Need help with POST infomation from tables

jules789

I'm horrible with php and not sure what to do, this is what i have so far for a program to calculate weekly pay after income tax:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<html>
<head>
	<title>Income Tax Form</title>
    <style type="text/css">
<!--
.style2 {font-size: 24px}
-->
    </style>
</head>
<body>
<form action='formproc.php' method='POST' id='phpform'>
<p align="center" class="style2">Shoeworkz - Pay Calculator</p>

<p>Employee's Name: 
  <input type="text" id="name" name="name" />
  </p>
	<p><br />
	Hours Worked: 
	  <input name="hrs" type="text" id="hrs" value="0" maxlength="8" />
    </p>
	<p><br />
    Pay Rate/Hr ($): 
      <input name="rate" type="text" id="rate" value="0" maxlength="8" />
    </p>
	<p><br />
      <input type="submit" value="Calculate Net Weekly Pay" />
  </p>
</form>

</body>
</html>

<?php
		$name=$_POST['name'];
    		$hrs=$_POST['hrs'];
    		$rate=$_POST['rate'];
		$income = $hrs*$rate;
		if $income<6000 = tax0
		if $income>35000<80000 = tax1
		if $income>80000<180000 = tax2
		if $income>180000<180000 = tax3
		if $income>80000<180000 = tax4
		$tax0 = $income
		$tax1 = ($income-6000)*0.15
		$tax2 = ($income-39200)*0.3
		$tax3 = ($income-98000)*0.38
		$tax4 = ($income-238000)*0.45
?>
<html>
<head>
	<title>Form Processing with PHP</title>
</head>
<body>
	<?php
		echo $name=$_POST['name'];
</body>
</html>

Second part isn't anywhere finished.

dagon

And the question is?

jules789

well, could someone please help me fix the start of the code so I can get the data from the table to post to another page?

bradgrafelman

jules789;10952130 wrote:
so I can get the data from the table to post to another page?

What do you mean 'table' ? Do you mean from the HTML form?

If so, you've got several parse errors in the PHP code you posted. If you're developing these scripts locally in a development environment, you should have display_errors set to On and error_reporting set to E_ALL in your php.ini file. Here's a quick rundown of some of the errors I see at a glance:

        $name=$_POST['name']; 
            $hrs=$_POST['hrs']; 
            $rate=$_POST['rate'];

This isn't so much of an error but rather a bad coding practice. You should always check that external data exists before referencing it. In other words, either do something like this:

$name = (isset($_POST['name']) ? $_POST['name'] : NULL);

for each variable, or at least add a 'name' attribute to your submit button and do something like:

if(isset($_POST['submit'])) {
    // process form here
} else {
    // no form submitted; display error, redirect user, etc. etc.
}

```
        if $income<6000 = tax0 
        if $income>35000<80000 = tax1 
        if $income>80000<180000 = tax2 
        if $income>180000<180000 = tax3 
        if $income>80000<180000 = tax4 
```
I'm not even sure how to correct this block of code. For starters, that's not the correct syntax for an [man]if/man statement.

Next, I'm not sure what you're trying to do with those numbers and <>'s... are you trying to create inequalities such as "if income is greater than 35000 and less than 80000" ? If so, you'll have to split those up into two separate statements in the code. It should be obvious how to do this if you read the above sentence carefully (notice the 'and' ?).

Then you have '= tax0' (etc.). First of all, what is supposed to be assigned 'tax0' ? There's no variable name in there. Then there's the matter of what you mean by tax0; if it's supposed to be a string, then it needs to be surrounded with quotes.

Finally, none of those lines have ending semicolons (an error you repeat throughout the rest of the code).

       $tax0 = $income 
        $tax1 = ($income-6000)*0.15 
        $tax2 = ($income-39200)*0.3 
        $tax3 = ($income-98000)*0.38 
        $tax4 = ($income-238000)*0.45

See my last comment above about the missing semicolons.

```
echo $name=$_POST['name']; 
```
While this will technically work, I'm not sure why you've written it like this. Why are you creating another variable, $name, when you can just as easily echo out $_POST['name'] ?