check if rows exist in mysql, update. if no, create new row

[farahZ]

hello
i am trying to submit data in a table in phpmyadmin

the database is called "inshapewebsite"

the table is "caloriescounter"

the connection works fine for inserting news rows

but what i'm trying to do now is to update the table if a row with the same ID and Date exsists
i.e: primary keys are ID and Date

 here's the code am working on

 

 

 
// Create connection
$con=mysqli_connect("localhost","root","","inshapewebsite");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}  
$clid=111;
$date=date("m.d.y");
foreach ($_SESSION['foodTypes'] as $food)
{
$foodS= $foodS . ",". $food;
}
$result = mysql_query("Update caloriescounter set Lunch='$foodS' where ID='$clid' and Date='$date'");       
if (mysql_affected_rows()==0) {
    $result = mysql_query("insert into caloriescounter (ID, Date, Lunch)
VALUES ('$clid', '$date','$foodS')");
}
;
mysqli_close($con);
 
    // session exists and has content
    // process the array list here.
// after the processing, empty the session
    $_SESSION['foodTypes'] = array();

[Barand]

A single query will suffice

INSERT INTO caloriescounter (ID, Date, Lunch)
VALUES ('$clid', '$date','$foodS')
ON DUPLICATE KEY UPDATE Lunch = '$foodS'

[farahZ]

thanks

am getting this error:

Warning: mysql_query() expects parameter 1 to be string,

[farahZ]

thats the code

<?php
 
session_start();
 
if (isset($_POST['add']))
{
  // check if an option has been selected
  if (empty($_POST['foodType']))
  {
    echo 'You need to select some food!'; 
  }
  else
  {
    if (!isset($_SESSION['foodTypes']))
    {
 // if the session is not yet created, create it now
      $_SESSION['foodTypes'] = array();
    }
    // check to see if the newly added food type is not already in the array
    if (in_array($_POST['foodType'], $_SESSION['foodTypes']) === false) 
    {
      // The selected food item is not in the array
      // add the selected food item to total food array
      $_SESSION['foodTypes'][] = $_POST['foodType'];
    }
 
  }
echo "Food Added:" . '<br>';
  // display the current food list (in a really ugly PHP way)
  foreach ($_SESSION['foodTypes'] as $food)
{
  echo $food . '<br>';
}
} 
else if (isset($_POST['submit']))
{
  // check if the session exists, or if its empty
  if (!isset($_SESSION['foodTypes']) || empty($_SESSION['foodTypes']))
  {
    echo 'No food in the list to submit!';
  }
  else
  {
  
// Create connection
$con=mysqli_connect("localhost","root","","inshapewebsite");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}  
$clid=111;
$date=date("m.d.y");
foreach ($_SESSION['foodTypes'] as $food)
{
$foodS= $foodS . ",". $food;
}
    $result = mysql_query($con,"INSERT INTO caloriescounter (ID, Date, Lunch)
VALUES ('$clid', '$date','$foodS')
ON DUPLICATE KEY UPDATE Lunch = '$foodS'");
 
//mysqli_query($con,"INSERT INTO caloriescounter (ID, Date, Lunch)
//VALUES ('$clid', '$date','$foodS')");
mysqli_close($con);
 
    // session exists and has content
    // process the array list here.
// after the processing, empty the session
    $_SESSION['foodTypes'] = array();
  }
}
 
?>

Edited May 4, 2013 by ignace
Correct \ to /

[Barand]

you used mysql_query() instead of mysqli_query()

[Yohanne]

Yes you miss Mysql.

[farahZ]

thank you guys!