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minhankyaw's avatar

Warning: mysqli_num_rows() in php

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in. This is my php code

 $query = "SELECT * FROM `posts`";
 $result = mysqli_query($conn,$query);
 if(mysqli_num_rows($result) != 0):

anyone help me please.

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6 replies
Sinnbeck's avatar

Well mysqli_query returns false on failure. So you are getting an error for some reason

From the official docs

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

1 like
minhankyaw's avatar
minhankyaw
OP
Best Answer
Level 1

error solve

 $query = "SELECT * FROM `posts`";
            $result = mysqli_query($conn,$query);
            if (!$result){
               die();
            }
            if(mysqli_num_rows($result) != 0):
zeekash's avatar

how to resolve ? Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, array given in

if (mysqli_num_rows($result)) { echo " match_id country league match_status match_datetime play_time team1 team2 team1score team2score team1scoreHT team2scoreHT match_static_id "; while($row = mysqli_fetch_array($result) <= 1) { echo " edit ".$row['match_id']." ".$row['country']." ".$row['league']." ".$row['match_status']." ".$row['match_datetime']." ".$row['play_time']." ".$row['team1']." ".$row['team2']." ".$row['team1score']." ".$row['team2score']." ".$row['team1scoreHT']." ".$row['team2scoreHT']." ".$row['match_static_id']." "; } echo " "; foreach($result as $good_id){ mysqli_free_result($good_id) <= 1; } } else{ echo "No records matching your query were found."; } mysqli_close($conn); ?>

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