为什么在尝试从此代码中获取数据时出错？

Question

It says the line of it has an error like mysqli_num_rows() expects parameter 1 to be mysqli_result,it points to this line $resultCheck = mysqli_num_rows($result);

My code is:

<?php

$sql = "SELECT FROM videos ORDER BY video_id ASC LIMIT 2,1;";
$result = mysqli_query($con, $sql);
$resultCheck = mysqli_num_rows($result);

if (mysqli_num_rows($result) > 0) {
    while ($row = mysqli_fetch_assoc($result)) { ?>
<li class="jp-playlist-current"><div><a href="javascript:;" class="jp- 
playlist-item-remove" style="display: none;">×</a><a href="javascript:;" 
class="jp-playlist-item jp-playlist-current" tabindex="0">1. <?php echo 
$row['video_name']; ?> <span class="jp-artist">by <?php echo 
$row['video_artist']; ?></span></a></div></li>

<?php   }

} else {
    echo "there are no songs!";
}

?>

Answer 1

mysqli_num_rows() function expects a mysqli_result object as it's parameter.

It can fail if there were any error on the query and no mysqli_result object returned. A query can fail, if there ware syntax error, or wrong tablename-fieldnames used in it and also the important one is that the DB connection is active.

You need to check them all.

If the DB connection is Ok, done and checked previously, then you don't need the following code:

<?php
$con = mysqli_connect("{DB_HOST}", "{db_user}","{db_password}","{db_name}");
if (mysqli_connect_errno()) die("DB Connect Error: " . mysqli_connect_error());
?>

Then please check if the query successful too before trying to fetch result from it (mysqli_query() function returns false if the query fails).

<?php
$result = mysqli_query($con, "SELECT * FROM videos ORDER BY video_id ASC LIMIT 2,1");
if (!$result) die('Error on the query...');
$resultCheck = mysqli_num_rows($result);

if ($resultCheck) {
    while ($row = mysqli_fetch_assoc($result)) { ?>
...<?php echo $row['video_name']; ?>...
...<?php echo $row['video_artist']; ?>...
<?php   }

} else {
    echo "there are no songs!";
}
?>