php - 准备好的语句获取行不返回任何内容

Question

我想获取这一行并将其保存到 $notescheck 中，但是当我尝试执行此操作时，当我想要回显时 $notescheck 为空，并且有没有错误。对于未准备好的语句，它可以正常工作。

代码:

if($user_ok == true) {
    $sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
    $stmt = $conn->prepare($sql);
    $stmt->bind_param("s",$log_username);
    $stmt->execute();
    $row = $stmt->fetch();
    $notescheck = $row[0];
    $stmt->close();
}

对于未准备好的语句，它看起来像这样:

 if($user_ok == true) {
    $sql = "SELECT notescheck FROM users WHERE username='$log_username' LIMIT 1";
    $query = mysqli_query($conn, $sql);
    $row = mysqli_fetch_row($query);
    $notescheck = $row[0];
    mysqli_close($conn);
}

Answer 1

这不是 fetch() 处理准备语句的方式，您并没有像您想象的那样获取数组。您还需要将选择的结果绑定(bind)到变量中，然后使用它们来显示。如果有多个记录，您可以使用 while($stmt->fetch){ echo $notescheck };

if($user_ok == true) {
    $sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
    $stmt = $conn->prepare($sql);
    $stmt->bind_param("s",$log_username);
    $stmt->execute();
    $stmt->bind_result($notescheck);
    $stmt->fetch();
    $stmt->close();
}
echo $notescheck;

您应该阅读以下内容:

http://php.net/manual/en/mysqli-stmt.fetch.php

匹配 username=x 的多个记录将如下所示:

if($user_ok == true) {
        $sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
        $stmt = $conn->prepare($sql);
        $stmt->bind_param("s",$log_username);
        $stmt->execute();
        $stmt->bind_result($notescheck);
        $stmt->store_result()
        while($stmt->fetch()){
           echo $notescheck;
        }
        $stmt->close();
    }