Php SQL查询会有什么问题?
如何修复此SQL查询Php SQL查询会有什么问题?,php,Php,如何修复此SQL查询 $q = $db->sqlQuery("SELECT * FROM project WHERE email = " . $email . ""); $q=$db->sqlQuery(“从项目中选择*,其中email='”“$email.”); $q=$db->sqlQuery(“从项目中选择*,其中email='”“$email.”); 转义参数 $q = $db->sqlQuery("SELECT * FROM project WHERE email = '
$q = $db->sqlQuery("SELECT * FROM project WHERE email = " . $email . "");
$q=$db->sqlQuery(“从项目中选择*,其中email='”“$email.”);
$q=$db->sqlQuery(“从项目中选择*,其中email='”“$email.”);
转义参数
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '" . $email . "'");
转义参数
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '" . $email . "'");
你可以试试这个:
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '" . $email . "'");
缺少应包含$email
值的单引号”
,因为它是一个字符串。您可以尝试以下操作:
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '" . $email . "'");
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '$email'");
缺少应包含$email
值的单引号”
,因为它是一个字符串
$q = $db->sqlQuery("SELECT * FROM project WHERE email = '$email'");
您需要将“”用于字符串变量
错误:从电子邮件=email@email.email
确定:SELECT*FROM project WHERE email='1!'email@email.email“
您需要将“”用于字符串变量
错误:从电子邮件=email@email.email
确定:
SELECT*FROM project WHERE email='1!'email@email.email“
我猜您需要在$email
字符串周围添加引号。我希望字符串已清理…我猜您需要在$email
字符串周围添加引号。我希望那根绳子经过消毒。。。