donglinyi4313 2014-06-30 15:42
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INNER JOIN阻止用户和朋友表删除用户

I know that PHP can become quite complex when using INNER JOIN. I've had a play around with a few ideas and nothing seems to be working.

So this is what I'm working towards When user A Blocks user B user B gets removed from their friends.

So I have this and have an error

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'user1_id=15 AND user2_id=3 AND status=1 INNER JOIN friends ON block. DELETE fro' at line 1

Could anyone explain inner join in full and show me where I've gone wrong for future usage.

$query = mysqli_query($mysqli,"INSERT INTO block user1_id=$user1_id AND user2_id=$user2 AND status=1
INNER JOIN friends ON block. 
DELETE from friends where user1_id=$user1_id AND user2_id=$user2 AND status=2 ")
or die (mysqli_error($mysqli));
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2条回答 默认 最新

  • doujianwei8217 2014-06-30 15:59
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    You should use two different statements and you have to use parameterized queries.

    $sql_insert = "INSERT INTO block (user1_id, user2_id, status) values (?,?,1)";    
    $stmt = $mysqli->prepare($sql_insert);
    $stmt->bind_param("s", $user1_id);
    $stmt->bind_param("s", $user2_id);
    if (!$stmt->execute()) {
        echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
    }
    
    $sql_delete = "DELETE FROM friends where user1_id=? AND user2_id=? AND status=2 "
    $stmt = $mysqli->prepare($sql_delete);
    $stmt->bind_param("s", $user1_id);
    $stmt->bind_param("s", $user2_id);
    if (!$stmt->execute()) {
        echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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