Question

 $sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
 $result = mysql_query($sql);
 $value = mysql_fetch_object($result);
 $teacheremail2 = $value->temail;
 echo $teacheremail2;

echo $teacheremail2不返回任何内容。

$teachername有效，并且我已多次检查。

Answer 1

它应该是一个二维数组，您需要 $value[0]->temail

Answer 2

mysql_fetch_object($result)的结果是一个对象（stdClass）。

对此link

可以找到关于object(stdClass)的解释

$sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
$result = mysql_query($sql);
while ($value = mysql_fetch_object($result))
    {
    $teacheremail2 = $value->temail;
    echo $teacheremail2;
    }

Answer 3

首先，您将要直接对数据库运行查询，以确保查询返回某种结果。其次，如果可行，您将希望直接回显$ value以检查是否在网页上获得了结果。

然后您可以检查temail是否为$ value字段

Answer 4

$sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
 $result = mysql_query($sql);
while ($value = mysql_fetch_object($result))
    {
    $teacheremail2 = $value->temail;
    echo $teacheremail2;
    }

希望获得帮助

选择查询以获取单个值不起作用

4 个答案: