CH 5 pg 160

wadesmart · November 10th, 2003, 08:26 PM

When the third script runs I get a unexpected T_Variable on line 3, which is $Count = 0; line. Why is this unexpected? It is set to 0 and is local.

<html><head></head>
<body>
    <form method = "POST" action = "dynamic.php">
    <p> How many children do you have? </p> <input name = "number" type = "text">
    <br />
    <input type = "submit">
    </form>
</body></html>

--------------
<html><head></head>
<body>
    <form method = "POST" action = "dynamic2.php">
    <?php
        for ($Counter = 0; $Counter< $_POST["number"]; $Counter++)
            {
                $Offset = $Counter + 1;
                echo "<br /> Please enter the name of child number $Offset<br />";
                echo "<input name = child[] type = text>";
            }
        if ($Counter == 0) echo "Press the button to move on.";
    ?>
    <br />
    <input type = "submit">
    </form>
</body></html>

------------------------
<html><head></head>
<body>
    <?pho $Count=0;
        echo "Your childrens names are: ";
    do
    {
        echo "<br />$_POST[ child[$Count]] ";
        $checkempty = "$_POST[child[$Count]]";
        $Count = $Count + 1;
    } while ($checkempty!="");
    if ($Count==1) echo "Not Applicable";
    ?>
</body>
</html>

wadesmart · November 10th, 2003, 08:35 PM

Ok. scratch the last message. I realized in the third script the <?php was actually <?pho. But when I changed that I received this:

Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\Documents and Settings\Administrator\My Documents\My Work\PHP\Beg_PHP4\CH5\dynamic2.php on line 7

Im sure that has to do with the array inside the Post but I dont know why.

wadesmart · November 10th, 2003, 08:48 PM

I fixed the last error to. The $_POST[ child there was a space between { and child.

Fixed that and have still another error.

Parse error: parse error, unexpected '[', expecting ']' in C:\Documents and Settings\Administrator\My Documents\My Work\PHP\Beg_PHP4\CH5\dynamic2.php on line 7

I tried () and [] and I tried \[ \] but nothing worked. Im not sure why its saying this.

nikolai · November 10th, 2003, 09:11 PM

Because this is invalid:

  $_POST[child[$count]]

What exactly are you trying to access? See, in your code, you have child[$count], which means you're treating the word 'child' like an array variable. It's not a variable, it's just a string literal.

If $child is an array, and the value of $child[$count] is the index you want to access in the $_POST array, your code would make sense.

The way I figure it, $_POST['child'] is an array, and you want to access the $count index of that array.

That's written like this:

$_POST['child'][$count]

Bear in mind that you cannot access nested array indexes within double-quoted strings using the regular variable substitution. Suppose $count is 3 in the following line of code:

  echo "it is $_POST[child][$count]."

This will print "it is Array[3]."

The reason is that $_POST['child'] is evaluated as an array. Since it's not a string (or can be converted to string), PHP substitutes the word "Array" for it's value. The next square bracket is echoed as a literal character, then $count is substituted with it's value, 3, and finally, the last square bracket character is printed.

You either need to use curly-brace syntax in your string, or close your string and concatenate your variable:

1) echo "It is {$_POST['child'][$count]}.";
2) echo "It is " . $_POST['child'][$count] . ".";

I prefer the first way. Notice that within curly-braces, you must quote all string indexes just as you would when accessing an array index from outside of a string.

Here are some links to the manual relevant to your problems:
  http://www.php.net/types.string
  http://www.php.net/types.array

Take care,

Nik
http://www.bigaction.org/

richard.york · November 10th, 2003, 09:18 PM

When you create an array of input elements using the empty brackets, you are creating a multidimensional array.

So... echo "<br />$_POST[ child[$Count]] ";
Isn't going to work.

The correct syntax is:
echo "<br />{$_POST["child"][$Count]} ";

Curly brace syntax is necessary here because PHP does not recognize more than a single layer of array nesting within strings. The curly braces essentially tells PHP to drop out of normal string parsing mode and treat the object between braces like a normal variable. Because you are no longer in normal string parsing mode, you won't be able to leave the quotes off of the indices, as doing so would cause PHP to look for a named constant first and the plain string indice second.

This will look for a constant named 'child' and throw an error notice under error_reporting = E_ALL
echo "<br />{$_POST[child][$Count]} ";

This will print the literal string child, as sadly, curly brace syntax does not do for constants what it does for variables, when there is only a constant between the braces.
echo "{child}";

Or you can just concatenate:
echo "<br />".$_POST["child"][$Count];

: )
Rich

:::::::::::::::::::::::::::::::::
Smiling Souls
http://www.smilingsouls.net
:::::::::::::::::::::::::::::::::

wadesmart · November 11th, 2003, 10:13 AM

Ok. I got that. Thanks for the help.

I got it to work but I get some extra stuff:

Your childrens names are:

wade .
natalie .
maria .
Notice: Undefined offset: 3 in C:\Documents and Settings\Administrator\My Documents\My Work\PHP\Beg_PHP4\CH5\dynamic2.php on line 7

.
Notice: Undefined offset: 3 in C:\Documents and Settings\Administrator\My Documents\My Work\PHP\Beg_PHP4\CH5\dynamic2.php on line 8

These two notices go on for hundreds of iterations.

nikolai · November 11th, 2003, 03:10 PM

Your loop is a little screwy. It's more readable if you do it this way:

if (isset($_POST['child']) && !empty($_POST['child']))
{
    echo "Your children's names are: <br />\n";
    foreach($_POST['child'] as $child)
    {
        echo "{$child}.<br />\n";
    }
}

First of all, you're iterating the loop once without ever checking whether there ARE any children. That's bad.

Second, and most importantly, you echo the value of an array index, THEN create a string to hold that value, THEN check if that string is empty. That is completely backwards.

Also, the error messages suggest that you're not creating the $checkempty string properly, since it's ALWAYS != "".

If you're going to check whether a variable or an array index exists, you MUST use isset() to avoid getting undefined variable warnings when your error_reporting setting includes the E_NOTICE level.

It's best to write code with error_reporting set to E_ALL, the most string setting, because your code should then work on any other server setup, regardless of their error_reporting settings.

Take care,

Nik
http://www.bigaction.org/

wadesmart · November 11th, 2003, 03:25 PM

Nik,

just so your know, Im coping this straight out of the book and changing it the best I can with Globals off. The code is teaching a Do loop. But, I will use the If.

Wade

nikolai · November 12th, 2003, 11:23 PM

That's a horrible way to teach a do-while loop. Sigh. I guess the author(s) can justify it saying that you wouldn't GET to the do-while loop unless you submitted the form that created children, but still!!!

Hope you're at least picking up how do-while, while, and for loops work. Let me know if you have any questions!

Take care,

Nik
http://www.bigaction.org/