如何将来自mysql数据库的数据与内联if语句进行比较以返回true或false

Question

This is my code, the first function. fun.php

function post_exists($post_id){
    global $con;
    $post_id = (int)$post_id;
    return (mysqli_fetch_array(mysqli_query($con, "SELECT COUNT(post_id) FROM home_post WHERE  post_id='$post_id'"), MYSQLI_NUM) != 0) ? true : false;
}

This is the function call in index.php

if(post_exists(23) === true){
    echo "<script>alert('working oo')</script>";
}

function usually return true when the argument is false. what i want is, if the argument is true(which means if the data exist in my database)return true if is false return false .

Answer 1

What you are doing is checking if the array is not equal to 0 - not the actual result inside the array. Change the true/false ternary to this:

return (mysqli_fetch_array(mysqli_query($con, "SELECT COUNT(post_id) FROM home_post WHERE post_id='$post_id'"), MYSQLI_NUM)[0] != 0) ? true : false;

as you can see, we check the first value in the array (using the [0] key after the array is assigned). This kind of syntax, however, is only available in newer versions of PHP, but should be useable in 5.1 and up.