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我有以下代码,它可以正确地将数据插入到mysql数据库中,但是当我上传文件时,该特定文件将永远不会包含某个名为'date'的列存在于mysql表中。我想把它放在每一行...所以基本上,我如何使最后一列变量$ dadate?如何在LOAD DATA INFILE中使用变量MYSQL,PHP
下面是代码:
<?php
require_once('header2.php');
require_once('access.php');
?>
<head>
<title>Upload</title>
</head>
<body>
<?php
$filedir = addslashes($_FILES['file']['tmp_name']);
$dadate = $_POST['datadate'];
echo $dadate;
if (isset($_POST['submit'])) {
$sql="LOAD DATA LOCAL INFILE '$filedir' INTO TABLE employeehourstest FIELDS TERMINATED BY ',' LINES TERMINATED BY '\r\n' IGNORE 18 LINES (@dummy, @dummy, @dummy, @dummy, employeeid, @dummy, @dummy, paycode, @dummy, @dummy, @dummy, employeehours, @dummy, @dummy, @dummy, @dummy, @dummy, @dummy, @dummy, @dummy)";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
if ($result) {
echo "<h1>" . "File ". $_FILES['file']['name'] ." uploaded
successfully." . "</h1>";
}
}
?>
</body>
</html>
这里是我的表是如何上传与此代码:
这里是显示为$ dadate php的上传结果:
将你'echo'的'$ _ POST [“datadate”]'和后导致它可能是有帮助的也回答显示文本目前在该文件中,以帮助您完成 – gvgvgvijayan
!请看上面,谢谢 – atomapps
你会显示那个csv的文件内容 – gvgvgvijayan