使用结果中的 ID 创建 MySQL 查询

Question

I have aa form which let's the user select what they want to display. Now the results of this looks like that: 9, 10, 11 . These are IDs from the table.

These are the IDs of the type they what to show. I have my query already, but I want to add this part at the end of my query.

So in this case:

$query = "type_ID = $result1 or type_ID = $result2 or type_ID = $result3"

if printed out with echo:

type_ID = 9 or type_ID = 10 or type_ID = 11

How can I achieve this?

I tried to loop and it,however this did not work and I am confused how to do add the MySQL code to this.

$result = $result . $_GET['type_ID'][$i]

Answer 1

I'm a tad unsure of what it is you are trying to achieve but from what I understand this should do it:

<?php
$tID = $_GET['type_ID'];

$query = 'SELECT * FROM table WHERE';

$i = 0;
foreach($tID AS $id){
    if($i == 0){
        $query .= ' type_ID = ' . $id;
        $i++;
    }else{
        $query .= ' OR type_ID = ' . $id;
    }
}

Although if you are only looking for the type_ID I'd still recommend using IN() like so:

<?php
$tID = $_GET['type_ID'];

$query = "SELECT * FROM table WHERE type_ID IN (" . implode(',',$tID) . ")";

In works just like OR, just instead of having to write multiple OR you can just use a single IN() :)

Answer 2

你试过这个吗？

$sql = "SELECT * FROM table WHERE ".$query;