How to use LIKE statement to check if each value in an array is present in a SQL table?

Question

I have an array of words and I want to iterate over this array to check if any of these words exist in my sql table. If they exist, it should return the particular row.

I have tried using a sql statement inside a for loop that iterates over a variable i to and checks if each word in the array is present in the SQL table.

/*This code produces the separate tags or words from the 'tag' which contains comma separated values*/
if($result == True){
    while($row = $result->fetch_assoc()){
        $tag_string = $row['tag'];
        echo $row['tag'];

    }
}

$tag_arr = explode(",", $tag_string);

/*This is the part where there is an error (Undefined index i in the sql query)*/

for($i=0; $i <= count($tag_arr); $i++){
    $sql1 = "select count from skills where skills.skill like '%'$tag_arr[i]'%' or '%'tag_arr[i] or tag_arr[i]'%'";
    $result1 = mysqli_query($conn, $sql1);      
    if($result1 == True){
        while($row = $result1->fetch_assoc()){
            //$tag_string = $row['tag'];
            echo $row['count'];

        }
    }
}

There was an error in the SQL query statement.

'Undefined index: i'

This is the structure of my 'skills' table:

You're probably better off using a single query with a JOIN. — Qirel, Jul 10 '19 at 06:49
I think it should be `$tag_arr[$i]`, you lack `$` in the `i`. Plus the other `tag_arr` also don't have `$` — Carl Binalla, Jul 10 '19 at 06:49

How to use LIKE statement to check if each value in an array is present in a SQL table?

0 Answers0