Question

$selectra_result = mysql_query("SELECT * FROM `my_selectra` WHERE user_id='$user_id'") or die(mysql_error()); 

while($row = mysql_fetch_array( $selectra_result )) { 
     $product_id = $row['product_id']; 
     $namep[] = $product_name= $row['product_name'];
     $linkp[] = $product_permalink = $row['product_permalink'];
     $imgpp[] = $product_image_path = $row['product_image_path'];    
}

     foreach($namep as $pname => $prodname){
        foreach($linkp as $plink => $prodlink) {
         foreach($imgpp as $ppath => $prodpath) {

        $res1 = "<a href='$prodlink'><p>$prodname</p></a>";
        $res2 = "<a href='$prodlink'><img src='$prodpath'/>";

            }
        }     
      }
      echo $res2.''.$res1;

}

I want to display data from my_selectra table using foreach, unfortunately it appears only one data when echo.

Answer 1

Your code is pretty convoluted… You don’t need 3 arrays to grab the data you want, nor do you need a triple-nested foreach loop structure. Try something like this:

$results = array();
while($row = mysql_fetch_array( $selectra_result)) 
{ 
     $product_id = $row['product_id']; 
     $results[] = array( $row['product_name'], $row['product_permalink'], $row['product_image_path']);
}

$res1 = $res2 = '';
foreach( $results as $row)
{
    $res1 .= "<a href='" . $row[1] . "'><p>" . $row[0] . "</p></a>";
    $res2 .= "<a href='" . $row[1] . "'><img src='" . $row[2] . "'/>";
}
echo $res1 . $res2;