Php脚本按名称过滤数据库

Question

我想通过两种方式显示数据库中的医院名称：

1. 选择城市。
2. 在搜索栏中输入医院名称。

我在下面写了php脚本。第1部分工作正常，显示所选医院的所有医院，但第2部分对我不起作用。没有显示错误。需要帮助。我没有把'如果＆＃39;条件在正确的地方？或者我错过了其他什么？

if (isset($_POST['search'])) {
    if (isset($_POST['search-by-city'])) {
        $city_id = $_POST['search-by-city'];
        $query = "SELECT * FROM `hospitals` WHERE `City_ID` LIKE '%$city_id%'";
        $result = mysqli_query($con,$query);

        if (mysqli_num_rows($result) == 0) {
                echo '<h2>No recod Found</h2>' ;
            }
    }
    if (isset($_POST['search-by-name'])) {
            $hospital_name = $_POST['search-by-name'];
            $query = "SELECT * FROM `hospitals` WHERE `Name` LIKE '%$hospital_name%'";
            $result = filterTable($query); {

            if (mysqli_num_rows($result) == 0) {
                echo '<h2>No recod Found</h2>' ;
            }
    }
    while($row = mysqli_fetch_array($result)){
        $city_id = $row[3];
        $query = "SELECT `Name` FROM `cities` WHERE `ID` LIKE '$city_id'";
        $result2 = mysqli_query($con,$query);
        $row2 = mysqli_fetch_row($result2);
        $city_name = $row2[0];

        echo '
            <h3>'.$row[1].'</h3>
            <h4>'.$city_name.'</h4> 
        ';
    }
}