PHP Array option to get new file

Question

I've been struggling getting an array information to match and get posted/get onto another page.

How is it possible to match the $name so that it will display the rest of the array to the appropriate name on the new page?

I have spent several days doing various combinations and tries without success, any assistance will be much appreciated!

This is the array I'm retrieving the information:

<?php  //This is the p6_Assignment2.php
$students = array(
    "Jim, Carrey" => array( "0001"  => array("Class1" => array("grade1" => 33, "grade2" => 33, "grade3" => 34))),
    "John, Connor" => array( "0002" => array("ClassB" => array("grade1" => 20, "grade2" => 60, "grade3" => 20))),
    "Anderson, Silva" => array( "0003"  => array("ClassC" => array("grade1" => 40, "grade2"=> 30, "grade3" => 30)))
);

?>

The html data

<html>
<head>
    <title>test</title>
</head>
<body>
    <form action="p8_Assignment2.php" method="get">
        <select name="selectPage">
            <?php
            include ('p6_Assignment2.php');
            foreach ($students as $name => $id) {
                ?>
                <option value="<?php echo $name ?>">  <?php echo $name ?>  </option>;
                <?php
            }
            ?>




        </select>
        <input type="submit" value="Submit">

    </form>
</body>

The 3rd File Where the data should show

<html>
<head>
    <title>Results</title>
</head>
<body>  
        <?php 
    foreach($_POST as $name){
        echo $name;
    }
            ?>
</body>

apologies for missing this Expected Output

Jim Carrey (0001)

ClassC : 81% , ClassC : 44% , ClassC : 55% ,

UPDATE I have readjusted the problem above.

Now i still cant seem to access the array to display the information.

how is it possible to use the $_POST data, (in this case lets assume: "Jim, Carrey") and get the rest of the array for the specified user?

Post the expected output

Narendrasingh Sisodia
– Narendrasingh Sisodia

2015-07-14 06:12:26 +00:00
Commented Jul 14, 2015 at 6:12 — Narendrasingh Sisodia
– Narendrasingh Sisodia, Commented Jul 14, 2015 at 6:12

Sougata Bose · Accepted Answer · 2015-07-14 06:16:56Z

1

Without the name of the input field you cannot access them in php.

Add name of the select -

<select name="nameOfFoeld">

And also add the value of option -

echo "<option value='theValueOfOption'>".$name ." </option>";

answered Jul 14, 2015 at 6:16

Sougata Bose

31.8k8 gold badges52 silver badges89 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Daniel Krom · Accepted Answer · 2015-07-14 06:20:30Z

0

you should add your select tag also a name attribute

<select name="test">
    <option value=1>option name to show to user</option>
    <option value=2>another option name to show to user</option>
</select>

this will allow you the get the VALUE of the selected option. if the user selected option 2

you'll be able to see this by

<?php
    echo $_GET['test']; //this will print 2
?>

btw, why do you need a structure of array=>array=>array=>array?

answered Jul 14, 2015 at 6:20

Daniel Krom

10.1k3 gold badges47 silver badges45 bronze badges

1 Comment

Semachka Ukr Over a year ago

I appreciate the help, its an assignment I had but couldn't figure out.

Collectives™ on Stack Overflow

PHP Array option to get new file

Jim Carrey (0001)

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Jim Carrey (0001)

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related