【问题标题】:PHP PDO Update Statement not workingPHP PDO 更新语句不起作用
【发布时间】:2017-09-27 21:26:52
【问题描述】:

这对你们这些天才来说可能是一件容易的事,但我已经尝试了所有我知道的方法,但无法让这个 UPDATE 语句起作用。问题在于更新语句或执行绑定。我希望语句将 2 点添加到用户的积分列。

<?php
$dbConnection = new PDO('mysql:dbname=App;host=localhost;charset=utf8', '*', '*');
$dbConnection->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$dbConnection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$points = 2;
$username = $_POST["username"];
$password = $_POST["password"];
$response = array();
$stmt->$dbConnection->prepare("UPDATE user SET points = points + ? WHERE username = ? AND password = ?");
$stmt->execute(array($points, $username, $password));
$hi = $dbConnection->prepare("SELECT username, password, points FROM user WHERE username = ? AND password = ?");
$hi->execute(array($username, $password));
$red = $hi->fetchAll(PDO::FETCH_ASSOC);
if (count($red) > 0){
    $response["success"] = true;
    foreach($red as $item) {
        $response["username"] = $item["username"];
        $response["password"] = $item["password"];
        $asd = $item["points"];
        $response["points"] = (string)$asd;
    }
}else{
    $response["success"] = false;
}
    echo json_encode($response);
?>

【问题讨论】:

    标签: php mysql arrays pdo


    【解决方案1】:

    您需要将 $stmt 分配给连接中的准备好的语句

    $stmt = $dbConnection->prepare("UPDATE user SET points = points + ? WHERE username = ? AND password = ?");
    

    【讨论】:

    • 是的,你明白了!愚蠢的错误,完全没有看到!我现在感觉很笨!我必须等待 3 分钟才能接受您的回答。感谢您的帮助!
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