Question

每当我尝试使用php连接到数据库时，我都会不断收到此错误：Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /homepages/12/d441172468/htdocs/Organizer/dashboard/index.php on line 258如何解决此问题？我的代码是

            $email = 'redacted@me.com_classes';

            echo $email;

            $result = mysqli_query($con,"SELECT * FROM $email");

            $classcount = 1;

            while($row = mysqli_fetch_array($result))
              {

              $period = $row['period'];
              $teacher = $row['teacher'];
              $subject = $row['subject'];
              $subjecto = strtolower($subject);
              $subjecto = str_replace(' ', '', $subjecto);
              $grade = $row['grade'];

              echo "<li id='button" . $classcount . "' onclick='" . $subjecto . "(),homework" . $classcount . "()'>" . $classcount . ". " . $subject . "-" . $grade . "</li>\n";

              $classcount += 1;

              }

当我回显它时，$email变量工作正常。第128行是while($row = mysqli_fetch_array($result))部分

Answer 1

如果$email的值确实是您的表名，则需要引用它，因为它包含不带引号的table names中不允许的字符（@）：

 $result = mysqli_query($con,"SELECT * FROM `$email`");

请注意，您需要使用mysql中的反引号引用表名和列名。

Answer 2

ASCII：http://dev.mysql.com/doc/refman/5.1/en/identifiers.html

允许使用表格名称

    Permitted characters in unquoted identifiers:

    ASCII: [0-9,a-z,A-Z$_] (basic Latin letters, digits 0-9, dollar, underscore)

    Extended: U+0080 .. U+FFFF 

    Permitted characters in quoted identifiers include the full Unicode Basic Multilingual Plane (BMP), except U+0000:

    ASCII: U+0001 .. U+007F

    Extended: U+0080 .. U+FFFF 

    ASCII NUL (U+0000) and supplementary characters (U+10000 and higher) are not permitted in quoted or unquoted identifiers.

    Identifiers may begin with a digit but unless quoted may not consist solely of digits.

    Database, table, and column names cannot end with space characters.

    Before MySQL 5.1.6, database and table names cannot contain “/”, “\”, “.”, or characters that are not permitted in file names.

很奇怪，你是否已经手动添加了字符串@的表格。我认为这不能作为表名处理？

跳过下面的

您的查询无效，请将代码更改为示例：

mysqli_query($con,"SELECT * FROM TABLE_NAME WHERE 'email' = $email");

使用表名更改TABLE_NAME。使用表TABLE_NAME上的列电子邮件名称更改email。

你是mysql注入的受害者，所以不会检查entites和specialchars！

如果您的VALUE $电子邮件是TABLE_NAME，则该字符串应为限制字符！

显示mysql数据库表时出现php错误

2 个答案: