无法从MySQL数据库更新数据

无法从MySQL数据库更新数据

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无法从MySQL数据库更新数据

本文介绍了无法从MySQL数据库更新数据的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

这是我的update.php代码。我已经检查过其他的php文件，发现没有问题。这些代码没有错误，但它无法更新数据库中的数据。 

  require_onceconn.php; 
 $ conn = connect（）; 
 $ db = connectdb（）; 
 
 $ id =; 
 $ parcelno =; 
 $ items =; 
 if（isset（$ _ REQUEST ['id']））{$ id = $ _REQUEST ['id']; } 
 if（isset（$ _ REQUEST ['parcel']））{$ parcel = $ _REQUEST ['parcel']; } 
 if（isset（$ _ REQUEST ['items']））{$ items = $ _REQUEST ['items']; } 
 
 
 mysql_select_db（$ db，$ conn）或die（mysql_error（）。\\\
）; 
 $ sql =UPDATE parcel SET parcelno ='$ parcelno'，items ='$ items'where id ='$ id'; 
 $ result = @ mysql_query（$ sql）或die（mysql_error（）。\\\
）;

这是用于更新形式的edit.php代码：

 <？php 
 $ ic = $ _REQUEST [ic]; 
 require_onceconn.php; 
 $ conn = connect（）; 
 $ db = connectdb（）; 
 
 mysqli_select_db（$ conn，$ db）或die（mysqli_error（$ conn）。\\\
）; 
 $ query_usr =select * from parcel; 
 $ usr = mysqli_query（$ conn，$ query_usr）或die（mysqli_error（$ conn）。\\\
。$ query_usr）; 
 $ row_usr = mysqli_fetch_assoc（$ usr）; 
？> 
< html>< head> 
< meta http-equiv =Content-Typecontent =text / html; charset = utf-8> 
< style type =text / css> 
 body {
 background-color：#CCC; 
} 
< / style> 
< / head>< body><中心>< p>< a href =adminselect.php>< img src =image / header.pngwidth =800 height =200>< / a>< / p>< / center> 
< form action =update.phpmethod =get> 
<？php 
 $ sql =SELECT * FROM parcel where ic = $ ic; 
 $ result1 = mysqli_query（$ conn，$ sql）; 
 while（$ row = mysqli_fetch_assoc（$ result1））{
？> 
< center>< table border =0> 
< tr> 
< td colspan =3bgcolor =＃0066FF>< strong>< center>更新注册< / strong>< / td& 
< / tr> 
< / tr> 
< tr> 
< td bgcolor =＃0099FF>包裹号< / td> 
< td>：< / td> 
< td bgcolor =＃FFFFFF>< input name =parcelnotype =textid =parcelnovalue =<？php echo $ row [parcelno] ;? > size =50>< / td> 
< / tr> 
 
< tr> 
< td bgcolor =＃0099FF> Items< / td> 
< td>：< / td> 
< td colspan =2bgcolor =＃FFFFFF> 
< p> 
< label for =select>< / label> 
< select name =itemssize =1id =items> 
< option><？php echo $ row [items];？>< / option> 
< option> Pos Laju< / option> 
< option> Pos Ekspress< / option> 
< option> Skynet< / option> 
< option> GDEX< / option> 
< option>全国快递< / option> 
< option> FedEx< / option> 
< option> UPS< / option> 
< / select> 
< / p>< / td> 
< / tr> 
< tr> 
< td>& nbsp;< / td> 
< td>& nbsp;< / td> 
 
< td colspan =2bgcolor =＃CCCCCC> < input name =type =submitvalue =Update>< / td> 
< / tr> 
<？php}？> 
< / table>< / center> 
< / form>< / body>< / html>

我研究这个问题几乎半天，没有什么工作：/

解决方案

您使用两个不同的变量：



$ parcelno 在您的UPDATE查询中



和 $ parcel = $ _REQUEST ['parcel']; / p>

两个变量必须匹配。





将错误报告添加到文件顶部）

 >  error_reporting（E_ALL）; 
 ini_set（'display_errors'，1）;

您现在的代码可以打开。使用，或和。





此外，我在评论 +1 （如果可能）中引用 nkchandra ：



与您的问题无关，但是FYI，PHPMyAdmin不是数据库，而是与您的案例中的mysql数据库交互的工具





修改：阅读您的评论后，您需要切换到功能。



在学习使用预编译语句之前，这只是一个快速修复。

 <？php 
 
 error_reporting（E_ALL）; 
 ini_set（'display_errors'，1）; 
 mysqli_report（MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT）; 
 
 
 $ DB_HOST =xxx; // replace with yours 
 $ DB_USER =xxx; // replace with yours 
 $ DB_PASS =xxx; // replace with yours 
 $ DB_NAME =xxx; // replace with yours 
 
 
 $ conn = new mysqli（$ DB_HOST，$ DB_USER，$ DB_PASS，$ DB_NAME）; 
 if（$ conn-> connect_errno> 0）{
 die（'Connection failed ['。$ conn-> connect_error。']'）; 
} 
 
 $ id =; 
 $ parcelno =; 
 $ items =; 
 if（isset（$ _ REQUEST ['id']））{
 $ id = mysqli_real_escape_string（$ conn，$ _ REQUEST ['id']）; } 
 if（isset（$ _ REQUEST ['parcel']））{
 $ parcelno = mysqli_real_escape_string（$ conn，$ _ REQUEST ['parcel']）; } 
 if（isset（$ _ REQUEST ['items']））{
 $ items = mysqli_real_escape_string（$ conn，$ _ REQUEST ['items']）; } 
 
 
 $ sql =UPDATE parcel SET parcelno ='$ parcelno'，items ='$ items'where id ='$ id'; 
 $ result = mysqli_query（$ conn，$ sql）或die（mysqli_error（）。\\\
）; 
 
 if（！$ result）
 {
 throw new Exception（$ conn-> error）; 
} 
 
 else {echoSuccess; } 
 
 mysqli_close（$ conn）; //关闭连接





code> r3wt's 注释：您也可以使用：

  $ result = $ conn- > query（$ sql）或die（mysqli_error（）。\\\
）;

而不是

  $ result = mysqli_query（$ conn，$ sql）或die（mysqli_error（）。\\\
）;


This is my update.php codes. I already checked others php file and found no problem. This codes have no error but it cant updating data in database. 

require_once "conn.php";
$conn=connect();
$db=connectdb();

$id= "";
$parcelno = "";
$items = "";
if(isset($_REQUEST['id'])){ $id= $_REQUEST['id']; }
if(isset($_REQUEST['parcel'])){ $parcel = $_REQUEST['parcel']; }
if(isset($_REQUEST['items'])){ $items = $_REQUEST['items']; }


mysql_select_db($db,$conn) or die (mysql_error()."\n");
$sql="UPDATE parcel SET parcelno='$parcelno', items='$items' where id='$id'";
$result=@mysql_query($sql) or die(mysql_error()."\n");

This is edit.php code which used for update form :

<?php
$ic = $_REQUEST["ic"];
require_once "conn.php";
$conn = connect();
$db = connectdb();

mysqli_select_db($conn,$db) or die (mysqli_error($conn)."\n");
$query_usr = "select * from parcel";
$usr = mysqli_query($conn,$query_usr) or die (mysqli_error($conn)."\n".$query_usr);
$row_usr = mysqli_fetch_assoc($usr);
?>
<html><head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<style type="text/css">
body {
background-color: #CCC;
}
</style>
</head><body><center><p><a href="adminselect.php"><img src="image/header.png" width="800" height="200"></a></p></center>
<form action="update.php" method="get">
<?php
$sql = "SELECT * FROM parcel where ic = $ic";
$result1 = mysqli_query($conn,$sql);
while ($row=mysqli_fetch_assoc($result1)){
?>
<center><table border="0">
<tr>
<td colspan="3" bgcolor="#0066FF"><strong><center>Update Registration </strong></td>
</tr>
</tr>
<tr>
<td bgcolor="#0099FF">Parcel Number</td>
<td>:</td>
<td bgcolor="#FFFFFF"><input name="parcelno" type="text" id="parcelno" value="<?php echo $row["parcelno"];?>" size="50"></td>
</tr>

<tr>
<td bgcolor="#0099FF">Items</td>
<td>:</td>
<td colspan="2" bgcolor="#FFFFFF">
<p>
<label for="select"></label>
<select name="items" size="1" id="items">
<option><?php echo $row["items"];?></option>
<option>Pos Laju</option>
<option>Pos Ekspress</option>
<option>Skynet</option>
<option>GDEX</option>
<option>Nationwide Express</option>
<option>FedEx</option>
<option>UPS</option>
</select>
</p></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>

<td colspan="2" bgcolor="#CCCCCC"> <input name="" type="submit" value="Update"></td>
</tr>
<?php }?>
</table></center>
</form></body></html>

I researched this problem almost half a day and nothing works :/

解决方案

You're using two different variables:

$parcelno in your UPDATE query

and $parcel = $_REQUEST['parcel'];

both variables must match. If one doesn't, then your entire query will fail.

Add error reporting to the top of your file(s) right after your opening <?php tag, which will help during pre-production testing.

error_reporting(E_ALL);
ini_set('display_errors', 1);

Your present code is open to SQL injection. Use prepared statements, or PDO with prepared statements.

Plus, I quote nkchandra in a comment +1 (if I may):

"Irrelevant to your question, but FYI, PHPMyAdmin is not a database but a tool to interact with database like mysql in your case"

Edit: After reading your comment, it seems like you will need to switch to mysqli_ functions.

This is just a quick fix before you learn to use prepared statements.

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);


$DB_HOST = "xxx"; // replace with yours
$DB_USER = "xxx"; // replace with yours
$DB_PASS = "xxx"; // replace with yours
$DB_NAME = "xxx"; // replace with yours


$conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($conn->connect_errno > 0) {
  die('Connection failed [' . $conn->connect_error . ']');
}

$id= "";
$parcelno = "";
$items = "";
if(isset($_REQUEST['id'])){
$id= mysqli_real_escape_string($conn,$_REQUEST['id']); }
if(isset($_REQUEST['parcel'])){
$parcelno = mysqli_real_escape_string($conn,$_REQUEST['parcel']); }
if(isset($_REQUEST['items'])){
$items = mysqli_real_escape_string($conn,$_REQUEST['items']); }


$sql="UPDATE parcel SET parcelno='$parcelno', items='$items' where id='$id'";
$result=mysqli_query($conn,$sql) or die(mysqli_error()."\n");

if (!$result)
    {
        throw new Exception($conn->error);
    }

else { echo "Success"; }

mysqli_close($conn); // close the connection

Plus, as per r3wt's comment: You can also use:

$result= $conn->query($sql) or die(mysqli_error()."\n");

instead of 

$result=mysqli_query($conn,$sql) or die(mysqli_error()."\n");

这篇关于无法从MySQL数据库更新数据的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

07-22 14:46
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