Question

我已经过了30次这个剧本，我不能为我的生活找到我的问题。这是代码：

function redeem() {

        $case = $_POST["case"]; 
        $name = $_POST["name"]; 
        $profession = $_POST["profession"];
        $city = $_POST["city"];
        $country = $_POST["country"];
        $totalpercent = $_POST["totalpercent"];
        $pretest = $_POST["pretest"];
        $posttest = $_POST["posttest"];
        $investigationspercent = $_POST["investigationspercent"];
        $timesreset = $_POST["timesreset"];
        $creditsspent = $_POST["creditsspent"];
        $timescompleted = $_POST["timescompleted"];

        //Add the information to the learnent_cases_leaderboard table
        $stmt = $this->db->prepare("INSERT INTO learnent_cases_leaderboard (case, name, profession, city, country, totalpercent, pretest, posttest, investigationspercent, creditsspent, timescompleted, timesreset, timestamp) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, CURRENT_TIMESTAMP)");

        $stmt->bind_param("sssssiiiiiii", $case, $name, $profession, $city, $country, $totalpercent, $pretest, $posttest, $investigationspercent, $creditsspent, $timescompleted, $timesreset); //the quotations specify the type of variable;
        //See http://php.net/manual/en/mysqli-stmt.bind-param.php for more information on bind_param
        $stmt->execute();
        $stmt->close();

当我查看错误日志时，它会显示以下错误消息：

第105行是这一行：

PHP致命错误：在第105行的非对象上调用成员函数bind_param（）

代码：

$stmt->bind_param("sssssiiiiiii", $case, $name, $profession, $city, $country, $totalpercent, $pretest, $posttest, $investigationspercent, $creditsspent, $timescompleted, $timesreset);

Answer 1

您从未检查$stmt是否为对象。在这种情况下，它更有可能是FALSE，这是PDO::prepare在您的查询中出现错误时返回的内容。

您的查询中有错误，因为您没有在反引号中划分字段名称，timestamp是关键字。

从第三方API 调用函数后检查错误，并修复您的查询。

Answer 2

第一个;始终在localhost中运行您的查询，以查看您的查询是否正确执行。接下来，请务必确保您的字段和数据类型的名称与代码中的名称相对应

PHP致命错误：调用成员函数bind_param（）

2 个答案: