PHP动态用户输入到php表中

Question

I am trying to insert user input into a php table. For example if I define a 2x3 table (2 rows and three columns) from a previous page, an input screen displays with 6 inputs for you to fill in whatever you would like to. The 6 content variables are then inserted into a table. I believe an array and possibly a foreach loop may be necessary. However, I am not sure how to get the user input content to carry over to the next page and inserted into a table properly. I also trying to get it into a list, but I assume if the table problem can be solved, it should be the same for the lists. Any help would be greatly appreciated.

//Page where defined inputs are displayed

$row = (isset($_POST['rows']) ? $_POST['rows'] : null);
$col = (isset($_POST['columns']) ? $_POST['columns'] : null);
$list = (isset($_POST['listitems']) ? $_POST['listitems'] : null);
$list2 = (isset($_POST['listitems2']) ? $_POST['listitems2'] : null);

array('[content[]');//Not sure if array should be here or on next page

$y = $list;
$z = $list2;
$x = $row * $col;

        for ($i = 1; $i <= $x; $i++) {

            echo "$i<input type='text' name='content[]' required> <br>";

        }
        for ($t = 1; $t <= $y; $t++) {

            echo "$t<input type='text' name='content[]' required> <br>";
        }

        for ($s = 1; $s <= $z; $s++) {

            echo "$s<input type='text' name='content[]' required> <br>";
        }

         echo   "<input type='submit' name='submit2' value='Submit'/>
            <input type='hidden' name='method' value='post' />
            <input type='hidden' name='unorderedinput' value='$list' />
            <input type='hidden' name='orderedinput' value='$list2' />
            <input type='hidden' name='rowsinput' value='$row' />
            <input type='hidden' name='columnsinput' value='$col' />
            <input type='hidden' name='cont' value='content[]' />";

echo "</form>";

//page where the table is displayed.

        $trow = $_POST['rowsinput'];
        $tcol = $_POST['columnsinput'];
        $ulist = $_POST['unorderedinput'];
        $olist = $_POST['orderedinput'];
        $tcontent = $_POST['cont'];

        echo "<table border='1'>";

        for ($tr = 1; $tr <= $trow; $tr++) {
            echo "<tr>";
                for ($td = 1; $td <= $tcol; $td++) {
                    echo "<td align='center'>". "$tcontent". "</td>";//I'm not sure if this is correct
                }

                    echo "</tr>";
            }

        echo "</table>";

Answer 1

First, you don't need this line:

    array('[content[]');

It does not do anything, because it is not assigned to a variable. If the purpose was to make a content array, that is something php will do for you on the second page. See below.

Also, you don't need this line either:

    <input type='hidden' name='cont' value='content[]' />";

That is because you already have input tags with name content[], so use those directly in the next page, where you had this line:

    $tcontent = $_POST['cont'];

That line just made $tcontent = "content[]", i.e. the literal string, not the content.

Instead put:

    $tcontent = $_POST['content'];

... and now you will have an array of values.

Then, also on the second page, you could iterate over these values using the current and next functions:

    echo "<table border='1'>";

    $value = current($tcontent);
    for ($tr = 1; $tr <= $trow; $tr++) {
        echo "<tr>";
        for ($td = 1; $td <= $tcol; $td++) {
            echo "<td align='center'>". $value . "</td>";
            // get next element
            $value = next($tcontent);
        }
        echo "</tr>";
    }

    echo "</table>";

You should not forget to make your code deal with missing arguments in $_POST, using the isset function, like you did for the first page.