Weird result in PHP form

[bukwus]

Hi

I'm beginning a form that asks questions and has a drop down list of options to choose from. I'm going to use PHP to check for errors, loops to cut down on repetition, etc., but right now I'm just testing code at a basic level.

Here's the PHP:

if(isset($_POST['submit'])){
	$value[1] = $_POST['choice1'];
	$value[2] = $_POST['choice2'];
	$value[3] = $_POST['choice3'];
}
else {
	$value[1] = 'Choose one...';
	$value[2] = 'Choose one...';
	$value[3] = 'Choose one...';
}

 

Here's the basic form element:

<select name="choice1">
	<option value="0">'.$value[1].'</option>
    <option value="Addressed">Addressed</option>
    <option value="Does not apply">Does not apply</option>
    <option value="Needs attention">Needs attention</option>
    </select>

 

My problem is that the text displayed in the select box is only the first letter of whatever it's supposed to be. For example: if it's the first time visiting the form, each box only has "C" in it instead of "Choose one..." OR if "Does not apply" was chosen and the form is submitted, "D" is all that shows up in that box.

 

Any idea what's happening here?

 

Many thanks,

Andy

[Ken2k7]

Did you ever change the value of $value before it gets printed out?

[aeroswat]

You did not define $value as an array before you assigned values to it. It thinks it is a string or character array maybe? I believe it is also better to assign it like this $value[] = "Choose one..."

[bukwus]

You did not define $value as an array before you assigned values to it. It thinks it is a string or character array maybe? I believe it is also better to assign it like this $value[] = "Choose one..."

 

Should this work, or am I incorrectly defining $value as an array?

$value = array();
if(isset($_POST['submit'])){
	$value[1] = $_POST['choice1'];
	$value[2] = $_POST['choice2'];
	$value[3] = $_POST['choice3'];
}
else {
	$value[] = 'Choose one...';
}

 

[aeroswat]

You should be able to do this and it will automatically assign key values 0,1,2,etc for the stuff

 

$value = array();
if(isset($_POST['submit'])){
	$value[] = $_POST['choice1'];
	$value[] = $_POST['choice2'];
	$value[] = $_POST['choice3'];
}
else {
	$value[] = 'Choose one...';
}

[bukwus]

You should be able to do this and it will automatically assign key values 0,1,2,etc for the stuff

 

$value = array();
if(isset($_POST['submit'])){
	$value[] = $_POST['choice1'];
	$value[] = $_POST['choice2'];
	$value[] = $_POST['choice3'];
}
else {
	$value[] = 'Choose one...';
}

 

Okay, here's what it looks like now:

<a name="result"></a>
<?php
$value = array();
if(isset($_POST['submit'])){
	$value[] = $_POST['choice1'];
	$value[] = $_POST['choice2'];
	$value[] = $_POST['choice3'];
}
else {
	$value[] = 'Choose one...';
}
?>

 

And the form:

<form action="page-name.html#result" method="POST">
<select name="choice1">
	<option>'.$value[0].'</option>
    <option value="Addressed">Addressed</option>
    <option value="Does not apply">Does not apply</option>
    <option value="Needs attention">Needs attention</option>
    </select>

 

I made sure the first $value key was 0.

 

When I fill out the form and submit, the values carry over just fine. Unfortunately, when you first come to the page, only the first select box is populated with "Choose one...". The others are blank.

 

Any thoughts why?

 

Thank you

Andy

[aeroswat]

You need to write

$value[] = 'Choose one...';

 

for each instance