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php - php mysql中的多行插入错误

问题描述

以下查询中的问题是什么......我经常收到此错误。错误：INSERT INTO Myguests (firstname,lastname,email)VALUES 您的 SQL 语法有误；检查与您的 MariaDB 服务器版本相对应的手册，以获取在第 1 行的 '' 附近使用的正确语法

<?php  include("new_db.php");?>

<?php
if(!empty($_POST['f_name'])&&!empty($_POST['l_name'])
&&!empty($_POST['email']))
{       $sql = "INSERT INTO Myguests (firstname,lastname,email)VALUES";
        for($i=0;$i<$_POST['num'];$i++){
        $first_name = mysqli_real_escape_string($conn,$_POST['f_name'][$i]);
        $last_name  = mysqli_real_escape_string($conn,$_POST['l_name'][$i]);
        $Email      = mysqli_real_escape_string($conn,$_POST['email'][$i]);

  $sql.="('".$first_name."','".$last_name."','".$Email."')";

}
$sql =rtrim($sql, ',');
if (mysqli_query($conn, $sql)) {
    echo "Records Created";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}


}
    mysqli_close($conn);

    ?>

标签： phpmysqli

解决方案

我使用了这个创建表语句

CREATE TABLE `Myguests` (
 `firstname` varchar(11) ,
 `lastname` varchar(11) ,
 `email` varchar(11));

而这段代码

<?php

$_POST['f_name'][0] = "john";
$_POST['f_name'][1] = "will";
$_POST['f_name'][2] = "jane";
$_POST['l_name'][0] = "doe";
$_POST['l_name'][1] = "smith";
$_POST['l_name'][2] = "2x";
$_POST['email'][0] = "mail@mail.com";
$_POST['email'][1] = "mail2@mail.com";
$_POST['email'][2] = "mail3@mail.com";
$sql = "INSERT INTO Myguests (firstname,lastname,email) VALUES ";
for($i=0;$i<3;$i++){
$first_name = ($_POST['f_name'][$i]);
$last_name  = ($_POST['l_name'][$i]);
$Email      = ($_POST['email'][$i]);
  $sql.="('".$first_name."','".$last_name."','".$Email."'),";

}
$sql =rtrim($sql, ',');

它起作用了，在添加“，”之后，您是否仍然遇到相同的错误$sql.=？

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