当我使用此php时，结果显示所有表而不是table1.id = table2.id的特定条件

Question

• table 1

|---------------------|------------------|
|   User_id Primary   |      Email       |
|---------------------|------------------|
|         1           |      a@a.com     |
|---------------------|------------------|
|         2           |      b@b.com     |
|---------------------|------------------|

• table 2

|---------------------|------------------|
|   Post_id Primary   |  User_id Foreign |
|---------------------|------------------|
|         1           |       1          |
|---------------------|------------------|

Here's the code

$conn1 = @mysqli_connect('127.0.0.1','root','','signup');

$sql1 = "SELECT post_pic ,post_text FROM table2 INNER JOIN table1
    ON (table2.User_id = table1.User_id) " ;

$result = mysqli_query($conn1, $sql1);

if(mysqli_num_rows($result)>0){
     while ($row = mysqli_fetch_array($result)){

so when im logged in user id=2 user id=1 posts are shown

Answer 1

You need to specify the user id in your query, why you are getting images for user 1 is that your code at the moment runs as expected.

If you want to get data only for user1, you need to add a where clause specifying the user ID.

Like so:

//For Arguments Sake
$user_id = $_SESSION['loggedin_user_id'];

$sql1 = "SELECT post_pic ,post_text
         FROM table2
         INNER JOIN table1
         ON (table2.id = table1.id)
         WHERE table2.User_id = '$user_id'";

EDIT: Actually I just noticed that @04FS already mentioned it in a comment.