PHP-如何选择多个复选框并显示所选复选框中的所有值？

Question

I have one html table with all the values from a specific table, and i want to put an update button to update specific rows

And i want to update those rows using checkboxes.

The user just need to check the selectboxes and click at a button, and this button will display all the information from the rows that were selected.

I have here a bit of code that i created, but i just can read the last selected box value.

I'm using MVC structure, but with my own rules.

Here the model, I created the function inside the model:

$editjo_query = ("SELECT * FROM my_table");

$print_editjo = db_array($editjo_query, 'a+'); //this is a function that i created to    select, ignore please

if(!empty($_POST['editcheck'])){

    $editcheckbox = $_POST['editcheck'];
    $countCheckedit = count($_POST['editcheck']);

    for($i=0;$i<$countCheckedit;$i++) {
        $edit_id = $editcheckbox[$i];

        $editjo_query = ("SELECT * FROM my_table WHERE id=$edit_id");

        $print_editjo = db_array($editjo_query, 'a+'); //this is a function that i created to select, ignore please
     }
}

And this function is being called at the view

function editjoview($print_editjo){
     foreach($print_editjo as $check){
        echo $check['id'];
     }
}

Answer 1

you are always writing to $print_editjo with $print_editjo = db_array($editjo_query, 'a+'); and you you do not process it. So in ever for loop you overwrite it, leaving you with the last entry. Try this:

$print_editjo = array();
for($i=0;$i<$countCheckedit;$i++) {
  $edit_id = $editcheckbox[$i];
  $editjo_query = ("SELECT * FROM my_table WHERE id=$edit_id");
  $print_editjo[] = db_array($editjo_query, 'a+');
}

and in your view, iterate over the array $print_editjo rather than just having $print_editjo as the last entry.

Note: If you would add how exactly the function in your view is called, the help could be more precise.