Php MySQLi加载数据填充到加载.CSV文件时出现问题

Php MySQLi加载数据填充到加载.CSV文件时出现问题,php,mysql,sql,mysqli,Php,Mysql,Sql,Mysqli,我正试图使用MySQLi和PHP将一个40 mg.CSV文件加载到MYSQL数据库中,但在加载页面后,我只得到用户更新失败:消息(Witdout错误消息!) <?PHP define ( 'DB_HOST', 'localhost' ); define ( 'DB_USER', 'root' ); define ( 'DB_PASS', '' ); define ( 'DB_NAME', 'map' ); $con = new mysqli(DB_HOST

我正试图使用MySQLi和PHP将一个40 mg.CSV文件加载到MYSQL数据库中,但在加载页面后,我只得到
用户更新失败:
消息(Witdout错误消息!)

<?PHP   
   define ( 'DB_HOST', 'localhost' );
   define ( 'DB_USER', 'root' );
   define ( 'DB_PASS', '' );
   define ( 'DB_NAME', 'map' );


$con = new mysqli(DB_HOST,DB_USER,DB_PASS,DB_NAME);

if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
};
$sql = "LOAD DATA INFILE 'C:/wamp/www/UP/Modified_Single.csv'
       INTO TABLE  `single-tbl` 
       FIELDS TERMINATED BY ','
       OPTIONALLY ENCLOSED BY '\"' 
       LINES TERMINATED BY '\n' 
       IGNORE 1 LINES;";

$result = mysqli_query($con, $sql);

if (mysqli_affected_rows($con) == 1) {
  $message = "The data was successfully added!";
} else {
  $message = "The user update failed: ";
  $message .= mysqli_error($con); 
};

echo $message;
mysqli_close($con);

您的成功测试是错误的
mysqli\u infected\u rows()
返回插入到表中的行数,该行数应与CSV文件中的行数相同。我怀疑一个40兆的文件只有一行,所以测试
==1
是错误的

如果您想知道查询是否成功,请测试
$result

if ($result) {
    $message = "The data was successfully added!";
} else {
    $message = "The user update failed: " . mysqli_error($con);
}

需要多长时间,超时了吗?也应该
(mysqli\u受影响的行($con)==1)
(mysqli\u受影响的行($con)>=1)
,否则即使成功,它也会返回该错误。没有超时!我一刷新页面,消息就会显示CSV文件中有多少行
mysqli\u infected\u rows()
应该与行数相同,因此它不会是
1
,除非文件只有一行。@Bramar,有82000多行!所以你认为我必须将`(mysqli_受影响的行($con)>=1`更改为
(mysqli_受影响的行($con)>=8200