PHP Submit Form Error

Question

Error below: I'm pretty much trying to see what the user chose from the checkboxes.

Notice: Undefined variable: ch1 in /Applications/XAMPP/xamppfiles/htdocs/ProjectOne/index.php on line 42 ch1

HTML CODE :

<form action='submit.php' method="GET">

<div id="self">
<input type='text' name='Name' value='Name' />
    <br>
<input type='text' name='Cwid' value='CWID' />
    <br>
</div>

<div id='gender'>
<strong>Gender:</strong><input type="radio" name="sex" value="male"checked>Male
Or
<input type="radio" name="sex" value="female">Female
<br>
<div>

<div id='class'>
<strong>Class:</strong> <select name='class'>
<option value='Freshman'> Freshman </option>
<option value='Sophomore'> Sophomore </option>
<option value='Junior'> Junior </option>
<option value='Senior'> Senior </option>
</select>
</div>
<br>
<div id='pref'>
<strong>Student Preferences</strong>
<br>
<!-- line 42 -->
<input type="checkbox" name="ch1" value="ch133" /> <?PHP echo $ch1; ?>  ch1 <br />
<input type="checkbox" name="ch2" value="Laundry on Premise" /> Laundry on Premise<br />
<input type="checkbox" name="ch3" value="Fully Equipped Kitchen" /> Fully Equipped Kichen<br />

</div>
<div id='submit'><input type="submit"  name='submit' value="Submit" />   </div>

</form>

PHP CODE BELOW :

<?php
$ch1 = 'unchecked';
$ch2 = 'unchecked';
$ch3 = 'unchecked';
if(isset($_GET['submit'])) 
{
$name = $_GET['Name'];
$cwid = $_GET['Cwid'];
$sex = $_GET['sex'];
$class = $_GET['class'];
$ch1 = $_GET['ch1'];
$ch2 = $_GET['ch2'];
$ch3 = $_GET['ch3'];
    if (isset($ch1)) {
    $ch1 = $_GET['ch1'];

    if ($ch1 == 'ch1') {
        $ch1 = 'checked';
    }
}
}
?>

Possible duplicate of Reference - What does this error mean in PHP? — Brian
– Brian, Commented Oct 9, 2015 at 16:47

mlaribi · Accepted Answer · 2015-10-09 16:58:22Z

1

You call an echo on a inexistant variable. When you submit your form, $ch1 does not exist yet. Add this :

<?php $ch1 = 'unchecked'; ?>

in the line before :

<input type="checkbox" name="ch1" value="ch133" /> <?PHP echo $ch1; ?>  ch1 <br />

answered Oct 9, 2015 at 16:58

mlaribi

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Comments

Mm-Art-In · Accepted Answer · 2015-10-09 19:21:21Z

You have several inconsistencies in your code.

Notice: Undefined variable: ch1 in /Applications/XAMPP/xamppfiles/htdocs/ProjectOne/index.php on line 42 ch1

Read this, it says a NOTICE, which is Not an error, that the variable ch1 ($ch1) is undefined. so there has been no value set as $ch1.

So, what you need to do is set a value for $ch1 before the script reaches line 42.

Your html says:

 <input type="checkbox" name="ch1" value="ch133" />

Which means when you submit the data to the page, the $_POST/$_GET/$_REQUEST system you are using will return a value of ch133 for variable ch1 .

Because it's a checkbox you can get either this value, or nothing, so your code in your PHP will never be the output you want because:

$ch1 = $_GET['ch1'];

    if (isset($ch1)) {
    //you can only reach this point if the value is already set, 
    //Below, so there's no point setting the value twice. 
    $ch1 = $_GET['ch1']; 
    if ($ch1 == 'ch1') {
        //$ch1 can only be the value of $_GET['ch1'] and so it can 
       // never ever be 'ch1' as it's value, it can be NULL or ch133
       //only. So this IF statement will never ever run. 
        $ch1 = 'checked'; 
    }
}

Also note that your echo $ch1 statement comes outside of the HTML checkbox, so the checkbox will never be marked as ticked. You need to adjust your code to be:

  <input type="checkbox" name="ch1" value="ch133"  <?php print $ch1;?> />

I hope this helps clarify your issues and your approach. :-)

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