在加载数据本地填充PHP中使用变量

我一直在编写一个脚本，该脚本将数据从CSV文件导入phpMyAdmin数据库表

使用此代码工作正常：

$query = <<<eof
    LOAD DATA LOCAL INFILE '$file'
     INTO TABLE c170720
     FIELDS TERMINATED BY ',' 
     OPTIONALLY  ENCLOSED BY '"'
     LINES TERMINATED BY '\n'
     IGNORE 1 LINES
    (ccode,custarea,ttlpkt)
eof;

---

$query=将
更改为表“$mu cfile”
到
转换为表$mu cfile
您好，$mu cfile表是在其他地方创建的吗？您所说的“它工作完全正常，但我需要变量”是什么意思？@Lambda7 Hi，Ops！我把剧本放错了。我会编辑它。稍等。您正在尝试使用herdoc生成查询，在双引号字符串中创建查询将非常方便。然后，在语句
exit（mysqli_error（$con））之前打印您的查询对你有好处，写下你的答案，然后接受它
$query = <<<eof
    LOAD DATA LOCAL INFILE '$file'
     INTO TABLE '$m_cfile'
     FIELDS TERMINATED BY ',' 
     OPTIONALLY  ENCLOSED BY '"'
     LINES TERMINATED BY '\n'
     IGNORE 1 LINES
    (ccode,custarea,ttlpkt)
    eof;

<?php
$host       = "localhost";
$user       = "root";
$password   = "";
$db         = "smposf";

$con        = mysqli_connect($host,$user,$password,$db);

$message = "";
if (isset($_POST['submit3'])) 
{
   $m_date=date($sdate);
   $m_yy=substr($m_date,2,2);
   $m_mm=substr($m_date,5,2);
   $m_dd=substr($m_date,8,2);

   $m_cfile='c'.$m_yy.$m_mm.$m_dd;

   $allowed = array('csv');
   $filename = $_FILES['file']['name'];
   $ext = pathinfo($filename, PATHINFO_EXTENSION);
   if (!in_array($ext, $allowed)) 
   {
    // show error message       
      $message = 'Invalid file type, please use .CSV file!';
   } 
   else 
   {

       move_uploaded_file($_FILES["file"]["tmp_name"], "CSV/" . 
       $_FILES['file']['name']);

       $file = "CSV/" . $_FILES['file']['name'];

       $query = <<<eof
                   LOAD DATA LOCAL INFILE '$file'
                   INTO TABLE '$m_cfile'
                   FIELDS TERMINATED BY ',' 
                   OPTIONALLY  ENCLOSED BY '"'
                   LINES TERMINATED BY '\n'
                   IGNORE 1 LINES
                   (ccode,custarea,ttlpkt)
        eof;

        if (!$result = mysqli_query($con, $query)) 
        {
             exit(mysqli_error($con));
        }
        $message = "CSV file successfully imported!";
    }
}
?>