尝试根据所选复选框更改查询[重复]

Question

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• How to display errors for my MySQLi query? [duplicate] 3 answers

I'm trying to make a contact search option, right now i'm trying to change a query based on whichever checkbox is selected. Here is my code, please tell me what I'm doing wrong.

FORM

<form method = 'post'> 
<input type="checkbox" name="contactfilter[]"  value="all">All Contacts<br> 
</input>
<input type="checkbox" name="contactfilter[]"  value="mine">My Contacts<br> 
</input>
<button type="submit" value="submit"></button>
</form> 

PHP CODE

<?php
if(isset($_POST['submit'])) {
$contactfilter = $_POST['contactfilter']; 
$service_list = array(); 

if(in_array('all', $contactfilter)) { 
$service_list[] = 'all';
$mine = '';
}
if(in_array('mine', $contactfilter)) {
$service_list[] = 'mine';
$mine = " where contact_owner = '".$_SESSION ["username"]."'";
}
}

$query = "select contact_id, first_name, last_name, company_name, 
company_description, position, company_address, company_phone, home_phone, 
cell_phone, work_phone, email, how_they_heard, date, time from contact 
//im attempting to fill the query with the $mine variable
".$mine."";
$result = mysqli_query($connect, $query);

//this section echo's all data that matches the query
while($row = mysqli_fetch_array($result))
{
echo '
<tr id ="table">
<td data-url="contact_interface.php?contact_id=' . $row['contact_id'] . 
'">'.$row["first_name"].'</td>
<td data-url="contact_interface.php?contact_id=' . $row['contact_id'] . '" 
type="text" name="last_name">'.$row["last_name"].'</td>
<td data-url="contact_interface.php?contact_id=' . $row['contact_id'] . '" 
type="text" name="home_phone">'.$row["home_phone"].'</td>
etc.......
  ';      
 }  
 ?>

</div>