我没有发现任何错误.此代码很好.完美地更新了我的数据.但显示了1个错误
问题描述：
if ($_SERVER["REQUEST_METHOD"]=="POST") {
$updatedate=$_POST['date'];
$updateday=$_POST['day'];
$updateplace=$_POST['place'];
$updatehighlight=$_POST['highlight'];
$updatediscription=$_POST['discription'];
$sqlupdate="UPDATE $tableselect SET entrydate='$updatedate',day='$updateday',place='$updateplace',highlight='$updatehighlight',discription='$updatediscription' WHERE id ='$getid'";
$sqlquery=mysqli_query($db,$sqlupdate);
if (!mysqli_query($db,$sqlquery)) {
echo "error " .$sqlquery. "<br>" . mysqli_error($db);
}
}
它显示此错误:
您的SQL语法有错误；检查与您的MariaDB服务器版本相对应的手册以获取正确的语法，以在第1行的'1'附近使用
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '1' at line 1
答
很简单，您要运行两次查询功能，这就是为什么要获取1的原因.
Real simple, you're running the query function twice, that's why you're getting 1.
- 1为布尔值TRUE.
- (MySQL)布尔文字
所以
$sqlquery=mysqli_query($db,$sqlupdate);
if (!mysqli_query($db,$sqlquery)) {...}
需要更改为公正
$sqlquery=mysqli_query($db,$sqlupdate);
if(!$sqlquery){...}
第一个被执行，并且(if)!运算符也将触发查询函数，因为它为TRUE，如(if)not failing".
The first gets executed, and the (if)! operator will also trigger the query function since it was TRUE, as in "(if)not failing".
对于其他成功的查询，mysqli_query()将返回 TRUE ."
"For other successful queries mysqli_query() will return TRUE."
还对查询进行参数设置，您可以进行SQL注入.
Parametrize your query also, you're open to an SQL injection.