SQL错误:You have an error in your SQL syntax;急求帮助，万分感谢。

Question

$query=mysql_query("update student set stu_classone=$stu_classone,stu_classtwo=$stu_classtwo,stu_classthree=$stu_classthree,stu_classfour=$stu_classfour,stu_classfive=$stu_classfive,stu_class6=$stu_class6,stu_class7=$stu_class7,stu_class1_value=$stu_class1_value,stu_class2_value=$stu_class2_value,stu_class3_value=$stu_class3_value,stu_class4_value=$stu_class4_value,stu_class5_value=$stu_class5_value,stu_class6_value=$stu_class6_value,stu_class7_value=$stu_class7_value where stu_num=$stu_num");SQL错误：You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'stu_classtwo=,stu_classthree=,stu_classfour=,stu_classfive=,stu_class6=,stu_clas' at line 1

Answer 1

$stu_classfive,外面都加上''吧，像这样'".$stu_classfive."'
从报错情况来看，$stu_classtwo，$stu_classthree，$stu_classfour，$stu_classfive值都为空，而你SQL里这几个数据列，都是not null吧。所以加上''，默认为空。

Answer 2

你这么多字段中应该有不少是字符串类型的吧，对于字符串类型，值要加上引号啊，像这样：
$query=mysql_query("update student set stu_classone='{$stu_classone}', stu_classtwo='{$stu_classtwo}', ....

Answer 3

$query=mysql_query("update student set stu_classone=".$stu_classone.",stu_classtwo=".$stu_classtwo.",stu_classthree=".$stu_classthree.",stu_classfour=".$stu_classfour.",stu_classfive=".$stu_classfive.",stu_classsix=".$stu_classsix.",stu_classseven=".$stu_classseven.",stu_classone_value=$stu_classone_value,stu_classtwo_value=$stu_classtwo_value,stu_classthree_value=$stu_classthree_value,stu_classfour_value=$stu_classfour_value,stu_classfive_value=$stu_classfive_value,stu_classsix_value=$stu_classsix_value,stu_classseven_value=$stu_classseven_value where stu_num=$stu_num");
echo $stu_classone;
//if (!$query) {echo "SQL错误：".mysql_error();} if($query){
echo "<script>
alert(\"添加成功！\");
</script>";
}
   else{
   
   echo "<script>
alert(\"添加失败！\");
</script>";
   }
}//query无法执行，上述方法用过了，还是不行

Answer 4

应该修改成stu_classtwo='".$stu_classtwo."','".xxxxx."','".xxxx ....在""之间是有两个单引号的.

Answer 5

在phpmyadmin里面，修改下这个表，打印出来sql语句，直接贴过来就不会有错了

Answer 6

echo $sql放到数据库里，一直行。看看啥错误，

Answer 7

打印SQL语句出来看看。因为看代码是很难判断的。