Mysql从声明变量插入插入 [英] Mysql insert from declare variable error

问题描述

我正在尝试将数据插入表中。我要插入的一些字段来自该表。所以我创建一个临时变量来定义数据，然后在查询中调用该变量。当我尝试从sqlyog执行它没关系，但当我尝试使用PHP执行它我得到一个错误消息

Hi, I'm trying to insert a data to a table. some of the field that i want to insert are come from that table. so i create a temporary variable to define the data, then call the variable in the query. when i try to execute from sqlyog it's fine but when i try to execute it using php i get an error message

Quote：

错误你有一个错误在你的SQL语法中;检查与MariaDB服务器版本对应的手册，以便在eslii_logicalframework中使用'SELECT @lvl：= level + 1'附近使用正确的语法，其中id_framework ='5'; '第2行

ERROR 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 'SELECT @lvl := level+1 from eslii_logicalframework where id_framework='5'; ' at line 2

我的PHP代码

my php code

$sq_insert="SELECT @kdfrm := CONCAT(kode_framework,'.".$nomor_urut."') from eslii_logicalframework where id_framework='".$lookup_id."';
							SELECT @lvl := level+1 from eslii_logicalframework where id_framework='".$lookup_id."';
							insert into eslii_logicalframework
							(
								kode_framework,
								no,
								level,
								id_jenis,
								logical_framework,
								indikator_framework,
								unitkey
							)values
							(
								@kdfrm,
								'".$nomor_urut."',
								@lvl,
								'".$data_post_select."',
								'".$data_post."',
								'".$data_post_ind."',
								'".$unitkey_login."'
							);";

	             
			    if (!$result = mysqli_query($conmysql,$sq_insert)) {
			        echo"<div class='alert alert-mini alert-danger margin-bottom-30'> ERROR  ".mysqli_error($conmysql)."</div>";
			    }

我的尝试：



i只尝试执行php代码

What I have tried:

i only try to execute the php code

推荐答案

sq_insert =SELECT @kdfrm：= CONCAT（kode_framework，'。。

sq_insert="SELECT @kdfrm := CONCAT(kode_framework,'.".

nomor_urut。'）来自eslii_logicalframework，其中id_framework ='。

nomor_urut."') from eslii_logicalframework where id_framework='".

lookup_id。';
SELECT @lvl：= level + 1来自eslii_logicalframework其中id_framework ='。

lookup_id."'; SELECT @lvl := level+1 from eslii_logicalframework where id_framework='".

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