Over a year ago

i would also say to maybe run your query in a sql window (as opposed to just within your code). this will give u a real understanding of what data is in there and what your query actually returns

AAA Over a year ago

ok if i use print("$field1 and $field2"); then all etnires show up, but as you can see i am not doing that, i want to use the if (empty($field1)) { echo ""; } else { echo "$field1 and $field2"; } instead of print("");

owen gerig Over a year ago

see my edits and if that fits your needs. what i dont get is a null field is a null field. meaning printing "" is the equivalent to printing a values thats set to null.

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