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I am currently working on a few simple PHP form examples and am coming across a lot of errors. I am very new to PHP so errors are usually expected but I am very curious as to how I can overcome some of these errors. On form #1 I am running into a 'Notice: Undefined Index' error. I know that the solution for this is usually isset() but I don't know how to implement it in this part of my code that I will display below.

<p class="label">Reason for Contacting Us*</p>

<select class="" name="dropDwn" id="dropdown">
<option value="dummy" disabled selected>Select One...</option>

<option value="Server Issues" <?php if($_POST['dropDwn'] == "Server Issues") 
echo "Server Issues"; ?>>Server Issues</option>

<option value="Domain Problems" <?php if($_POST['dropDwn'] == "Domain 
Problems") echo "Domain Problems"; ?>>Domain Problems</option>

<option value="Website Creation" <?php if($_POST['dropDwn'] == "Website 
Creation") echo "Website Creation"; ?>>Website Creation</option>

<option value="Billing Issues" <?php if($_POST['dropDwn'] == "Billing 
Issues") echo "Billing Issues"; ?>>Billing Issues</option>

</select>
<span class="error-messages"><?php echo $dropdownErrMsg;?></span>
</p>

Here is a link to the form above: http://mujomusic.net/WDV341/FormProject/index.php

P.S. The error shows up on when you click on the dropdown!

The error on form #2 seems a lot more complex to me... the error being thrown is 'in_array() expects parameter 2 to be array, null given' and also 'Undefined Variable' which is one I am not at all familiar with. Here is the code(I am going to display just one checkbox set instead of all four):

<p><b>Tuesday/Thursday 10:10am-Noon</b></p>
<span class="error"><?php echo $optFourErr; //place error message on form  ?
></span>
<p>

    <label for="">Worst</label>
    <input type="checkbox" name="option4[]" value="Worst" <?= 
    (in_array('Worst', $inFour)) ? 'checked' : ''; ?>>

    <label for="">Decent</label>
    <input type="checkbox" name="option4[]" value="Decent" <?= 
    (in_array('Decent', $inFour)) ? 'checked' : ''; ?>>

    <label for="">Best</label>
    <input type="checkbox" name="option4[]" value="Best" <?= 
    (in_array('Best', $inFour)) ? 'checked' : ''; ?>>

  </p>

Here is the link to Form #2: http://mujomusic.net/WDV341/ClassProject/index.php

I am just looking for some possible solutions to these errors so I can advance my PHP skills! Any help is well appreciated!

M97
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  • It's all there really, you need to check if the variable exists before you can compare its value. And you are the only one that know what `$inFour` contains. – jeroen Nov 15 '17 at 08:34
  • Why was this marked as a duplicate...? – M97 Nov 15 '17 at 08:46

0 Answers0