Php 准备好的语句在绑定变量中返回 0，SQL 语句在 DB 中工作正常

Question

        // Load Filters
        $stmt = $mysqli->prepare("
        SELECT 
        display_showCancelled,
        display_showCompleted,
        display_showNotPossible
        FROM `ticketing_settings`
        WHERE user = ?
        "); 
        $stmt->bind_param('i', $_SESSION['userID']);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($displayCancelled, $displayCompleted, $displayNotPossible);

如果我运行数据库中的查询，它工作正常，但不知何故这些值没有绑定到变量...它返回“0”而不是“1”有什么想法吗？

在这个stmt im 运行ning

之后

            $stmt = $mysqli->prepare("SELECT 
        ticketing_tickets.id, 
        ticketing_tickets.description,
        ticketing_type.responsible,
        ticketing_tickets.date_created,
        users.firstname,
        users.lastname,
        ticketing_type.name,
        ticketing_tickets.status,
        ticketing_status.name
        FROM `ticketing_tickets`
        LEFT JOIN ticketing_status ON ticketing_tickets.status = ticketing_status.id
        LEFT JOIN ticketing_type ON ticketing_tickets.type = ticketing_type.id
        LEFT JOIN users ON ticketing_tickets.creator = users.id
        WHERE ticketing_tickets.status NOT LIKE ?
        AND ticketing_tickets.status NOT LIKE ?
        AND ticketing_tickets.status NOT LIKE ?
        AND ticketing_tickets.status NOT LIKE ?
        ");
        $ticketStatusReq = 1;
        $stmt->bind_param('iiii', $ticketStatusReq, $displayCancelled, $displayCompleted, $displayNotPossible); 
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($ticketId, $ticketDescription, $ticketResponsible, $ticketDate, $ticketFirstname, $ticketLastname, $ticketTypeName, $ticketStatus, $ticketStatusName);

完全没问题。

Answer 1

哈哈，我的天啊，不敢相信这个问题有多愚蠢，只是忘了获取...很抱歉哈哈但有时迷失在数千行代码中你没有注意到最明显的问题...

Php 准备好的语句在绑定变量中返回 0，SQL 语句在 DB 中工作正常

Php Prepared Statement Returning 0 in Binded Variables, SQL Statement works fine in DB

php

mysql

variables