Question

<?php
    include 'config.php'; //connect to db
    if(isset($_REQUEST["pwd"]) && isset($_REQUEST["name"])) {
        $password = $_REQUEST['pwd']; //pass from previous page
        $name = $_REQUEST['name']; //pass from previous page
                $checkUserPass = mysql_query("SELECT * FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn); //check if the user exist
                if(mysql_num_rows($checkUserPass) == 1) {
                    $personnelId = mysql_query("SELECT PersonnelID FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn); //query user id
                    while($row = mysql_fetch_assoc($personnelId)) {
                        echo $row['PersonnelD']; // print user id
                    }
                    mysql_close($conn);
                    //echo "<br/><br/>";
                    //echo "<script>alert('Logged In.')</script>";
                    //header("Refresh: 1; url=profile/profile.php?id="'.$id.');
                    //header('Refresh: 1; url=test.php?id=$personnelId');
                } else {
                    echo "<br/><br/>";
                    echo "<script>alert('Wrong Password.')</script>";
                    header('Refresh: 1; url=personnelselect.php');
                }
    }
?>

我无法echo $row['PersonnelD']页面显示为空白。我无法理解我哪里出错了。这个页面的问题已经解决了

Answer 1

看起来你的代码有误：

echo $row['PersonnelD'];

不应该跟随吗？

echo $row['PersonnelID'];

Answer 2

检查mysql_fetch_assoc（）函数可能是其参数为空因此无法进入while循环

Answer 3

尝试使用var_dump()函数调试并检查变量中的值。例如：var_dump($row);在while循环中。

Answer 4

在您的两个查询中，您都有

"SELECT * FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'"

应该是：

"SELECT * FROM validPersonnel WHERE Passkey = '".$password."' and Name = '".$name."';"

除非您关闭引号，否则PHP无法识别$ var。句点将$ var添加到字符串中。

我怎么能在PHP中显示SQL查询？关闭

4 个答案: