php SQL 语法错误 MariaDB 服务器版本，用于在第 1 行的“WHERE ID = 4”附近使用正确的语法
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SQL syntax error MariaDB server version for the right syntax to use near 'WHERE ID = 4' at line 1
提问by Sep
I have this line of SQL:
我有这行 SQL：
$sql = "SELECT ID, ListStID, ListEmail, Title FROM $entry_database WHERE ID = '". $ReqBookID ."'";
$result = mysqli_query($conn, $sql);
As you can see, I am selecting an entry's ID, ListStID, ListEmail and Title Column if ID is equal to a string of numbers (or text), which is given by user in a form.
如您所见，如果 ID 等于用户在表单中给出的一串数字（或文本），我将选择条目的 ID、ListStID、ListEmail 和标题列。
Everything is ok, and I don't get any syntax error when I write the code (I am using a code editor software. However, when I use it online, I get this error:
一切正常，我写代码时没有出现任何语法错误（我使用的是代码编辑器软件。但是，当我在线使用它时，我得到了这个错误：
Error: SELECT ID, ListStID, ListEmail, Title FROM WHERE ID = '4' 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 'WHERE ID = '4'' at line 1
错误：SELECT ID, ListStID, ListEmail, Title FROM WHERE ID = '4' 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，了解在第 1 行的“WHERE ID = '4'”附近使用的正确语法
I am very new to PHP, and I'm sure I am either adding extra ' or ", so I would really appreciate it if you could help me with this issue. I have tried the answers for similar questions, but no success yet.
我对 PHP 很陌生，我确定我要添加额外的 ' 或 "，所以如果你能帮助我解决这个问题，我将不胜感激。我已经尝试过类似问题的答案，但还没有成功。
回答by Forien
You have empty $entry_databasevariable. As you see in error: ListEmail, Title FROM WHERE IDbewteen FROMand WHEREshould be name of table. Proper syntax of SELECT:
你有空$entry_database变量。正如你在错误看到：ListEmail, Title FROM WHERE IDbewteen FROM和WHERE应该是表的名称。SELECT 的正确语法：
SELECT columns FROM table [optional things as WHERE/ORDER/GROUP/JOIN etc]
which in your way should become:
以您的方式应该成为：
SELECT ID, ListStID, ListEmail, Title FROM some_table_you_got WHERE ID = '4'
回答by Tim Lewis
You're missing your database name:
您缺少数据库名称：
$sql = "SELECT ID, ListStID, ListEmail, Title FROM ".$entry_database." WHERE ID = ". $ReqBookID .";
And make sure that $entry_database isn't null or empty:
并确保 $entry_database 不为 null 或为空：
var_dump($entry_database);
Also notice that you don't need to have $ReqBookID in '' as if it's an Int.
另请注意，您不需要将 $ReqBookID 放在 '' 中，就好像它是 Int 一样。
回答by Sundeep Singh
This problem also arise when we don't give the single or double quotes to the database value.
当我们不给数据库值提供单引号或双引号时，也会出现这个问题。
Wrong way:
错误的方法：
$query ="INSERT INTO tabel_name VALUE ($value1,$value2)";
As database inserting values must be in quotes ' '/" "
由于数据库插入值必须用引号' '/" "
Right way:
正确的方法：
$query ="INSERT INTO STUDENT VALUE ('$roll_no','$name','$class')";
