php - 插入 mysqli 不工作

Question

我似乎无法插入。什么都没有插入。我现在已经尝试了几种方法，但都不起作用。与数据库的连接存在，但没有插入。

$insert = "INSERT INTO 
              friendshipTable
             ( user_a_memberID , user_b_memberID )
             VALUES ($member_id,$addFriendID)";  
 $resultx = mysqli_query($connExt, $insert);  

JSfiddle

Answer 1

查询实际上将 $connExt 视为未定义是没有问题的，因为它是一个范围问题。您可以在本地函数中使用 global $connExt; 或者代码如下...

<form action="friendship.php" method="post">
    <input type="text" name="addfriend" placeholder="add friend">
    <input type="text" name="member_id" placeholder="member id">
    <input type="submit">
</form>

<?php


$connExt = mysqli_connect("localhost", "root", "root", "test");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: ".mysqli_connect_error();
}


if ( isset($_POST['addfriend']) && $_POST['addfriend']) {
    $addFriend = $_POST['addfriend'];
    $member_id = $_POST['member_id'];
    addFriend($addFriend, $member_id, $connExt);
}

function addFriend($addFriendID, $member_id, $connExt) { //add friend acts as a follow based system
    //create db query

    $insert = "INSERT INTO friendshipTable( user_a_memberID , user_b_memberID )
                              VALUES ('$member_id','$addFriendID');";

    $resultx = mysqli_query($connExt, $insert);



    echo "addfriend: ".$addFriendID;
    echo '<br />';
    echo "member_id: ".$member_id;
    echo '<br />';
    echo "woop!";
    echo '<br />';
    echo $resultx;;
    //return status

}

function unFriend() {

}




?>