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每次上传 csv 文件时如何更新 mysql 数据库表 - 这里的问题是我无法更新“日期列”

[英]How to update mysql database table each time csv file get uploaded - problem here is I cannot update the "date column"

 原文  2022-01-18 10:19:55  8   1    php/ mysql/ database/ csv

问题描述

我的代码中的主要问题是我无法使用 php 将“日期”从 csv 文件更新到 mysql 数据库表。代码行 $date = mysqli_real_escape_string($connect, $data[1]); 是这里的主要问题。 我正在寻找此特定代码行的任何替代查询。

这是 csv 文件： https://drive.google.com/file/d/1EdMKo-XH7VOXS5HqUh8-m0uWfomcYL5T/view?usp=sharing

这是完整的代码：

<?php
//index.php
$connect = mysqli_connect("localhost", "root", "1234", "ml_database");
$message = '';

if(isset($_POST["upload"])) {
    if($_FILES['product_file']['name']) {
        $filename = explode(".", $_FILES['product_file']['name']);
        if(end($filename) == "csv") {
            $handle = fopen($_FILES['product_file']['tmp_name'], "r");
            while($data = fgetcsv($handle)) {
                $data_id = mysqli_real_escape_string($connect, $data[0]);
                $date = mysqli_real_escape_string($connect, $data[1]); //My Problem
                $births = mysqli_real_escape_string($connect, $data[2]);
                $query = "UPDATE my_table 
                            SET date = '$date', 
                                births = '$births', 
                          WHERE data_id = '$data_id'";
                mysqli_query($connect, $query);
            }
            fclose($handle);
            header("location: index.php?updation=1");
        } else {
            $message = '<label class="text-danger">Please Select CSV File only</label>';
        }
    } else {
        $message = '<label class="text-danger">Please Select File</label>';
    }
}

if(isset($_GET["updation"])) {
    $message = '<label class="text-success">Product Updation Done</label>';
}

$query = "SELECT * FROM my_table";
$result = mysqli_query($connect, $query);
?>
<!DOCTYPE html>
<html>
 <head>
  <title>Update Mysql Database through Upload CSV File using PHP</title>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
 </head>
 <body>
  <br />
  <div class="container">
   <h2 align="center">Update Mysql Database through Upload CSV File using PHP</a></h2>
   <br />
   <form method="post" enctype='multipart/form-data'>
    <p><label>Please Select File(Only CSV Formate)</label>
    <input type="file" name="product_file" /></p>
    <br />
    <input type="submit" name="upload" class="btn btn-info" value="Upload" />
   </form>
   <br />
   <?php echo $message; ?>
   <h3 align="center">Birthss</h3>
   <br />
   <div class="table-responsive">
    <table class="table table-bordered table-striped">
     <tr>
      <th>Date</th>
      <th>Births</th>
     
     </tr>
     <?php
     while($row = mysqli_fetch_array($result))
     {
      echo '
      <tr>
       <td>'.$row["date"].'</td>
       <td>'.$row["births"].'</td>
     
      </tr>
      ';
     }
     ?>
    </table>
   </div>
  </div>
 </body>
</html>

1 个解决方案

解决方案1
 0 已采纳  2022-01-18 10:52:52

首先，您需要读取并丢弃 CSV 中的 Title 行。 然后使用适当的、准备好的和参数化的查询，您可以正确更新数据库。 由于 .csv 文件中的日期格式正确，因此无需在此处进行任何操作，但其他 CSV 文件可能并非总是如此，通常必须重新格式化日期才能正确存储在表中。

<?php
//index.php
$connect = mysqli_connect("localhost", "root", "1234", "ml_database");
$message = '';

if(isset($_POST["upload"])) {
    if($_FILES['product_file']['name']) {
        $filename = explode(".", $_FILES['product_file']['name']);

        if(end($filename) == "csv") {
            $handle = fopen($_FILES['product_file']['tmp_name'], "r");
            // read and ignore the TITLE line
            $data = fgetcsv($handle, 1000, ",");

            // prepare the query ONCE
            $query = "UPDATE my_table 
                            SET date = ?, 
                                births = ?
                          WHERE data_id = ?";
            $stmt = $connect->prepare($query);

            // loop over the remaining rows of the csv
            while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
                $stmt->bind_param('sss', $data[0],$data[1],$data[2]);
                $stmt->execute();
            }
            fclose($handle);
            header("location: index.php?updation=1");
            exit;   // always after a header('Location:.....')
        } else {
            $message = '<label class="text-danger">Please Select CSV File only</label>';
        }
    } else {
        $message = '<label class="text-danger">Please Select File</label>';
    }
}

注意：我假设所有 3 列都是绑定中的文本...

$stmt->bind_param('sss', $data[0],$data[1],$data[2]);
                   ^^^

如果date_id是 integer 您可以将其更改为'ssi'尽管 3 s几乎肯定会起作用。

参考：

fgetcsv
mysqli_prepare
mysqli_bind_param

解决方案2
 0  2022-01-18 12:04:53

这是我在工作区@@RiggsFolly 中的当前代码，也许我把代码放错了/排列错误，这就是为什么我不能反映我的表的任何更新。 你能通过先生看到吗？

<?php
//index.php
$connect = mysqli_connect("localhost", "root", "1234", "ml_database");
$message = '';

if(isset($_POST["upload"])) {
    if($_FILES['product_file']['name']) {
        $filename = explode(".", $_FILES['product_file']['name']);

        if(end($filename) == "csv") {
            $handle = fopen($_FILES['product_file']['tmp_name'], "r");
            // read and ignore the TITLE line
            $data = fgetcsv($handle, 1000, ",");

            // prepare the query ONCE
            $query = "UPDATE my_table
                            SET date = ?, 
                                births = ?
                          WHERE data_id = ?";
            $stmt = $connect->prepare($query);

            // loop over the remaining rows of the csv
            while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
                $stmt->bind_param('sss', $data[0],$data[1],$data[2]);
                $stmt->execute();
            }
            fclose($handle);
            header("location: index.php?updation=1");
            exit;   // always after a header('Location:.....')
        }
         else {
            $message = '<label class="text-danger">Please Select CSV File only</label>';
        }
    } else {
        $message = '<label class="text-danger">Please Select File</label>';
    }
}
if(isset($_GET["updation"]))
{
 $message = '<label class="text-success">Product Updation Done</label>';
}

$query = "SELECT * FROM my_table";
$result = mysqli_query($connect, $query);
?>
<!DOCTYPE html>
<html>
 <head>
  <title>Update Mysql Database through Upload CSV File using PHP</title>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
 </head>
 <body>
  <br />
  <div class="container">
   <h2 align="center">Update Mysql Database through Upload CSV File using PHP</a></h2>
   <br />
   <form method="post" enctype='multipart/form-data'>
    <p><label>Please Select File(Only CSV Formate)</label>
    <input type="file" name="product_file" /></p>
    <br />
    <input type="submit" name="upload" class="btn btn-info" value="upload" />
   </form>
   <br />
   <?php echo $message; ?>
   <h3 align="center">Birthss</h3>
   <br />
   <div class="table-responsive">
    <table class="table table-bordered table-striped">
     <tr>
      <th>Date</th>
      <th>Births</th>
     
     </tr>
     <?php
     while($row = mysqli_fetch_array($result))
     {
      echo '
      <tr>
       <td>'.$row["date"].'</td>
       <td>'.$row["births"].'</td>
     
      </tr>
      ';
     }
     ?>
    </table>
   </div>
  </div>
 </body>
</html>

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