为什么创建 table 然后插入多个值 return 会出错？

Question

所以我正在尝试创建一个 table，然后向其中插入多个值，如下所示：

        $servername = "localhost";
        $username = "root";
        $password = "";
        $dbname = "someDbName";
        $conn = mysqli_connect($servername, $username, $password, $dbname);
                if (!$conn) {
                    die("Connection failed: " . mysqli_connect_error());
                }
                    $sql = "CREATE TABLE IF NOT EXISTS someTableName(
                    someID INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
                    someVar1 VARCHAR(30) NOT NULL,
                    someVar2 INT NOT NULL);
                    INSERT INTO someTableName (someVar1 , someVar2 ) VALUES ('someString1', someInteger1),
                    ('someString2',someInteger2);";

其中 someInteger 位当然是整数。然后：

            $sql = mysqli_real_escape_string($conn, $sql);    
            if (mysqli_multi_query($conn, $sql)) {
                dtconsole("Tables populated successfully");
            } else {
                dtconsole("Error creating table: " . mysqli_error($conn));
            }

有了 dtconsole 功能，就可以输出到控制台来帮助我调试。

function dtconsole($data){
        $output=$data;
        if(is_array($output)){
            $output=implode(',',$output);
        }
        echo '<script>console.log("'.$output.'");</script>';
    }

每次我尝试运行这个，它returns以下错误：

Error creating table: 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 ' someID INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, ' at line 1

我这辈子都看不出自己做错了什么。

Answer 1

您的问题是您在整个查询中调用 mysqli_real_escape_string，而不仅仅是您插入的值。结果，它将 $sql 字符串中的 CR-LF 转换为 \r\n，MySQL 解析器无法解释。你应该这样做：

$someString1 = mysqli_real_escape_string($conn, 'someString1'); 
$someString2 = mysqli_real_escape_string($conn, 'someString2');
$sql = "CREATE TABLE IF NOT EXISTS someTableName(
        someID INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
        someVar1 VARCHAR(30) NOT NULL,
        someVar2 INT NOT NULL);
        INSERT INTO someTableName (someVar1 , someVar2 ) VALUES ($someString1, someInteger1),
        ($someString2,someInteger2);";
if (mysqli_multi_query($conn, $sql)) { 
    ...

为什么创建 table 然后插入多个值 return 会出错？

Why does creating a table and then inserting multiple values return an error?

php

sql

mysqli-multi-query