如何将表中的多行更新到MySQL

Question

First I tried to retrieve multiple rows of data from MySQL and display them in a table. 首先，我尝试从MySQL检索多行数据并将其显示在表中。

Then I tried to use foreach loop to update multiple rows into the MySQL but in vain. 然后，我尝试使用foreach循环将多个行更新到MySQL中，但没有成功。

If anyone has any solution would be greatly appreciated. 如果有人有任何解决方案，将不胜感激。

include('connect-db.php');

$result = mysql_query("SELECT vaccinedetail.id,vaccinedetail.vaccineid,vaccinedetail.vaccinename1,vaccinedetail.vaccinename2,vaccinedetail.vaccinename3, vaccinedetail.totalnoofinjection,vaccinedetail.nthinjection,vaccinedetail.skip 
FROM vaccinedetail 
WHERE vaccinedetail.vaccineid = '" . $_POST['vaccineid'] . "'")
or die(mysql_error());?>

<?php while($row = mysql_fetch_array( $result )){?>

<tr>
<td><input type="text" name = "id[]" value="<?php echo $row['id'] ?>"</td>

<td><input type="text" name = "vaccineid[]" value="<?php echo $row['vaccineid'] ?>"</td>
<td><input type="text" name = "vaccinename1[]" value="<?php echo $row['vaccinename1'] ?>"</td>
<td><input type="text" name = "vaccinename2[]" value="<?php echo $row['vaccinename2'] ?>"</td>
<td><input type="text" name = "vaccinename3[]" value="<?php echo $row['vaccinename3'] ?>"</td>
<td><input type="text" name = "totalnoofinjection[]" value="<?php echo $row['totalnoofinjection'] ?>"</td>
<td><input type="text" name = "nthinjection[]" value="<?php echo $row['nthinjection'] ?>"</td>
<td><input type="text" name = "skip[]" value="<?php echo $row['skip'] ?>"</td>

</tr>

<?php }?>

</table>
<input type="submit" name="submit" value="Save change">
</form>


</div>

</body>
</html>

<?php

include('connect-db.php');

if (isset($_POST["submit"])){
    foreach ($_POST['vaccineid'] as $index => $vaccineid) {
        $data1 = mysql_real_escape_string($vaccineid);
        $data2 = mysql_real_escape_string($_POST['vaccinename1'][$index]);
        $data3 = mysql_real_escape_string($_POST['vaccinename2'][$index]);
        $data4 = mysql_real_escape_string($_POST['vaccinename3'][$index]);
        $data5 = mysql_real_escape_string($_POST['totalnoofinjection'][$index]);
        $data6 = mysql_real_escape_string($_POST['nthinjection'][$index]);
        $data7 = mysql_real_escape_string($_POST['skip'][$index]);

        mysql_query("UPDATE vaccinedetail SET vaccineid ='$data1', vaccinename1 = '$data2',vaccinename2 = '$data3',vaccinename3 = '$data4',
        totalnoofinjection = '$data5', nthinjection ='$data6', skip ='$data7'") or die(mysql_error()); 
    header("Location: start.php");
    }
}
?>

<html>

<head>
  <title>Display Vaccine type</title>
</head>

<body>
  <?php
session_start();?>
    <form action="" method="POST">
      <!-- Create the table and its heading-->

      <table border='1' cellpadding='10'>
        <tr>
          <th>ID</th>
          <th>Vaccine ID</th>
          <th>疫苗名稱 (繁體)</th>
          <th>疫苗名称 (简体)</th>
          <th>Vaccine Name (Eng)</th>
          <th>Total no of injection</th>
          <th>Nth Injection</th>
          <th>Next Injection Skip</th>
          <th></th>
          <th></th>
        </tr>

Answer 1

You haven't got a WHERE clause in your update statement: 您的update语句中没有WHERE子句：

mysql_query("UPDATE vaccinedetail SET vaccineid ='$data1', vaccinename1 = '$data2',vaccinename2 = '$data3',vaccinename3 = '$data4',
        totalnoofinjection = '$data5', nthinjection ='$data6', skip ='$data7'")
or die(mysql_error());

So your update statement will update all records in the table with the values of the last update statement executed. 因此，您的update语句将使用最后执行的update语句的值来更新表中的所有记录。 You should pass through some kind of identifier for each record. 您应该为每条记录传递某种标识符。 So: 所以：

mysql_query("UPDATE vaccinedetail SET vaccineid ='$data1', vaccinename1 = '$data2',vaccinename2 = '$data3',vaccinename3 = '$data4',
        totalnoofinjection = '$data5', nthinjection ='$data6', skip ='$data7' WHERE ID = $id")

Answer 2

Well, there are a number of issues here 好吧，这里有很多问题

You're not updating at an index. 您不是在索引处更新。 This means every query will result in updating every single row. 这意味着每个查询将导致更新每一行。 To fix this, use a WHERE clause. 要解决此问题，请使用WHERE子句。 That will limit the number of rows. 那将限制行数。

foreach( $_POST['vaccineid'] as $index => $vaccineid ) {
    $id = mysql_real_escape_string($_POST['vaccineid'][$index]);
    /* Escape string queries */
    mysql_query(
    "UPDATE 
      vaccinedetail SET vaccineid ='$data1', 
      vaccinename1 = '$data2',
      vaccinename2 = '$data3',
      vaccinename3 = '$data4',
      totalnoofinjection = '$data5', 
      nthinjection ='$data6', 
      skip ='$data7' 
    WHERE id = $id") // <-- notice that I added a WHERE clause
        or die(mysql_error()); 
    header("Location: start.php");
}

In addition: 此外：

You're updating your data after you've already displayed your data. 显示数据后，您将更新数据。 Move if (isset($_POST["submit"])){ to right after include('connect-db.php'); 将if (isset($_POST["submit"])){移到include('connect-db.php'); . 。
- You're not putting an exit statement after the call to header (this will lead to output despite the fact that you don't want it to do that.) 您不会在对header的调用之后放置exit语句（尽管您不希望它执行此操作，但这仍会导致输出。）
- You have header called after you've already sent content back to the browser ( header must be called before any content is sent in a ?> <?php section). 你已经header叫你发出后的内容返回给浏览器（ header在发送任何内容之前必须调用?> <?php部分）。
- You're using mysql_ library. 您正在使用mysql_库。 That's deprecated. 不推荐使用。 Use mysqli_ . 使用mysqli_ 。

Answer 3

mysql_query("UPDATE vaccinedetail SET 
                            vaccineid ='".$data1."', 
                            vaccinename1 = '".$data2."',
                            vaccinename2 = '".$data3."',
                            vaccinename3 = '".$data4."',
                            totalnoofinjection = '".$data5."', 
                            nthinjection ='".$data6."', 
                            skip ='".$data7."' WHERE ID = $id;
                            ",$connectinObj) or die();

you have not add connection object. 您尚未添加连接对象。

如何将表中的多行更新到MySQL

问题描述

3 个解决方案

解决方案1
0 2017-06-20 14:09:16

解决方案2
0 已采纳 2017-06-20 14:10:39

解决方案3
0 2017-06-20 14:35:33

如何将表中的多行更新到MySQL

问题描述

3 个解决方案

解决方案1 0 2017-06-20 14:09:16

解决方案2 0 已采纳 2017-06-20 14:10:39

解决方案3 0 2017-06-20 14:35:33

解决方案1
0 2017-06-20 14:09:16

解决方案2
0 已采纳 2017-06-20 14:10:39

解决方案3
0 2017-06-20 14:35:33