Mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in

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<?php

$link = mysqli_connect("fdb25.awardspace.net","user","yc1234","user");
echo mysqli_connect_error();



if(isset($_GET['posts'])){
	$id=$_GET['posts'];
        $query="SELECT * FROM `landlord` WHERE ID='$id'";
        $query_run=mysqli_query($link,$query);
}


?>




	<table width='1500' border = '1'>
  <thead>
    <tr>
      <th>Remarks</th>
      <th>Image</th>
     
    </tr>
  </thead>
  <tbody>
  <?php  while($row=mysqli_fetch_array($query_run)) { ?>

    <tr>
      <td> <?php echo $row['Remarks'];  ?></td>
  
      <td>

      		<?php  $image_name="SELECT * FROM `landlord` as p join details as d 
      					on p.id =d.proid WHERE d.proid =".$row['id'];
      					$read1=$link->query($image_name);

      					foreach ($read1 as $value) { ?>

      						<img src="upload/<?php echo $value['images']; ?>" />
      						
      					<?php  } ?>
      					</td>
    </tr>
<?php   } ?>
  </tbody>
</table>

What I have tried:

The error of this project is

mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in

on

while($row=mysqli_fetch_array($query_run))

may i know how to solve it ? Thanks

Posted 27-Jun-20 22:39pm

YC Kwan

Updated 27-Jun-20 22:58pm

v2

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2 solutions

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OriginalGriff · Answer 1 · 2020-06-27T22:50:00

Don't do it like that! Never concatenate strings to build a SQL command. It leaves you wide open to accidental or deliberate SQL Injection attack which can destroy your entire database. Always use Parameterized queries instead.

When you concatenate strings, you cause problems because SQL receives commands like:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'Baker's Wood'

The quote the user added terminates the string as far as SQL is concerned and you get problems. But it could be worse. If I come along and type this instead: "x';DROP TABLE MyTable;--" Then SQL receives a very different command:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'x';DROP TABLE MyTable;--'

Which SQL sees as three separate commands:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'x';

A perfectly valid SELECT

SQL

DROP TABLE MyTable;

A perfectly valid "delete the table" command

SQL

--'

And everything else is a comment.
So it does: selects any matching rows, deletes the table from the DB, and ignores anything else.

So ALWAYS use parameterized queries! Or be prepared to restore your DB from backup frequently. You do take backups regularly, don't you?

Chances are that by the time you have finished fixing that throughout your whole app, you other problem will have gone away at the same time ...

Richard MacCutchan · Answer 2 · 2020-06-27T22:58:00

Solution 2

Mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in[^]

Posted 27-Jun-20 22:58pm

Richard MacCutchan