Question

我编写了以下代码，当它运行时，无论SQL查询是什么，我都会收到“错误：查询为空”。我知道生成的SQL很好，因为如果我将它粘贴到SQL数据库中它会返回行。记录集的PHP代码已经粘贴在其他工作正常的页面上，因此我无法查看错误的位置。

案例4：//检查用户之前是否已加载清单

    $mySQL = "SELECT * FROM tools_userChklists WHERE chklistID = '" . $_GET['chklistID'] . "'";
            echo $mySQL;
    $query_rsChecklists = $mySQL;
    $rsChecklists = mysql_query($query_rsChecklists) or die(mysql_error());
    $row_rsChecklists = mysql_fetch_assoc($rsChecklists);
    $totalRows_rsChecklists = mysql_num_rows($rsChecklists);

    if ($totalRows_rsChecklists <> 0){
        //the user has already opened this checklist
        $_SESSION['redirectorAction'] = "";
        $_SESSION['redirectorAction'] = "Location: actions.php?action=5&chklistID=" . $_GET['chklistID'];
    }else{
        //the user has never opened this checklist
        $_SESSION['redirectorAction'] = "";
        $_SESSION['redirectorAction'] = "Location: actions.php?action=6&chklistID=" . $_GET['chklistID'];
    }
break;

感谢任何帮助。感谢。

Answer 1

$ row_rsChecklists是结果资源，你在rsChecklists上有数字。我建议使用像'row'这样的简单单词来进行短块工作。

PHP Recordset返回“错误：查询为空”

1 个答案: