Warning:mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in?And after searching I rectified my code as below <p>DisplayLandlord PHP: I couldn't get the output.The query is not fetching the result</p> <pre><code> &lt;?php include("dbconfig.php"); ?&gt; &lt;?php $query = "SELECT * FROM rmuruge4_landlord"; $result = mysqli_query(mysqli_connect(),$query); if ($result === false) {die(mysqli_error(mysqli_connect())); } echo "&lt;div class='bs-example'&gt;"; echo "&lt;table class='table table-striped'&gt;"; echo "&lt;tr&gt;&lt;th&gt;LLID&lt;/th&gt;&lt;th&gt;LName&lt;/th&gt;&lt;th&gt;Phone&lt;/th&gt;&lt;th&gt;Address&lt;/th&gt;&lt;/td&gt;"; while($row = mysqli_fetch_array($result)){ echo "&lt;tr&gt;&lt;td&gt;" . $row['LLID'] . "&lt;/td&gt;&lt;td &gt;" . $row['LName'] . "&lt;/td&gt;&lt;td &gt;". $row['Phone'] . "&lt;/td&gt;&lt;td &gt;" . $row['Address'] . "&lt;/td&gt;&lt;/tr&gt;"; } echo "&lt;/table&gt;"; echo "&lt;/div&gt;"; mysqli_close(mysqli_connect()); ?&gt; dbconfig.php: Connection is successful </code></pre>