Php MySQLi加载数据填充到加载.CSV文件时出现问题
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
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($con)==1)
是(mysqli\u受影响的行($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