Php 呈现表格时会弹出错误“；“数据表警告”；

我有这个代码，将显示完美的项目，但有一个错误弹出窗口，我必须点击“确定”太查看页面

我曾在网上搜索过，据我所知，这与标题表的列数多于数据本身有关，我曾试图解决这一问题，但由于我是一名新手（noob），所以没有什么进展：/

如果有人能对此有所了解，那就太好了！：）

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错误

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我的代码

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为什么你认为这段代码与错误信息有关？我做了noob的事，谷歌把它删掉了。。。。为什么会是什么/
DataTables warning (table id = 'example'): Requested unknown parameter '0' 
from the data source for row 0

<?php
//include database connection
include 'db_connect.php';

//query all records from the database
$query = "select * from repair_jobs";

//execute the query
$result = $mysqli->query( $query );

//get number of rows returned
$num_results = $result->num_rows;

//this will link us to our add.php to create new record
if( $num_results > 0){ //it means there's already a database record

    //start table
    //creating our table heading
    echo "<table id='example' cellpadding='0' cellspacing='0' border='0' class='display' >";
    echo "<thead><tr>";
        echo "<th>Client</th>";
        echo "<th>Type</th>";
        echo "<th>Labour</th>";
        echo "<th>ourcosts</th>";
    echo "</thead></tr><tbody><tr class=\"gradeX\">";

    //loop to show each records
    while( $row = $result->fetch_assoc() ){
            //extract row
            //this will make $row['firstname'] to
            //just $firstname only
            extract($row);

            //creating new table row per record
            echo "<tr>";
                echo "<td>{$client}</td>";
                echo "<td>{$type}</td>";
                echo "<td>{$labour}</td>";
                echo "<td>{$ourcosts}</td>";
                                echo "</td>";
            echo "</tr>";
    }

    echo "</table>";//end table

}else{
    //if database table is empty
    echo "No records found.";
}

//disconnect from database
$result->free();
$mysqli->close();

?>