Php won't show requested data from database (sql)

Question

Im trying to display all the movies with specific hashtag.

The results are not showing but there are no errors.

However i should get movie poster image and title as result. :/

In my database, this is 'tags' row of one movie

#watchonline #movies-with-subs #hdmovies

and tag.php

<?php
require "db.inc.php";

if (isset($_GET['tag'])) {
$tag = preg_replace('#[^a-z0-9_-]#i', '', $_GET["tag"]);
$fulltag = "#" . $tag;
echo $fulltag;
$ASDsql = "SELECT * FROM movie WHERE tags LIKE '$fulltag' OR tags = 
'$fulltag'";
$tagQuery = mysqli_query($db, $ASDsql);

$count = mysqli_num_rows($tagQuery);
if($count > 0) {
while ($hastags = mysqli_fetch_array($tagQuery)) {
$id = $hastags['id'];

$movie_title = $hastags['movie_title'];

$movie_url = $hastags['movie_url'];

$movie_image = $hastags['movie_image'];


$movie_identity = $hastags['movie_identity'];

echo "<div class='item'><a href='movie.php?movie=$id'><img 
src='$movie_image'><p>$movie_title</p></a></div>";
}
}else {
echo "No movies under that tag";
}
}
?>

I've checked db.inc.php file it's all good. My output is "No movies under that tag"; Im sorry if this is a copy i really looked and could not find an answer

You need to explain what the error is, what the output is and the expected result. — Corey Taylor
– Corey Taylor, Commented Feb 10, 2018 at 16:58
I don't get any error, only the output "No movies udner that tag" — Sejoo
– Sejoo, Commented Feb 10, 2018 at 16:58
LIKE usually requires a % or _, like so: ...where tags like '%test%' — user8034901
– user8034901, Commented Feb 10, 2018 at 17:00
Try var_dump($tagQuery); This will let you see what is being returned from the Query — sam
– sam, Commented Feb 10, 2018 at 17:04

shashikant kuswaha · Accepted Answer · 2018-02-10 17:05:39Z

1

try this :

 $ASDsql = "SELECT * FROM movie WHERE tags LIKE '%$fulltag%' OR tags = 
'%$fulltag%'";

or may be the table is not having any value

answered Feb 10, 2018 at 17:05

shashikant kuswaha

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4 Comments

Sejoo Over a year ago

It worked. It seem "LIKE" was not necessary at all. The working output is $ASDsql = "SELECT * FROM movie WHERE tags = '%$fulltag%'";

shashikant kuswaha Over a year ago

does your column tags have value starting with "%" since you are equating with that

Sejoo Over a year ago

No they are like this #tag. however i have litle function that convert any hashtag string (#string) into clickable link

shashikant kuswaha Over a year ago

how can you get the result by condition where tags='%$fulltag%' ,since this contain a character % in the start hence the condition will match with % not with # and you are also not using like query.

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