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Php SQLSTATE[42S02]表示找不到基表或视图时出错1146_Php_Mysql_Pdo_Html Table - Fatal编程技术网

Php SQLSTATE[42S02]表示找不到基表或视图时出错1146

Php SQLSTATE[42S02]表示找不到基表或视图时出错1146,php,mysql,pdo,html-table,Php,Mysql,Pdo,Html Table,我得到这个错误 SQLSTATE[42S02]:找不到基表或视图1146表 “epiz_5453564654654_newdata.$assignmentsTable”不存在 但表存在于数据库中,当我用表名替换$assignmentsTable时,它会工作…$assignmentsTable不会在您的查询中计算 尝试: <?php $assignmentsTable = $ssubjectnamewithoutspace.'_assignments'; try

我得到这个错误

SQLSTATE[42S02]:找不到基表或视图1146表 “epiz_5453564654654_newdata.$assignmentsTable”不存在


但表存在于数据库中,当我用表名替换
$assignmentsTable
时,它会工作…

$assignmentsTable不会在您的查询中计算

尝试:

<?php
      $assignmentsTable = $ssubjectnamewithoutspace.'_assignments';

        try {
            $stmt = $db->query('SELECT contentID, contentTitle,contentAuthor, contentLink FROM .$assignmentsTable ORDER BY contentID ASC');
            while($row = $stmt->fetch()){

                    echo '<tr><td><a href="'.$row['contentLink'].'">'.$row['contentTitle'].'</a></td></tr>';            


            }

        } catch(PDOException $e) {
            echo $e->getMessage();
        }
    ?>

您有一个
。在
FROM.$assignmentsTable
中,不应该有。欢迎。另外,
$assignmentsTable
将不会被解析/替换,因为您在
$stmt=$db->query('SELECT contentID,…
感谢@kerbholzy您的答案帮助这就是为什么将变量直接放入字符串中是不好的做法的原因。
$stmt = $db->query('SELECT contentID, contentTitle,contentAuthor, contentLink FROM `'.$assignmentsTable.'` ORDER BY contentID ASC');