drop down list from mysql

[farahZ]

hey team
i'm trying to extract data from a table in xampp server, mysql

database name= inshapewebsite

table= food

and use this data to create a dropdown list that has 2 columns: Food, Calories

is that possible??
 

i have this code so far

 

 

 
<?
// Create connection
$con=mysqli_connect("localhost","root","","inshapewebsite");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}  
else 
{ 
 
<select name="foodType" onchange="autoSubmit();">
        <option VALUE="null"></option>
 $result = "SELECT  Food,Calories FROM food";
        $food = mysql_query($sql,$conn);
 
        while($row = mysql_fetch_array($food))
        {
            echo ("<option VALUE=\"$row[Food]\" " . ($food == $row["Food"] ? " selected" : "") . ">$row[Calories]</option>");
        }
 
}
</select>
?>
 

[computermax2328]

Are you getting any errors??

[farahZ]

yes!!
$result = "SELECT Food,Calories FROM food"; $food = mysql_query($sql,$conn); while($row = mysql_fetch_array($food)) { echo (""); } /*echo "Select food: \n" echo "";  while($row=mysqli_fetch_assoc($result)){ $Food=$row['Food']; $Calories=$row['Calories']; print "$Food\n"; }  print " \n"; print "

 

\n";*/ } ?>

[PaulRyan]

You have HTML inside of your PHP tags, you need to echo or print that data, or close the tags and open them when needed.

 

Something like this:

<?PHP
  // Create connection
  $con=mysqli_connect("localhost","root","","inshapewebsite");
  // Check connection
  if (mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    exit;
  }
?>
 
  <select name="foodType" onchange="autoSubmit();">
    <option VALUE=""></option>
<?PHP
  $result = "SELECT  Food,Calories FROM food";
  $food = mysql_query($sql,$conn);
 
  while($row = mysql_fetch_array($food)) {
    echo ("<option VALUE=\"{$row['Food']}\" " . ($food == $row['Food'] ? " selected" : "") . ">{$row['Calories']}</option>");
  }
?>
 
  </select>

Edited May 3, 2013 by PaulRyan

[farahZ]

no errors .. but getting an empty dropdown list
its not reading data from the table
i double checked the table name, attributes .. no error in that

Edited May 3, 2013 by farahZ

[davidannis]

You are mixing mysql and mysqli.

 

You can not do that.

 

Try

$query = "SELECT  Food,Calories FROM food";
  $result = mysqli_query($query,$conn);
while($row = mysqli_fetch_array($result))

Edited May 3, 2013 by davidannis

[davidannis]

I would also suggest using mysqli_fetch_assoc instead of mysqli_fetch_array

[farahZ]

still an empty list..

 

 <select name="foodType" onchange="autoSubmit();">
    <option VALUE=""></option>
<?PHP
  $query = "SELECT  Food,Calories FROM food";
  $result = mysqli_query($query,$conn);
while($row = mysqli_fetch_assoc($result)) {
    echo ("<option VALUE=\"{$row['Food']}\" " . ($food == $row['Food'] ? " selected" : "") . ">{$row['Calories']}</option>");
  }
?>

[computermax2328]

mysqli_query() is $connection, $query and not the other way around.

 

See reference link here

 

Should be like this....

$result = mysqli_query($conn,$query);

[farahZ]

Problem solved

i am getting now the list )

one last question, is it possible that in the drop down list, i get to show both column values?? i.e: Food, Calories ??

option: Cucumber, 30 ?

that's the correct code

 

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

Edited May 3, 2013 by farahZ

[farahZ]

Lol I found a way by just adding the calories in the food column

Example: cucumber-30

But keeping the calories column unchanged, which isbthe value I need for later usage

[Jessica]

Just concatenate the two values either in PHP or mysql.

[davidannis]

If you put them in the same column it makes it hard if you decide to do something like total the calories later. Try this instead:

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

[farahZ]

there's a parse error with the -

Parse error: syntax error, unexpected '-', expecting '}' in D:\xampp\htdocs\inshapewebsite\Untitled-1.php on line 24

 

i tried . - + ,
nothing works

[davidannis]

The following code:

<?php
$row['Calories']=90;
$food = $row['Food']="hot Dog";
echo ("<option VALUE=\"".$row['Calories']."\" " . ($food == $row['Food'] ? " selected" : "") . ">".$row['Food'].' - '.$row['Calories']."</option>");
?>

produces:

<option VALUE="90"  selected>hot Dog - 90</option>

I got rid of the curly braces and moved all the row array variables outside of the quotes. Perhaps there is a better way to do something like

 echo "ABC".$letters['next3']."DEF"; 

than to take the $letters variable out of the quotes but that's how I do it. Can anyone suggest a better solution?

[Jessica]

You can use curly braces

Echo "abc{$var['key']}123";

[Jessica]

Oh I see you did that - but you need them around each variable. So where you have

{$row['Food'] - $row['calories']}

 

It has to be

{$row['Food']} - {$row['calories']}

[farahZ]

thanks the two values are now being showed as one option in the drop down list .. 
the issue is that food chosen by the user are being saved in an array, and shown simultaneously.
but of course, only the calories value is shown in printing .. can it also be both..?

code for the list

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

// for saving the items chosen in an array

if (in_array($_POST['foodType'], $_SESSION['foodTypes']) === false) 
    {
      $_SESSION['foodTypes'][] = $_POST['foodType'];
    }
 

//code for printing the array simultaneously (after each add click)

 

 foreach ($_SESSION['foodTypes'] as $food)
{
  echo $food . '<br>';
}