Php Prepared语句fetch行不返回任何内容

我想获取此行并将其保存到

$notescheck

中，但当我尝试执行此操作时，

$notescheck

在我想要回显时为空，并且没有错误。对于未准备好的报表，它可以正常工作。

代码：

对于未准备好的声明，它将如下所示：

 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);
}

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();
    }

这不是fetch（）处理预处理语句的方式，您并没有像想象中那样获取数组。您还需要将select的结果绑定到变量中，然后使用这些变量来显示。如果有多条记录，则使用

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：


多条与username=x匹配的记录如下所示：

 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);
}

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();
    }