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view several values in a drop down list


farahZ
Go to solution Solved by taquitosensei,

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i have a table in mysql/phpmyadmin

Food, Size, Calories

i concatenated 2 values and its working successfully in the drop down list according to this code:

 

 

 
<?PHP
  $query = "SELECT  Food,Calories FROM food order by Food";
  $result = mysqli_query($con,$query);
while($row = mysqli_fetch_assoc($result)) {
    echo ("<option VALUE=\"".$row['Food']."\" " . ($food == $row['Food'] ? " selected" : "") . ">".$row['Food'].' - '.$row['Calories']."</option>");
  }
?>

 

i need to show the Size too!!
and this code isnt working any help??

 
while($row = mysqli_fetch_assoc($result)) {
    echo ("<option VALUE=\"".$row['Food']."\" " . ($food == $row['Food'] ? " selected" : "") . ">".$row['Food'].' - ' .$row['Size']. ' - '.$row['Calories']."</option>");
  }
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  • Solution

It looks like you're not selecting the size.

 

SELECT  Food,Calories FROM food order by Food

 

 

i have a table in mysql/phpmyadmin

Food, Size, Calories

i concatenated 2 values and its working successfully in the drop down list according to this code:

 
<?PHP
  $query = "SELECT  Food,Calories FROM food order by Food";
  $result = mysqli_query($con,$query);
while($row = mysqli_fetch_assoc($result)) {
    echo ("<option VALUE=\"".$row['Food']."\" " . ($food == $row['Food'] ? " selected" : "") . ">".$row['Food'].' - '.$row['Calories']."</option>");
  }
?>

i need to show the Size too!!
and this code isnt working any help??

 
while($row = mysqli_fetch_assoc($result)) {
    echo ("<option VALUE=\"".$row['Food']."\" " . ($food == $row['Food'] ? " selected" : "") . ">".$row['Food'].' - ' .$row['Size']. ' - '.$row['Calories']."</option>");
  }
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