如何将多个记录插入到MySQL wordpress数据库中？

Question

我真的很难在MySQL中插入多个记录。我使用的是这个参考资料：multiple.asp

这是我得到的错误：

错误:插入clk_9d0b4f116f_wp_comments (comment_post_ID，comment_author，comment_author_email，comment_author_url，comment_author_IP，comment_date，comment_date_gmt，comment_content，comment_karma，comment_approved，comment_agent，comment_type，comment_parent，user_id)值( '1'，'Ai'，‘ai@aipresscoach.com’，''，‘：1’，'2017-06-01 19:49:15'，在clk_9d0b4f116f_wp_comments (comment_post_ID，comment_author，comment_author_email，comment_author_url，comment_author_IP，comment_date，comment_date_gmt，comment_content，comment_karma，comment_approved，comment_agent，comment_type，comment_parent，user_id)中插入'2017-06-01 17:15‘，'0'，'1'，’1‘，''，'0'，’2‘)值( '25'，'Ai'，‘ai@aipresscoach.com’，''，‘：1’，'2017-06-01 19:49:15'，'2017-06-01 17:49:15'，''，'0'，'1'，'0'，'2')

您的SQL语法出现了错误；请检查与MySQL服务器版本对应的手册，以获得在第16行“插入到clk_9d0b4f116f_wp_comments (comment_post_ID，comment_author，注释”)附近使用的正确语法。

下面是代码：

<?php
$servername = "xxxx";
$server_username = "xxxx";
$server_password = "xxxx";
$dbname = "xxxx";

$comment_id = 5;
$comment_id += 1;
$content = $_POST["blogPostPost"];// "this is the client answer";
$clientCur = $_POST["clientCurrent"];// "this is the client current 
situation";
$clientAlt = $_POST["clientAlternative"];// "this is the client alternative 
answer";
$title = "unity blog post";
$timelocal = date('Y-m-d H:i:s');
$timegmt = gmdate("Y-m-d H:i:s");
$table = "clk_9d0b4f116f_wp_comments";

// Create connection
$conn = mysqli_connect($servername, $server_username, $server_password, 
$dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT INTO $table (comment_post_ID, comment_author, 
comment_author_email, comment_author_url, comment_author_IP, comment_date, 
comment_date_gmt, comment_content, comment_karma, comment_approved, 
comment_agent, comment_type, comment_parent, user_id)
        VALUES (
        '1',
        'Ai',
        'ai@aistresscoach.com',
        '',
        '::1',
        '".$timelocal."',
        '".$timegmt."',
        '".$content."',
        '0',
        '1',
        '',
        '',
        '0',
        '2')";

$sql .= "INSERT INTO $table (comment_post_ID, comment_author, 
comment_author_email, comment_author_url, comment_author_IP, comment_date, 
comment_date_gmt, comment_content, comment_karma, comment_approved, 
comment_agent, comment_type, comment_parent, user_id)
        VALUES (
        '25',
        'Ai',
        'ai@aistresscoach.com',
        '',
        '::1',
        '".$timelocal."',
        '".$timegmt."',
        '".$clientCur."',
        '0',
        '1',
        '',
        '',
        '0',
        '2')";

if (mysqli_multi_query($conn, $sql)) {
echo "New records created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>

任何意见和想法都是非常感谢的。亲切问候伊利亚斯

Answer 1

您的查询可以是：

$sql = "
INSERT INTO $table 
( comment_post_ID
, comment_author
, comment_author_email
, comment_author_url
, comment_author_IP
, comment_date
, comment_date_gmt
, comment_content
, comment_karma
, comment_approved
, comment_agent
, comment_type
, comment_parent
, user_id) VALUES 
(1,'Ai','ai@aistresscoach.com','','::1','$timelocal','$timegmt','$content',0,1,'','',0,2),
(25,'Ai','ai@aistresscoach.com','','::1','$timelocal','$timegmt','$clientCur',0,1,'','',0,2);
";

然后你就可以摆脱所有那些多查询的东西了。

Answer 2

尝试用';‘来完成您的第一个$sql

$sql = "INSERT INTO $table (comment_post_ID, comment_author, 
comment_author_email, comment_author_url, comment_author_IP, comment_date, 
comment_date_gmt, comment_content, comment_karma, comment_approved, 
comment_agent, comment_type, comment_parent, user_id)
        VALUES (
        '1',
        'Ai',
        'ai@aistresscoach.com',
        '',
        '::1',
        '".$timelocal."',
        '".$timegmt."',
        '".$content."',
        '0',
        '1',
        '',
        '',
        '0',
        '2');";