每当我尝试使用php连接到数据库时,我总是得到这个错误: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
我该如何解决这个问题?我的代码是显示mysql数据库表时出现php错误
$email = '[email protected]_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";
$classcount += 1;
}
的$email
变量正常工作时,我赞同它。 128线是while($row = mysqli_fetch_array($result))
部分
您正在传递'$ email'作为表名称。当然你的意思是''SELECT * FROM your_table WHERE email ='$ email''''通过mysqli_real_escape_string()转义'$ email'后 – 2013-03-08 01:52:49
您确定$ con正在工作吗? – 2013-03-08 01:53:00
'$ email'的值是我的表名,我确定我的'$ con'正在工作 – 2013-03-08 01:53:54