MySQL和PHP数据库？

Question

I am doing an exercise with PHP and MySQL database management, and I'm trying to return an amount from a field in a table with PHP. I am trying to get the value from the CalsPerServ in the foods table where the foodsID is equal to the foodsID the user has selected. All I'm getting now though is an array, when I should be returning one result. Any idea what went wrong?

if($tableName == "meals"){
        $foodsID = filter_input(INPUT_POST, "foodsID");
        $servings = filter_input(INPUT_POST, "Servings");
        $foodsID = mysql_real_escape_string($foodsID);
        $servings = mysql_real_escape_string($servings);

        $query = "SELECT CalsPerServ FROM foods WHERE foodsID=$foodsID";
        $result = mysql_query($query, $connect);
        $CalsPerServ = mysql_fetch_assoc($result);

        print $CalsPerServ;

        $calories = ($CalsPerServ * $servings) * 100;

        $sql .= "(foodsID, Servings, Calories) VALUES ('$foodsID', '$servings', '$calories')";

        $sqltwo = "INSERT INTO daily_diary (foodsID, CaloriesPlus) VALUES ('$foodsID', '$calories')";
        $resulttwo = mysql_query($sqltwo, $connect);
}

Answer 1

The problem lies here in this line

$calories = ($CalsPerServ * $servings) * 100;

$CalsPerServ is an array containing the result from the query. In order to get single column value out of it you have to use it like this

$calories = ($CalsPerServ['CalsPerServ'] * $servings) * 100;

Answer 2

Even though you are querying for one thing, mysql_fetch_assoc() returns an array.

This way even if you have selected more than one thing, you will always know you will get back an associative array, with the selected items as its keys.

$CalsPerServ = mysql_fetch_assoc($result);

 print $CalsPerServ['CalsPerServ'];